MathCuer

Geometry Chapter 5 - Congruence based on Triangles

2010-02-12T09:59:00.001-10:00

Geometry Chapter 5 - Congruence based on Triangles
5.1 - Line Segments Associated with TrianglesVocabulary:
1. Altitude of a triangle - is a line segment drawn from any vertex of the triangle, perpendicular to and ending in the line that contains the opposite side.
2. Median of a triangle - is a line segment that joins any vertex of the triangle to the midpoint of the opposite side.
3. Angle bisector of a triangle - is a line segment that bisects any angle of the triangle and terminates in the side opposite that angle.
5.2 - Using Congruent triangles to prove line segments congruent and angles congruent.
1. SAS = SAS Congruence Postulate - if two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the two triangles are congruent.
Example: Given Triangle ABC and Triangle DEF,if AB = DE, BC = EF and angle B = angle E, then triangle ABC = triangle DEF.
2. SSS = SSS Congruence Postulate - if three sides of one triangle are congruent to three sides of a second triangle, then the two triangles are congruent.
Example: Given triangle ABC and triangle DEF,if AB = DE, BC = EF and AC = DF, then triangle ABC = triangle DEF.
3. ASA = ASA Congruence Postulate - if two angles and the included side of one triangle are congruent to two angles and the included side of a second triangle, then the two triangles are congruent.
Example: Given triangle ABC and triangle DEF,if angle A = angle D, AB = DE, and angle B = angle E, then triangle ABC = triangle DEF.
4. AAS = AAS Congruence Postulate - if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of a second triangle, then the two triangles are congruent.
Example: Given triangle ABC and triangle DEF,if angle A = angle D, angle C = angle F, and BC = EF , then triangle ABC = triangle DEF.
5. HL = HL Congruence Postulate - if the leg and hypotenuse of one right triangle is congruent to the corresponding leg and hypotenuse of another right triangle, then the two triangles are congruent by hypotenuse - leg postulate.
Example: Given right triangle ABC and right triangle DEF, if angle B and angle E are both right angles and leg AB = leg DE and hypotenuse AC = hypotenuse DF, then triangle ABC = triangle DEF.
5.3 Isosceles and Equilateral Triangles:
Theorem: If two sides of a triangle are congruent, the angles opposite these sides are congruent.
Theorem: The median from the vertex angle of an isosceles triangle bisects the vertex angle.
Theorem: The median from the vertex angle of an isoscles triangle is perpendicular to the base.
Theorem: Every equilateral triangle is equiangular.
5.4 Using two pairs of Congruence Triangles:
5.5 Proving Overlapping Triangles Congruent:
5.6 Perpendicular Bisector of a line segment.
Definition: the perpendicular bisector of a line segment is any line or subset of a line that is perpendicular to the line segment at its midpoint.
Definition: Equidistant - equal distance from the endpoints.
Theorem: If two points are each equidistant from the endpoints of a line segment, then the points determine the perpendicular bisector of the line segment.
Theorem: If a point is equidistant from the endpoints of a line segment, then it is on the perpendicular bisector of the line segment.
Theorem: If a point is on the perpendicular bisector of a line segment, then it is equidistant from the endpoints of the line segment.
Theorem: A point is on the perpendicular bisector of a line segment if and only if it is equidistant from the endpoints of the line segment.
A. Methods of Proving lines or line segments perpendicular:Prove one of the following statements is true:1. The two lines form right angles at their point of intersection.2. The two lines form congruent adjacent angles at their point of intersection.3. Each of two points on one line is equidistant from the endpoints of a segment of the other.
B. Intersection of the Perpendicular Bisectors of the sides of a triangle.- the perpendicular bisectors of the sides of a triangle are concurrent (they intersect in one point).- The point where the three perpendicular bisectors of the sides of a triangle intersect is called the circumcenter.

Geometry Chapter 4 - Congruence of line segments, angles, and triangles

2008-10-22T09:19:00.009-10:00

Chapter 4 - Congruence of line segments, angles, and triangles - notes:
4-1 Postulates of lines, line segments, and angles –
Postulates:

4.1 A line segment can be extended to any length in either direction.
4.2 Through two given points, one and only one line can be drawn.
4.3 Two lines cannot intersect in more than one point.
4.4 One and only one circle can be drawn with any given point as it's center and the length of any given line segment as the radius.
4.5 At a given point on a given line, one and only one perpendicular can be drawn to the line.
4.6 From a given point not on a given line, one and only one perpendicular can be drawn to the line.
4.7 For any two distinct points, there is only one positive real number that is the length of the line segment joining the two endpoints.
4.8 The shortest distance between 2 points is the length of the line segment joining these two points.
4.9 A line segment has one and only one midpoint.
5.0 An angle has one and only one bisector.

4.2 Using Postulates and Definitions in Proofs:
We use the laws of logic to combine definitions and postulates to prove a theorem.
We will assume unless otherwise stated, that lines that appear to be straight lines in a figure actually are straight lines and that points that appear on the line actually are.

Example: Given: and AB = CD.
Prove: AC = BD
Statements Reasons
1. AB = CD 1. Given
2. BC = BC 2. Reflexive Postulate
3. AB + BC = CD + BC 3. Addition Postulate
4. AB + BC = AC; CD + BC = BD 4. partition postulate
5. AC = BD 5. Substitution postulate

4.3 Proving Theorems About Angles
Theorem 4.1 If 2 angles are right angles, then they are congruent.
Theorem 4.2 If two angles are straight angles, then they are congruent.

Definitions involving Pairs of Angles:

Definition:
Adjacent Angles are two angles in the same plane that have a common vertex and a common side but do not have any interior points in common.
Definition: Complementary angles are two angles, the sum of whose degree measure is 90 degrees.
Definition: Supplementary angles are two angles, the sum of whose degree measure is 180 degrees.
Theorem 4.3 If two angles are complements of the same angle or congruent angles, then the two angles are congruent.
Theorem 4.4 If two angle are congruent, then their complements are congruent.
Theorem 4.5 If two angles are supplements of the same angle, then they are congruent.
Theorem 4.6 If two angles are congruent, then their supplements are congruent.
If two angles are congruent, then they have the same supplement.

Definition: A linear pair of angles are two adjacent angles whose sum is a straight angle.
Theorem 4.7 If two angles form a linear pair, then they are supplementary.
Theorem 4.8 If two lines intersect to form congruent adjacent angles, then they are perpendicular.

Definition: Vertical angles are two angles in which the sides of one angle are opposite rays to the sides of the second angle.
Theorem 4.9: If two lines intersect, then the vertical angles are congruent.

How to present a proof in geometry using deductive reasoning.
1. As an aid, draw a figure that pictures the data of the theorem or the problem. Use letters to label points in the figure.
2. State the given, which is the hypothesis of the theorem, in terms of the figure.
3. State the Prove, which is the conclusion of the theorem, in terms of the figure.
4. Present the proof, which is a series of logical arguments used in the demonstration. Each step in the proof should consist of a statement about the figure. Each statement should be justified by the given, a definition, a postulate, or a previously proved theorem. The proof may be presented in a two-column format or in a paragraph form.

4.4 Congruent Polygons and Corresponding Parts
Corresponding Parts of Congruent Polygons

Two Polygons are congruent if and only if there is a one-to-one correspondence between their vertices such that corresponding angles are congruent and corresponding sides are congruent.
Therefore: Corresponding parts of congruent polygons are congruent.

Congruent Triangles:
Corresponding parts of congruent triangles are congruent (CPCTC).

if two triangles are congruent, then the corresponding parts are congruent.

Postulates:
Reflexive Postulate: any geometric figure is congruent to itself.
Symmetric Postulate: a congruence may be expressed in either order.
Transitive Postulate: Two geometric figures congruent to the same geometric figure are congruent to each other.

4.5 - 4.7 Proving Triangles Congruent using
5 ways to show Triangle Congruence:

1. SAS = SAS Congruence Postulate - if two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the two triangles are congruent.Example: Given ΔABC and ΔDEF,if AB = DE, BC = EF and angle B = angle E, then ΔABC = ΔDEF.

2. SSS = SSS Congruence Postulate - if three sides of one triangle are congruent to three sides of a second triangle, then the two triangles are congruent.Example: Given ΔABC and ΔDEF,if AB = DE, BC = EF and AC = DF, then ΔABC = ΔDEF.

3. ASA = ASA Congruence Postulate - if two angles and the included side of one triangle are congruent to two angles and the included side of a second triangle, then the two triangles are congruent.Example: Given ΔABC and ΔDEF,if angle A = angle D, AB = DE, and angle B = angle E, then Δ ABC = Δ DEF.

4. AAS = AAS Congruence Postulate - if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of a second triangle, then the two triangles are congruent.Example: Given Δ ABC and Δ DEF,if angle A = angle D, angle C = angle F, and BC = EF , then Δ ABC = Δ DEF.

5. HL = HL Congruence Postulate - if the leg and hypotenuse of one right triangle is congruent to the corresponding leg and hypotenuse of another right triangle, then the two triangles are congruent by hypotenuse - leg postulate.Example: Given right Δ ABC and right Δ DEF, if angle B and angle E are both right angles and leg AB = leg DE and hypotenuse AC = hypotenuse DF, then Δ ABC = Δ DEF.

Geometry Chapter 3 - Sections 3.5 - 3.8 - 2008 - 2009

2008-10-10T09:05:00.004-10:00

3.5 - 3.8 Postulates, Theorems and Proof:

I. Postulate or axiom - a true obvious statement and accepted without proof.
- is a statement whose truth is accepted without proof.

II. Theorem - is a statement that is proved by deductive reasoning.

III. Postulates:

A. Reflexive Postulate: angle A = angle A : everything is congruent to itself

B. Symmetric Postulate: if angle A = angle B then angle B = angle A (somewhat like the converse)

C. The Substitution Postulate: a quantity may be substituted for its equal in any statement of equality.

Example: if x = y and y = 8, then we can conclude by substitution that x = 8.

D. Partition Postulate: a whole is equal to the sum of the parts (part + part = whole)

E. Addition Postulate: if a = b and c = d, then a + c = b + d

F. Subtraction Postulate: if a = b and c = d, then a - c = b - d

G. Multiplication Postulate: if a = b and c = d, then ac = bd

H. Division Postulate: if a = b and c = d, then a/c = b/d.

I. Power Postulate: if a = b then a² = b²

One of my students thought this would be a better way:

if a = b and c = d then a^c = b^d

Does anyone see a flaw in this?

J. Roots Postulate: if a = b and a is positive, then the square root of a = the square root of b

Example using a proof:

Chapter 3.4 Proofs

2008-10-06T09:51:00.010-10:00

3 - 4 Direct Proofs: a proof that starts with the given statements and uses the laws of logic to arrive at the statement to be proved.

Examples we did in class:

Example1:

Example 2:

Example 3:

Example 4:

Example 5:

Unit 3 - Proving Statements in Geometry

2008-10-02T04:46:00.016-10:00

Vocabulary - see blue sheet
A. Definitions:
1. Midpoint of a segment - divides it into two congruent segments.
2. Bisector of a segment or an angle - divides it into two congruent parts
3. Complementary Angles - are two angles whose sum is 90 degrees.
4. Supplementary Angles - are two angles whose sum is 180 degrees.
5. Perpendicular Lines - form right angles.
6. Perimeter of a polygon - is the sum of the lengths of the sides of a polygon.
7. Scalene triangle - has no congruent sides.
8. Isosceles Triangle - has at least two congruent sides.
9. Equilateral Triangle - has three congruent sides.
10. Altitude of a triangle - is a line drawn perpendicular from a vertex to the opposite side.
11. Median of a triangle - is a line drawn from a vertex to the midpoint of the opposite side and divides the opposite side into two congruent parts.
12. Parallelogram - is a quadrilateral with both pairs of opposite sides parallel.

B. Postulates:
1. Reflexive Postulate - any quantity is congruent to itself. angle A @ angle A
2. Transitivity Postulate - If angle A @ angle B, and angle B @ angle C, then angle A @ angle C.
3. Substitution Postulate - If angle A @ angle B, and angle A + angle C = 150 degrees, then by substitution, angle B + angle C = 150 degrees.
4. Partition Postulate - the part + the part = the whole. AB + BC = AC
5. Addition Postulate - If congruent quantities are added to congruent quantities, then their sums are congruent.
angle A @ angle B and angle C @ angle C, (angle C does not equal zero degrees)
then angle A + angle C = angle B + angle C.
6. Subtraction Postulate - If congruent quantities are subtracted from congruent quantities, then their differences are congruent.
angle A @ angle B and angle C @ angle C, (angle C does not equal zero degrees)
then Angle A - angle C = angle B - angle C.
7. Doubles Postulate - if two quantites are equal, then double their quantities are equal.
angle A = angle B, then 2(angle A) = 2(angle B).
8. Halves Postulate - if two quantities are equal, then half their quantities are equal.
angle A = angle B, then (1/2)(angle A) = (1/2)(angle B).

C. Theorems:
1. All right angles are congruent.
2. If two angles form a linear pair, then they are supplementary.
3. If two angles are supplements of the same angle, then they are congruent to each other.
3b. If two angles are complements of the same angle, then they are congruent to each other.
4. If two angles are congruent, then their supplements are congruent.
4b. If two angles are congruent, then their complements are congruent.
5. Vertical angles are congruent.
6. Corresponding parts of congruent triangles are congruent (CPCTC).
7. If two sides of a triangle are congruent, then the angles opposite are congruent.
7b. If two angles of a triangle are congruent, then the sides opposite are congruent.
8. An equilateral triangle is equiangular.
8b. An euqiangular triangle is equilateral.
9. If two lines form congruent adjacent angles, then the lines are perpendicular.
10 The supplement fo a right angle is a right angle.
11. If two lines are cut by a transversal forming congruent alternate interior angles, then the lines are parallel.
11b. If two lines are cut by a transversal forming congruent corresponding angles, then the lines are parallel.
11c. If two lines are cut by a transversal forming supplementary interior angles on the same side of the transversal, then then the lines are parallel.
11d. If two parallel lines are cut by a transversal, then the alternate interior angles are congruent.
11e. If two parallel lines are cut by a transversal, then the corresponding angles are congruent.
11f. If two parallel lines are cut by a transversal, then the interior angles on the same side of the transversal are supplementary.
12. If two lines are perpendicular to the same line, then they are parallel.
13. If two lines are parallel to the same line, then they are also parallel.
14. Extensions and segments of parallel lines are parallel.
15. The sum of the measures of the interior angles of a triangle is 180 degrees.
16. The sum of the measures of the interior angles of a quadrilateral is 360 degrees.
17. The measure of the exterior angle of a triangle is equal to the sum of the two nonadjacent interior angles.
18. The sum of the measures of the interior angles of a polygon of n sides is 180(n - 2)
19. The measure of each interior angle of a regular polygon of n sides is (180(n-2))/n
20. The sum of the measures of the exterior angles of any polygon is 360 degrees.
21. The measure of each exterior angle of a regular polygon of n sides is 360/n.

3 - 1 Inductive Reasoning - uses a series of particular examples to lead to a general conclusion.

A. Inductive Reasoning is a powerful tool in discovering and making Conjectures - definition= generalizations arising from direct measurements of specific cases.

B. Care must be taken when applying inductive reasoning to ensure that all revelant examples are examined (no counterexample exists).

C. Inductive reasoning does not prove or explain the conjectures.

Based on these three triangles, what conjecture can you make about isosceles triangles?

Answer: If 2 sides of a triangle are congruent, then the opposite angles are congruent.

Example 2: Equilateral triangles - draw the 3 midpoints of the sides and connect them.
Make a conjecture about the 4 triangles that are formed.

Answer: It seems that it forms 4 congruent triangles

Example 3: If we draw parallelograms with opposite sides congruent by using both sides of the ruler, what conjectures can we form?

A. Opposite sides of a parallelogram are congruent.
B. It seems that all parallelograms have two acute angles and two obtuse angles.

The second conjecture is not true, if we make a parallelogram that is a rectangle, you can see all four angles are right angles, thus disproving conjecture B. This is called a counterexample - an example that shows the general conclusion is false.

You can see with 6 points, it should be 2^{6 - 1} = 2⁵ = 32 areas.

My students said that each time we added a point on the circle, it doubled the areas. They found out this is not true because they only had 30 or 31 areas.

In the left circle, there are 31 areas and in the right circle, there are only 30 areas. So this conjecture does not work.

This is why you have to be careful using inductive reasoning.

3 - 2 Deductive Reasoning
A. Deductive Reasoning - uses the laws of logic to combine definitions and general statements that we know to be true to reach a valid conclusion.

Every Good definition can be written as a true biconditional:

hypothesis if and only if conclusion

example:
Conditional statement: If a triangle has 2 congruent sides, then the angles opposite are congruent.

Converse: If a triangle has 2 congruent angles, then the sides opposite are congruent.

Since both the conditional and converse statements are true, then we can write the definition in biconditional form:

A triangle has 2 congruent sides if and only if the angles opposite are congruent.

Lets try another example:
Statement: A right triangle is a triangle with one right angle.

Conditional: If a triangle is a right triangle, then it has one right angle.
Converse: If a triangle has one right angle, then it is a right triangle.

Biconditional: A triangle is a right triangle if and only if it has one right angle.

3 - 3 Deductive Reasoning
1. A Proof in geometry is a valid argument that establishes the truth of a statement.
1. we use definitions, postulates and already proven theorems to show the truth of the statement.
Example: Recall the exercise with the equilateral triangle, we believed that it formed 4 congruent equilateral triangles. Using deductive reasoning, we can prove this conjecture to be true.

If 2 sides of a triangle are congruent, then the angles opposite are congruent. Each angle in an equilateral triangle measures 60 degrees so angle A = 60 degrees. That leaves angle AFB and angle BFA sum to be 180 - 60 = 120 degrees, (the sum of the angles in a triangle is 180 degrees). Since they are equal in measure, they each measure 60 degrees, forming a triangle with 3 angles measuring 60 degrees so this means it is an equiangular triangle (by definition of equiangular triangles) and equiangular triangles are equilateral triangles, so therefore each of the four triangles would have all three sides congruent (definition of equilateral triangle) forming 4 congruent equilateral triangles (all sides and angles are congruent).

How do we write a proof?

1. Two-Column Proof: has numbered statements and reasons that show the logivcal order of an argument.
2. Paragraph proof: a proof can be written in paragraph form. What we just did was a form of paragraph proof. You have to write the statements and give the reasons as part of your paragraph.
3. Flow proof: a chart that has arrows going from one statement to the next with the reasons written underneath the statement.

We will mainly use two-column proof:

Here are a few examples of algebraic proofs:

Just given some statement, we can conclude from the statement a new statement and have a reason for the statement.

Example:
Given: angle one and angle two are complementary, what can we conclude?
Conclusion: that the sum of angle one and angle two = 90 degrees by definition of complementary angles

Given: Ray BD bisects angle ABC
Conclusion: angle ABD is congruent to angle BDC by definition of angle bisector

Given: Line AB bisects line segment DE at point F
Conclusion: line segment DF is congruent to line segment FE by definition of segment bisector

Given: 0 is less than the measure of angle A which is less than 90 degrees.
Conclusion: angle A is an acute angle by definition of acute angle.

Given: B is between point A and C on line AC.
Conclusion: AB + BC = AC by partition postulate.

As you can see, you have been doing proofs, just not formally for awhile. We will continue these notes as we go along in our learning of proof writing. This is for a basic class in High School. There are many more ways to write proofs along with many more vocabulary words.

Geometry Chapter 2.5 - 2.8 Logic

2008-09-18T06:06:00.005-10:00

Geometry chapter 2.5 Conditional Statements:

I) Vocabulary:
A. Conditional Statements - (if-then) - has two parts:

1. Hypothesis - (if)
2. Conclusion - (then)

Example: If it is noon in Georgia, then it is 9 am in California.

B) Converse Statement: Switch the hypothesis and conclusion of a Conditional Statement.

Example: If it is 9 am in California, then it is noon in Georgia.

C) Inverse Statement: negate both the hypothesis and conclusion of a Conditional Statement.

Example: If it is not noon in Georgia, then it is not 9 am in California.

1) Negation - write the negative of the statement.

Example: Statement: Angle A is acute.
Negation: Angle A is not acute or It is not true that angle A is acute.

D) Contrapositive Statement: Switch and negate both the Hypothesis and Conclusion of a Conditional Statement.

Example: If it is not 9 am in California, then it is not noon in Georgia.

E) Equivalent Statements: 2 statements that are both true or both false, they have the same truth value. Contrapositive statements is always equivalent to its Conditional statement. The converse statement is always equivalent to its inverse statement.

F) Counterexample: An example that shows that a conditional statement is false.

Example: If x² = 25, then x = 5

A counterexample is x = (-5) because (-5)² = 25 but 5 is not equal to (-5)

II) Postulates:

#5) Postulate 5: Through any 2 points there exists one line.
#6) Postulate 6: A line contains at least two points.
#7) Postulate 7: If two lines intersect, then their intersection is exactly one point.
#8) Postulate 8: Through any three noncollinear points there exists exactly one plane.
#9) Postulate 9: A plane contains at least three noncollinear points.
#10) Postulate 10: If two points lie in a plane, then the line containing them lies in the plane.
#11) Postulate 11: If two planes intersect, then their intersection is a line.

Geometry Chapter 2.6 Definitions and Biconditional Statements
I) Vocabulary:

A) Perpendicular: Two lines that intersect to form right angles.

B) Line Perpendicular to a Plane: A line that intersects a plane in a point and is perpendicular to every line that includes that point in the plane that intersects it.

C) Biconditional Statement: A statement that contains the words "if and only if" (iff) and is equivalent to writing a statement combining a conditional statement and its converse.

For the truth value of a biconditional statement to be true, both the conditional statemene and its converse have to have the same truth value.

Example: Conditional Statement: If two sides of a triangle are congruent, then the angles opposite them are congruent.
Converse Statement: If two angles of a triangle are congruent, then the sides opposite them are congruent.

Biconditional Statement: Two sides of a triangle are congruent if and only if the two angles of the triangle are congruent.

2.7 The Laws of logic -
patterns that are frequently used in drawing conclusions.

I. The law of Detachment -
A valid argument - uses a series of statements called premises that have known truth values to arrive at a conclusion.

If a conclusion is true (p → q ) and the hypothesis (p) is true, then the conclusion (q) is true.

Example: Given the following true statements, what can we conclude?
1. If adjacent angles are supplementary, then the angles form a linear pair.
2. Angle ABC and angle CBD are adjacent supplementary angles.

Conclusion: angle ABC and angle CBD form a linear pair.

II. The Law of Disjunctive Inference -
A. If a disjunction (p V q) is true and the disjunction (p) is false, then the other disjunction (q) has to be true.
B. If a disjunction (p V q) is true and the disjunction (q) is false, then the other disjunction (p) has to be true.

Example: Given the following true statements, what is a valid conclusion?
1. Paul is tall or Mort is short.
2. Paul is not tall.

Since Paul is not tall, so Mort has to be short.
Conclusion: Mort is short.

Example: Given the following true statements, what is a valid conclusion?
1. Carl listens to the radio or he cannot do his homework.
2. Carl cannot do his homework.

Since Carl cannot do his homework is true, then we have (p V true) is true. p could be either true or false so therefore:

Conclusion: No conclusion

Example: Given the following true statements, what sport does each person play?
1. Zach, Steve and David each play a different sport: basketball, soccer, or baseball. Zach made each of the following true statements:
2. I do not play basketball.
3. If Steve does not play soccer, then David plays baseball.
4. David does not play baseball.

What sport does each person play?
David does not play baseball. So therefore Steve does not play soccer is false for the 3rd statement to be true. This means that Steve does play soccer. This leaves David to play basketball and Zach to play baseball.

Example: Given the following true statements, which stock did Victoria sell yesterday?
1. Victoria owns stock in 3 companies: Alpha, Beta and Gamma.
2. Yesterday, Victoria sold her shares of Alpha or Gamma.
3. If she sold Alpha, then she bought more shares of Beta.
4. Victoria did not buy more shares of Beta.

Since Victoria did not buy more shares of Beta, then she did not sell Alpha. Since she did not sell Alpha, she had to sell her shares of Gamma.

Conclusion: She sold her shares of Gamma.

Chapter 2 Logic

2008-09-17T04:53:00.014-10:00

Chapter 2 - Logic

Who are Leibniz, Boole, DeMorgan?

2 - 1 Sentences, Statements, and Truth Value

I. Logic - is the science of reasoning.
- help us to determine if a statement is true, false, or uncertain
- (the truth value of the statement)

The statements that we will use will be mathematic sentences

3 + 5 = 8 This is a true mathematical sentence

A midpoint of a line segment divides the line segment into 2 congruent parts.
This is a true mathematical sentence.

A pentagon is a six-sided polygon. This is a false mathematical sentence.

4 + 6 = 10. This is a false mathematical sentence.

II. Nonmathematical Sentences and Phrases
Sentences that do not state a fact, such as questions, commands, or exclamations
are not sentences that we use in the study of logic.

Example:

4 - 3 This is not a mathematical sentence

Go to your room! This is not a mathematical sentence.

III. Open Sentences - sentences that contain a variable.

Example:

x + 2 = 16 open sentence; variable is x

He ran the football in for a touchdown. open sentence; variable is he

17 - x = 9 open sentence; variable is x

Domain - is the input or the replacement set - the set of elements that are possible replacements for the variable.

Example:
x + 2 = 16 with the domain {5, 6, 7, 8, 9 }

Solution Set: the element or elements from the domain that make the open sentence true.

Therefore, the solution set is {8} because when x = 8, then 17 - 8 = 9 is true.

IV. Statement or closed sentence - can be judged to be true or false (no variables) - the truth value is either true (T) or false (F)

V. Negations of a statement always has the opposite truth value of the given statement.

Example: A frog is a snake. (This is a false statement).
Negation: A frog is not a snake. (This is a true statement).

Example: A triangle is not a polygon with 4 sides. (this is a true statement)
Negation: A triangle is a polygon with 4 sides. (this is a false statement)

to write this negation, you may have wanted to say:
A triangle is not not a polygon with 4 sides. As this doesn't make grammarical sense, we change the double negative to a positive.

VI. You can let a statement be represented by a lowercase letter.

Usually we use p, q, r, or s

If "p" is true, then not "p" or "~p" is false.

~p represents symbolically: not p.

Example:

Let p: Summer follows spring.

~p: Summer does not follow spring.

p is true so ~p is false

2 - 2 Conjunctions: is a compound statement formed by combining two simple statements using the word "and". From the table above, you see the symbol for and looks like an upside down V.

Example:

Let p: A dog is an animal.

Let q: A trumpet is a brass instrument.

What is the sentence for p and q?

A dog is an animal and a trumpet is a brass instrument.

What is the truth value for the following (we will negate the different parts of the sentence)

1. A dog is an animal and a trumpet is a brass instrument. (T and T = T)

2. A dog is an animal and a trumpet is not a brass instrument. (T and F = F)

3. A dog is not an animal and a trumpet is a brass instrument. (F and T = F)

4. A dog is not an animal and a trumpet is not a brass instrument. (F and F = F)

2 - 3 Disjunctions - is a compound statement formed by combining two simple statements using the word "OR" and the symbol for or is V.

Example:

Let p: January is the first month of the year.

Let q: Breakfast is a meal.

What is the sentence for p or q?

January is the first month of the year or breakfast is a meal.

What is the truth value for the following (we will again negate each part of the sentence)

January is the first month of the year or breakfast is a meal. (T or T = T)

January is the first month of the year or breakfast is not a meal. (T or F = T)

January is not the first month of the year or breakfast is a meal. (F or T = T)

January is not the first month of the year or breakfast is not a meal. (F or F = F)

Example:

Buffalo Bills is a football team.

Buffalo Sabres is a hockey team.

the or statement would be:

Buffalo Bills is a football team or Buffalo Sabres is a hockey team.

What is the truth value of this sentence?

True

Example:

Let our Set A = {1, 2, 3}

Set B = {2, 4, 6}

What is:

Set A V Set B = {1, 2, 3, 4, 6}

you include all the elements in both sets for "or"

Set A and Set B = {2}

you include only the elements that are in both set A and set B. This is what you called the intersection of the two sets.

I. Complement - is that which is not included in the set

Example:

the complement of Set A = {4, 6}

the complement of Set B = {1, 3}

II. Inclusive "or" - when we use the word "or" to mean that one or both of the simple sentences are true. This is the truth table we have used in this class.

2 - 4 Conditional - is a compound statement formed by using the words:

if ... then

symbolically: if p then q is p → q

"p" is the hypothesis or premise or antecedent

"q" is the conclusion or consequent

Example:

If today is Tuesday, then tomorrow is Wednesday.

Hypothesis: today is Tuesday

Conclusion: tomorrow is Wednesday

Other ways to write conditionals.

1. If today is Tuesday, then tomorrow is Wednesday.

2. Today is Tuesday implies that tomorrow is Wednesday.

3. When today is Tuesday, tomorrow is Wednesday.

Geometry - Chapter 1 - Essentials of Geometry

2008-09-10T10:48:00.022-10:00

Chapter 1 - Essentials of Geometry

1-1 Undefined Terms:
A. Undefined terms: their meaning is accepted without definition
1. Point: definition - is that which has no part, a dot.
2. Set: is a collection of objects
3. Line - is breathless length;
4. Straight line: is a line that lies evenly with the points on itself.
5. Plane: is a set of points that form a flat surface extending indefinitely in all directions.

1-2 The real numbers and their Properties
B. Every real number corresponds to a point on a number line and every point on the number line corresponds to a real number.

1-3 Definitions, lines, and line segments

A. Definition: is a statement of the meaning of the term

B. Collinear set of points: is a set of points all of which lie on the same straight line.

C. Noncollinear set of points: is a set of three or more points that DO NOT all lie on the same straight line.

D. Distance between two points - every point on a line corresponds to a real number called its coordinate. To find the distance between any two points, find the absolute value of the difference between the coordinates of the two points.

If point A is at -2 and point D is at 1, what is the distance between A and D?

take the absolute value of (-2 - 1) = the absolute value of (-3) = 3

E. Order of points on a line
1. Betweenness: B is between A and C if and only if A, B, and C are distinct collinear points and AB + BC = AC

also, this is the partition postulate: the part + the part = the whole

2. Line Segment: or segment, is a set of points consisting of two endpoints on a line, called endpoints, and all of the points on the line between the endpoints.

1-5 Rays and Angles:

Definition: two points, A and B, are on one side of a point P if A, B, and P are collinear and P is not between A and B.

Half-line: every point on a line divides the line into 2 opposite set of points called half-lines.

Definition: A ray consists of a point (endpoint) on a line and all points on one side of the point.

Definition: Opposite rays: are two rays of the same line with a common endpoint and no other point in common.

Point B is the vertex, BA is a side of the angle and BC is the other side of the angle.

Definitions:

The measure of an angle is the number of degrees in the angle.

A straight angle is an angle that is the union of opposite rays.

Its degree measure is 180 degrees

Acute Angles - is an angle whose degree measure is greater than 0 degrees and less than 90 degres.

Right angles - is an angle whose degree measure is 90 degrees.

Obtuse angle - is an angle whose degree measure is greater than 90 degrees and less than 180 degrees.

Note:If you bisect a straight angle, you get two 90 degree angles or 2 right anglesIf you bisect an obtuse angle, you get two acute angles.

1-6 More angle definitions

Congruent angles are angles that have the same measure.

Definition: A bisector of an angle is a ray whose endpoint is the vertex of the angle and that divides that angle into two congruent angles.

Given angle ABC with angle bisector BD, we can conclude that angle ABD is congruent to angle CBD.

If angle ABC measures 70 degrees, what is the measure of angle ABD? 35 degrees

Example: Given angle QRS with angle bisector RT, if the measure of angle QRS = 10x, and the measure of angle SRT = 3x + 30, what is the measure of angle QRS?

we know: measure of angle QRT = measure of angle SRT + measure of angle QRS

and since the angle QRS is bisected by ray RT, then angle QRT = angle SRT

so 3x + 30 = measure of angle SRT,

therefore

3x + 30 + 3x + 30 = 10x

6x + 60 = 10x

60 = 4x

15 = x

measure of angle QRS = 10x = 10 (15) = 150 degrees

1-7 Triangles:

I) Vocabulary:

A) Base angles - the two angles adjacent to the base of an isosceles triangle.

B) Vertex angles - the angle opposite the base of an isosceles triangle.

II) Theorems:

A) Base Angles Theorem: if two sides of a triangle are congruent, then the angles opposite them are congruent.

B) Converse of the Base Angles Theorem: if two angles of a triangle are congruent, the the sides opposite them are congruent.

Corollaries:

A) If a triangle is equilateral, then it is equiangular.

B) If a triangle is equiangular, then it is equilateral.

I) Vocabulary:

A) Triangle - is a figure formed by three segments joining three noncollinear points.

B) Classification of triangles:

1) By sides:

a) Equilateral Triangle - has 3 congruent sides

b) Isosceles Triangle - has at least 2 congruent sides

c) Scalene Triangle - has no congruent sides

2) By angles:

a) Acute Triangle - has 3 acute angles

b) Equiangular Triangle - has 3 congruent angles that measure 60 degrees each

c) Right Triangle - has one right angle and 2 acute angles

d) Obtuse Triangle - has one obtuse angle and 2 acute angles

C) Vertex - each of the three points joining the sides of a triangle (plural - vertices)

D) Adjacent Sides - in a triangle, two sides sharing a common vertex.

E) Right triangles have 2 sides that form the right angle called the legs. The side opposite the right angle is the hypotenuse of the triangle.

F) Interior angles - when the sides of a triangle are extended, other angles are formed. the three original angles are the interior angles.

G) Exterior angles - the angles that are adjacent to the interior angles.

H) Theorem:

1) Triangle sum theorem - the sum of the measures of the interior angles of a triangle is 180 degrees.

2) Exterior Angle Theorem - the measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles.

I) Corollary to a theorem - is a statement that can be proved easily using the theorem.

Example: The acute angles of a right triangle are complementary.

Example: given Triangle ABC with side BC extended through point D, if angle A = 65 degrees and angle ACD = 2x + 10 and angle B = x, solve for x.
angle A + angle B = angle ACD65 + x = 2x + 1055 = x

Geometry - Similar Triangle Proofs

2008-04-08T10:25:00.005-10:00

Similar Triangle Proofs:

If 2 angles of one triangle are congruent to two corresponding angles of a second triangle, then the third angles are congruent.

There are 3 ways to show triangle similarity:

Let's try a proof:

Let's try another:

Geometry HW #14

2008-04-01T11:03:00.004-10:00

For my Geometry Class: here is homework #14, questions 1 & 2 Quadrilateral Proofs

Quadrialteral Properties

2008-03-26T02:53:00.004-10:00

Quadrilateral "Family Tree" -

Quadrilateral - 4 sided polygon

A. Kite -
1. has consecutive sides congruent but opposite sides are not.
2. the diagonals are perpendicular
3. has exactly one pair of opposite angles congruent.

B. Trapezoid -
1. Has exactly one pair of opposite sides parallel

C. Isosceles Trapezoid -
1. Has exactly one pair of opposite sides parallel
2. the non-parallel sides are congruent (legs congruent)
3. the diagonals are congruent
4. the base angles are congruent

D. Parallelogram -
1. Definition: has both pairs of opposite sides parallel
2. has both pairs of opposite sides congruent
3. has one pair of opposite sides parallel and congruent
4. has both pairs of opposite angles congruent
5. has consecutive angles supplementary
6. the diagonals bisect each other
7. the diagonals divide the parallelogram into 2 congruent triangles

E. Rhombus -
1. All of the properties of a parallelogram
2. All sides are congruent (equilateral)
3. diagonals are perpendicular
4. diagonals bisect opposite angles

F. Rectangle -
1. All of the properties of a parallelogram
2. all angles are congruent (equiangular)
3. diagonals are congruent

G. Square -
1. All the properties of a parallelogram, rhombus and rectangle

check out this website:
http://regentsprep.org/Regents/math/quad/LQuad.htm

Coordinate Geometry

2008-03-26T02:06:00.009-10:00

Geometry - Coordinate Geometry:

check out this website:
http://regentsprep.org/Regents/mathb/1D/Coordinatelesson.htm

B(x₁, y₁) and A (x₂, y₂) is from the
Pythagorean Theorem which is:

c² = a² + b²

since a = (x₂ – x₁) and b = (y₂ – y₁)

we get

c² = (x₂ – x₁)² + (y₂ – y₁)²
so taking the square root of both sides we have our distance formula.

Here are the Slope Formula and the Midpoint Formula:

Using the following chart, if we have to prove congruent segments, we have to show they have equal length by using the distance formula.

Here is a web-site to hopefully help:

http://library.thinkquest.org/20991/geo/coordgeo.html

Geometry Proofs - unit 2 Triangles

2008-02-25T12:23:00.007-10:00

Triangles – Proofs – Unit 2
1. Δ ABC is congruent to Δ ABC , this is by the Reflexive Postulate
2. if Δ ABC is congruent to Δ DEF then Δ DEF is congruent to Δ ABC, this is by the Symmetric Postulate
3. if Δ ABC is congruent to Δ DEF and Δ DEF is congruent to Δ GHI, Then Δ ABC is congruent to Δ GHI by the Transitivity Postulate

5 ways to show Triangle Congruence:
1. SAS = SAS Congruence Postulate - if two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the two triangles are congruent.
Example: Given ΔABC and ΔDEF,
if AB = DE, BC = EF and angle B = angle E, then ΔABC = ΔDEF.

2. SSS = SSS Congruence Postulate - if three sides of one triangle are congruent to three sides of a second triangle, then the two triangles are congruent.
Example: Given ΔABC and ΔDEF,
if AB = DE, BC = EF and AC = DF, then ΔABC = ΔDEF.

3. ASA = ASA Congruence Postulate - if two angles and the included side of one triangle are congruent to two angles and the included side of a second triangle, then the two triangles are congruent.
Example: Given ΔABC and ΔDEF,
if angle A = angle D, AB = DE, and angle B = angle E, then Δ ABC = Δ DEF.

4. AAS = AAS Congruence Postulate - if two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of a second triangle, then the two triangles are congruent.
Example: Given Δ ABC and Δ DEF,
if angle A = angle D, angle C = angle F, and BC = EF , then Δ ABC = Δ DEF.

5. HL = HL Congruence Postulate - if the leg and hypotenuse of one right triangle is congruent to the corresponding leg and hypotenuse of another right triangle, then the two triangles are congruent by hypotenuse - leg postulate.
Example: Given right Δ ABC and right Δ DEF, if angle B and angle E are both right angles and leg AB = leg DE and hypotenuse AC = hypotenuse DF, then Δ ABC = Δ DEF.

I) Vocabulary:
A) When two figures are congruent, there is a correspondence between their angles and sides such that corresponding angles are congruent and corresponding sides are congruent.
Example: Given Δ ABC is congruent to Δ PQR then we know
1) angle A = angle P, angle B = angle Q, and angle C = angle R
2) AB = PQ, BC = QR, and AC = PR By Corresponding Parts of Congruent Triangles are Congruent (CPCTC) - which means that if 2 triangles are congruent, then their corresponding parts are congruent

Make sure that you list the corresponding angles in the same order with the triangle congruence.
Example: ΔABC = ΔDEF is not the same as ΔABC = ΔEFD because
ΔABC = ΔDEF has angle A = angle D, angle B = angle E and angle C = angle F
while ΔABC = ΔEFD has angle A = angle E, angle B = angle F and angle C = angle D

B) Third Angles Theorem - if two angles of one triangle are congruent to two angles of another triangle, then the third angles are also congruent.

C) Reflexive Postulate - Every triangle is congruent to itself

D) Symmetric Postulate - If ΔABC = ΔDEF, then ΔDEF = ΔABC

E) Transitive Postulate - If ΔABC = ΔDEF and ΔDEF = ΔJKL, then ΔABC = ΔJKL

Theorems to remember:

If 2 angles of one triangle are congruent then the sides opposite are congruent.
If 2 sides of one triangle are congruent then the angles opposite are congruent.
________________________________________________________________
Practice Proofs:

Precalculus Unit 8 - chapter 6.3, 6.4, 10.5, 10.6, 10.7

2008-02-15T03:46:00.003-10:00

PRECALCULUS
Unit 8
Vectors and Parametric/Polar Equations

Section 6.3A HW# 53; Pg.453; #1 – 45 by 4’s (1, 5, 9, …)

Section 6.3B; HW # 54; *Pg.453; #3, 11, 19, 23, 31, 39, 43
Pg.454; #49, 53, 57, 61, 65, 69, 73, 74, 82

Section 6.4; HW # 55; *Pg. 454; #51, 55, 59, 63, 67, 71,
Pg. 464; #1, 5, 9, 13, 17, 21, 25, 29, 33, 37
Quiz on Sections 6.3A – 6.3B Next Class

Section 10.5; HW# 56; *Pg.464; #3, 7, 11, 15, 19, 23, 27, 31
Pg.736; #1, 5, 9, 13, 21, 29, 41, 47

Section 10.6; HW# 57; *Pg.736; #3, 7, 11, 17, 27, 43
Pg.743; #5 – 41 by 4’s, 45, 50, 53 – 57 odds
Quiz on Sections 6.4 – 10.5 Next Class

Section 10.7; HW# 58; *Pg.743; #7, 11, 15, 23, 27, 39, 47, 59
Pg.752; #21 – 33 odds, 55, 59

Review HW #59
Pg.480; 39, 43, 47, 51, 59, 69, 79, 83, 85
Pg.762; 47, 49, 59, 65, 69, 73, 75, 81, 85, 89

HW #60; Unit 8 Test

Unit One for Proofs - Angles and Lines

2008-02-05T08:06:00.000-10:00

The addition Postulate:

If a = b and c = d, then a + c = b + d

The Partition Postulate:

AB + BC = AC

These are two different concepts. Let's try a proof:

As you can see, in step 2, we added equal quantities to each other. This is Addition Postulate.

But in step 3, we added a part plus a part equals a whole so this is Partition.

Let's try another example:

Here we used the Subtraction Postulate: subtracting equal quantities from equal quantities.

In step 3, we had the whole minus a part equals a part. This is still Partition Postulate.

Geometry Vocabulary for Proofs

2008-01-31T10:50:00.000-10:00

Vocabulary - Euclidean Geometry:

1. Space - is a set of points that forms a completely flat surface extending indefinitely in all directions.

2. Collinear Points - is a set of points all of which lie on the same straight line.

3. Coplanar points - is a set of points all of which lie on the same plane.

4. Betweenness of Points or Segment/Angle Addition Postulate (we will call this Partition Postulate) - point B is between point A and point C, if A, B and C are distinct collinear points and AB + BC = AC. We shall say that the part plus the part equals the whole (part + part = whole)

5. Segment or line segment - is a set of points consisting of two points on a line, called endpoints, and all the points on the line between the endpoints.

6. Length of a line segment - is the distance between the endpoints.

7. Congruent segments - are segments that have the same measure.

Congruent angles - are angles that have the same measure.

8. Midpoint of a segment - is the point of the line segment that divides the segment into two congruent segments or divides the line segment in half.

9. Bisector of a Segment or angle - is any line, ray, or point that intersects the segment or angle at its midpoint.

10. Rays - is a part of a line that consists of a point on the line called the endpoint and all the points on one side of the endpoint

11. Opposite Rays - are two rays of the same line with a common endpoint and no other point in common.

12. Angle - is a set of points that is the union of two rays having the same endpoint or vertex.

13. Sides of an angle - are the rays that make up the angle

14. Vertex of an angle - is the common endpoint of the two sides of an angle.

15. Adjacent angles - are two angle in the same plane that have a common vertex and a common side but do not have any common interior points.

16. Exterior sides of adjacent angles - the two sides of adjacent angles that are not common to both angles.

17. Vertical angles - are two angles in which the sides of one angle are opposite rays to the sides of the second angle. Vertical angles are congruent.

18. Addition Postulate - if A = B, then A + C = B + C

19. Subtraction Postulate - if A = B, then A - C = B - C

20. Multiplication Postulate - if A = B, then AC = BC

21. Division Postulate - if A = B and C does not equal zero, then A/C = B/C

22. Halves Postulate - if A = B, then A/2 = B/2

23. Doubles Postulate - if A = B, then 2A = 2B

24. Substitution Postulate - if A = 5 and A + B = 12, then 5 + B = 12.

- you plug in the equal value

25. Reflexive Postulate - A = A

26. Symmetric Postulate - if A = B, then B = A.

27. Transitivity Postulate - if A = B AND B = C, then A = C

OR if A is less than B AND B is less than C, then A is less than C.

28. Distributive Postulate - if A(B + C) then AB + AC

Other Vocabulary:

Congruent Angles - angles that have the same measure.

Complementary Angles - are two angles that the sum of their measures is 90 degrees

Supplementary Angles - are two angles that the sum of their measures is 180 degrees.

Perpendicular Lines - form right angles.

Right angle - has a measure of 90 degrees.

Linear pair - are two adjacent angles where their noncommon side are opposite rays.

Here is one way how to write a proof:

1. You set up a t-table and write the word Statements in the left column and Reasons in the right column.

2. You are given a statement with another statement to prove based on the given statement.

3. You write the given statement in the left column with the word given for your reason. Then you write statements that support what you are given with logically reason using vocabulary, postulates or already proven theorems.

Here are a few examples of algebraic proofs:

As you can see, you have been doing proofs, just not formally for awhile. We will continue these notes as we go along in our learning of proof writing. This is for a basic class in High School. There are many more ways to write proofs along with many more vocabulary words.

Precalculus Midterm Requirements

2008-01-02T10:20:00.000-10:00

Requirements for Midterm:

Do 3 of the 6 projects from Chapters: Prerequisite, One, Two, Three, Four, or Five
These can be found at the end of each of the chapters.
Handwritten is fine. If you type the project out, one extra credit point per problem.
Show all work!! No work, no credit. Each question is worth 15 points.

Total points for midterm: 45 points.

The projects can be found on the following pages:
Prerequisite, page 70
Chapter 1, page 133
Chapter 2, page 213
Chapter 3, page 279
Chapter 4, page 373
Chapter 5, page 425

Answer all of the questions for each project chosen.

Precalculus Homework Unit 6 - chapter 5

2007-12-10T03:54:00.000-10:00

Precalculus Homework Unit 6 - Chapter 5

5.1 Homework #38; Pg.381; 5, 7, 25, 27, 37, 43, 55, 59,
67, 71, 75-77odd, 91, 101, 103

5.2 Homework #39; Pg.389; 5-9odd, 21-25odd, 34, 35, 43, 55, 71;
*Pg. 381; 39, 41, 57, 69, 73, 93

5.3a Homework #40; Pg.400; 3, 11, 15, 23, 31, 39, 43, 47, 51, 55
*Pg. 389; 3, 19, 27, 37, 49, 59
Quiz on Sections 5.1-5.2 Next Class

5.3b Homework #41; Pg.400; 5, 7, 19, 27, 35, 41, 45, 53, 57, 59, 75, 76

5.4 Homework #42; Pg.408; 3, 5, 11, 15, 19, 25-27odd, 37-41odd, 45, 47 – 55 odd
*Pg 400; 7, 19, 27, 35, 43

5.5a Homework #43; Pg.418; 5, 7, 17, 19, 21-23odd, 31, 39,
43, 47, 51-57odd, *Pg 408; 21, 23, 35, 43, 57
Quiz on Sections 5.3-5.4 Next Class

5.5b Homework #44; Pg.418; 9, 18, 22, 27, 35, 45, 65, 71,
83-87odd, 103

Review Homework #45; Pg.422; 1 – 107 every 4th odd (1, 5, 9, …)

Homework #46; Chapter 5 Test

Precalculus 2.6 Rational Functions and Asymptotes

2007-10-23T07:50:00.000-10:00

2.6 Rational Functions and Asymptotes

A rational function can be written in the form:

f(x) = N(x)/D(x)

where N(x) is the numerator and
D(x) is the denominator and both are polynomials but D(x) is not the zero polynomial (because you cannot divide by zero)

I. Asymptotes of a Rational Function:
Let "f" be the rational function f(x) where

f(x) = N(x)/D(x) where

N(x) = a_nxⁿ + a_n-1x^n-1 + ... + a₁x + a₀ and

D(x) = b_mx^m + b_m-1x^m-1 + ... + b₁x + b₀

Where N(x) and D(x) have no common factors.
1. The graph of "f" has vertical asymptotes at the zeros of D(x).
2. The graph of "f" has at most one horizontal asymptote determined by comparing the degrees of N(x) and D(x).
a. If n is less than m, the line y = 0 (x-axis) is a horizontal asymptote.
b. If n = m, the line y = (a_n)/(b_m) is a horizontal asymptote.
c. If n is greater than m, the graph of "f" has no horizontal asymptote.

Examples:
a) f(x) = (5x² + 3)/(-6x³ + 2x + 4)

n = 2 and m = 3 which means that n is less than m so the horizontal asymptote is y = 0

b) f(x) = (2x³ + 2x²)/(3x³ + 4x)

n = 3 and m = 3 so the line y = 2/3 is a horizontal asymptote.

c) f (x) = (3x⁴ + 2x² + 5)/(4x³ + 3x)

n = 4 and m = 3 so n is less than m, therefore there is not a horizontal asymptote.

II. Find the Domain and Asymptotes of f(x) = 3/ ((x - 2)³)

1. Find the vertical asymptotes by taking D(x) and setting it equal to zero.

(x - 2)³ = 0
x - 2 = 0
x = 2

therefore a vertical asymptote is the line x = 2

2. Find the horizontal asymptotes
n = 0 and m = 3 so n is less than m, therefore the line y = 0 is the horizontal asymptote.

3. Graph it using a graphing utility.

4. Domain (- ∞, 2) U (2, ∞)

5. Range (- ∞, 0) U (0, ∞)

6. check using a table:

{ (-3, -0.24), (-2, -.047), (-1, -.11), (0, -3.75), (1, -3), (2, error), (3, 3)}

Homework: Quiz (19, 20)
Pg. 187/ 7, 21, 23, 43*Pg. 195/ 13 – 19 odd, 27 – 31 odd, 35

Precalculus 2.5 The Fundamental Theorem of Algebra

2007-10-23T06:13:00.000-10:00

2.5 The Fundamental Theorem of Algebra –
Proved by Carl Friedrich Gauss

If f (x) is a polynomial of a degree “n”, where n is greater than 0,
“ f ” has at least one zero in the complex number system.

I. Linear Factorization Theorem:
If f(x) is a polynomial of degree “n” where “n is greater than zero”, “f” has precisely “n” linear factors.

f (x) = a_n (x – c₁)( x – c₂)( x – c₃) … ( x – c_n) where
c₁, c₂, c₃, … c_n are complex numbers.

Factors of a Polynomial:

Example 1: f (x) = t⁴ - 5t³ + 15t² - 45t + 54

4th degree n = 4, n is greater than zero is true

Step 1: find all the possible zeros of the function:
(factors of 54)/(factors of 1) = ±1, ± 2, ± 3, ±6, ± 9, ± 18, ± 27, ± 54

Step 2: Use synthetic division to find out which ones are factors

f (x) = (x – 2)(x – 3)(x² + 9)
0 = (x – 2)(x – 3)(x² + 9)

0 = x – 2
x = 2

0 = x – 3
x = 3

0 = x² + 9
- 9 = x²
±√(-9) = x
± 3i = x

So to write this function in linear form:

f(x) = (x – 2)(x – 3)(x + 3i)(x – 3i)

So the following zeros of “f” are:

x = {2, 3, 3i, and -3i}

II. Complex Zeros Occur in Conjugate Pairs

In the last example, “± 3i”, the pair is 3i and -3i so we can conclude:

Let f(x) be a polynomial function that has real coefficients.
If a + bi, where b ≠ 0, is a zero of the function,
the conjugate a – bi is also a zero of the function.

Example: Your zeros are {2, 4 + i, and 4 – i}, find the polynomial function:

f(x) = (x – 2)(x – (4 + i))(x – (4 – i))
= (x – 2)((x – 4) – i)((x – 4) + i)
= (x – 2)( (x – 4)² - (i)²)
= (x – 2)(x² - 8x + 16 – (-1))
= (x – 2)(x² - 8x + 17)
= x³ – 8x² + 17x – 2x² + 16x – 34
= x³ – 10x² + 33x – 34

III. If you are given one zero, can you find the rest?

g(x) = 4x³ + 23x² + 34x – 10 given zero: -3 + i

Recall both conjugates:
– 3 + i and – 3 – i so

(x + 3 – i) (x + 3 + i)
= ((x + 3) – i)((x + 3) + i)
= ((x + 3)² - (i)²)
= x² + 6x + 9 – (-1)
= x² + 6x + 10

So using long division:

Every polynomial of degree n is greater than zero with real coefficients can be written as the product of linear and quadratic factors with real coefficients, where the quadratic factors have no real zeros.

A quadratic factor with no real zeros is said to be irreducible over the real numbers.

x² + 1 = (x + i)(x – i) is irreducible over the real numbers
x² - 2 = (x + √2)(x - √2) is irreducible over the rational numbers but reducible over the real numbers.

Example: The complex number 4i is a zero of f(x) = x⁴ + 13x² - 48.

Find the remaining zeros of f(x), and write it in its linear factorization.

Since 4i is a zero, then -4i is also a zero so

(x-4i)(x+4i) = x² - (4i)²

= x² - 16i²

= x² + 16

therefore a factor of f(x) is x² + 16

f(x) = x⁴ + 13x² - 48
= (x² + 16)(x² - 3)
= (x + 4i)(x – 4i)(x + √3)(x - √3)

So the zeros are:
x = {-4i, 4i, -√3, √3}

Homework: Pg. 170/ 57, 65, 69
*Pg. 187/ 1 – 5odd, 9 - 19odd, 33, 37, 41, 45, 53, 55, 61, 65

Precalculus 2.4 Complex Numbers

2007-10-10T08:04:00.000-10:00

2.4 Complex Numbers

I. A complex number is written in standard form: a + bi
where a ε set of real numbers
and
bi is a pure imaginary number

Therefore a + bi is an imaginary number

√(-1) = i
(√(-1))² = i² = -1
(√(-1))³ = i³ = -i
(√(-1))⁴ = (√(-1))²(√(-1))² = (-1)(-1) = 1
(√(-1))⁰ = 1

Therefore you can see the pattern that:

i⁰ = 1
i ¹= √(-1) = i
i² = -1
i³ = -i
i⁴ = 1
so i⁵ = i
i⁶ = -1
i⁷ = - i
i⁸ = 1

so what would i⁶³ = ?

i⁶⁰i³
i⁶⁰ = 1 and so therefore
i⁶⁰i³ = (1) i³ = - i

B. Additive Identity is zero
what do you add to a number to get that number? of course, zero so

a + bi + ________ = 0

______ = -a - bi

C. Adding and Subtracting complex Numbers:

Example 1: (3 + 4i) + (7 + 2i)
= (3 + 7) + (4i + 2i) by grouping like terms
= 10 = 6i

Example 2: (3 + 4i) - (7 + 2i)
= (3 - 7) + (4i - 2i)
= (-4) + (2i)
= -4 + 2i

D. Multiplying Complex Numbers
Example 1: (3 + 4i)(7 + 2i)
= (3)(7) + (3)(2i) + (4i)(7) + (4i)(2i)
= 21 + 6i + 28i + 8i²
= 21 + 34i + 8(-1)
= 21 + 34i - 8
= 13 + 34i

E. Dividing Complex Numbers - multiply by the complex conjugate

Given "a + bi", the complex conjugate would be "a - bi"

Example 1: complex #1

but recall that "a + bi" has to be in standard form so...

29/53 + 22i/53 would be the answer!

F. Applications:
1. Fractal Geometry

Mandelbrot Set: to draw this, consider: Sequence of numbers

c, c² + c, (c² + c)²+ c, [(c² + c)²+ c ]² + c, ...

For some values, it is Bounded, which means that all elements in the sequence are less than some fixed number N. Therefore, complex number "c" is in the Mandelbrot Set.

For other values, it is Unbounded, which means that the elements in the sequence become infinitely large. Therefore, complex number "c" is not in the Mandelbrot Set.

Check out this website:

http://mathworld.wolfram.com/MandelbrotSet.html

Example 1: Let's let c = -1

So the sequence would be:

-1, (-1)² + (-1) = 0, (0)² + (-1) = -1, (-1)² + (-1) = 0

or -1, 0, -1, 0

so this is bounded!

Example 2: Let's let c = - i

-i, (-i)² + (-i) = -1 + -i, (-1 -i)² + (-i) = -3i, (-3i)² + (-i) = -9 - i

or -i, -1 - i, -3i, - 9 - i

so this is not bounded.

2. Impedance - the opposition to current in an electrical circuit.

Equation of 2 pathways:

1/z = 1/z₁ + 1/z₂

where

z₁ is the impedance of pathway 1

z₂ is the impedance of pathway 2

Precalculus 2.3 Real Zeros of Polynomial Functions

2007-10-06T05:02:00.000-10:00

Precalculus 2.3a

I. Long Division of Polynomials:

A. Recall: When you did long division

Example 1:

625/5 =

As you can see, 5 goes into 625 perfectly and there is no remainder so this means that 5 is a factor of 625.

Example 2:
625/4 =

As you can see, 4 does not go into 625 perfectly, there is a remainder of 1 so we would write the answer like

156 + 1/4

You write the answer or quotient plus the remainder over the divisor

B. Long Division of Polynomials:

Now try doing the same concept only using Polynomials:

Example 3:

So therefore since (x - 4) is a factor so

x - 4 = 0
x = 4

Therefore (x - 4) is a factor and since there is not a remainder, then (x - h) is a factor and x = h.

Example 4:

Therefore (x + 2) is a factor and since there is not a remainder, then (x - h) is a factor and x = h,
then x = -2

Example 5:

Since when you divide by (x + 1), it has a remainder, it is not a factor and x = -1 is not a zero.

Example 6:

So again, since (x - 2) is not a factor, x = 2 is not a zero.

Precalculus 2.3b Synthetic Division:
To divide ax³ + bx² + cx + d by (x - k), use the following pattern:

Example 1: (5x² - 17x - 12) / (x - 4)

Therefore, (5x² - 17x - 12) /(x - 4) = 5x + 3

since there is not a remainder!

Example 2: (x⁴ + 5x³ + 6x² - x - 2) / (x + 2)

Again, since the remainder is 0,

(x⁴ + 5x³ + 6x² - x - 2) / (x + 2) =

x³ + 3x² - 1

Example 3: (4x³ - 13x + 10) / (x + 1)

Since there is not an x², we put zero in that place:

Since 19 is the remainder:

(4x³ - 13 x + 10) / (x + 1) = 4x² - 4x - 9 + 19/(x + 1)

C. The Remainder Theorem: If a polynomial f (x) is divided by (x - k), the remainder is r = f (x)

Recall from Example 3, we saw the remainder = 19 so...

f (-1) = 4(-1)³ - 13 (-1) + 10= 4 (-1) + 13 + 10= -4 + 13 + 10= 19

this gives the same answer!!

Let's try another:

Example 4: (4x³ - 13x + 10) / (x - 2)

f(2) = 4(2)³ - 13(2) + 10 = 16, again, the same answer!

D. The Factor Theorem:
A polynomial f(x) has a factor (x - k) if and only if f(k) = 0.

In Summary:

The remainder "r", obtained in the synthetic division of f (x) by (x - k) provides the following information:
1. the remainder "r" gives the value of "f" at x = k. That is r = f (k)

2. If r = 0, then (x - k) is a factor of f (x)

3. If r = 0, then (k, 0) is an x-intercept of the graph of f (x).

II. The Rational Zero Test:
relates the possible rational zeros of a polynomial

Rational zero = p/q

where "p" and "q" have no common factors other than 1, "p" is a factor of the constant a₀ and q is a factor of the leading coefficient a_n
polynomial:

f(x) = a_nxⁿ + a_n-1x^n-1 + ... + a₂x² + a₁x + a₀

Possible rational zeros = (factors of constant)/(factors of leading coefficient)

Example 1: f (x) = 4x⁵ - 8x⁴ - 5x³ + 10x² + x - 2

So the possible rational zeros = (1, -1, 2, -2) / (1, -1, 2, -2, 4, -4) so

The possible zeros are : {1, -1, 1/2, -1/2, 1/4, -1/4, 2, -2}

Looking at the graph of the function, we see that the real zeros are:
(-1, 0), (-1/2, 0), (1/2, 0), (1, 0), and (2, 0)

You can also use synthetic Division to eliminate possible zeros:
f (x) = 4x⁵ - 8x⁴ - 5x³ + 10x² + x - 2

try x = -1,

so f(x) = (x + 1)(4x⁴ - 12x³ + 7x² + 3x - 2)

so f (x) = (x - 1)(x + 1)(4x³ - 8x² - 1x + 2)

Therefore f(x) = (x - 1)(x + 1)(x -2)(4x² - 1)

f(x) = (x -1)(x + 1)(x -2)(2x + 1)(2x - 1)

so x = -1, x = 1, x = 2, x = -1/2, x = 1/2

so you can see they are the same answers.

III. Third Test for zeros:

A real Number "b" is an upper bound for the real zeros of "f" if no zeros are greater than "b". Similarily, "b" is a lower bound if no real zeros of "f" are less than "b"

Example 1: f(x) = x⁴ - 4x³ + 15

1. degree is even

2. Leading coefficient is positive

so it rises to the left and rises to the right

3. points (-3, 204), (-2, 63), (-1, 20), (0, 15), (1, 12), (2, -1), (3, -12), (4, 15)

So from the points we chose, the lower bound will be 1 and the upper bound is 4

We can use synthetic division to verify these thoughts:

Let x = -1, you can see that the remainder is 20 and the last row is:

1, -5, 5, -5, 20

Let x = 4, you can see that the remaider is 15 and the last row is:

1, 0, 0, 0, 15

Let x = 1, you can see that the remainder is 12 and the last row is:

1, -3, -3, -3, 12

Therefore we can conclude, given x = c:

1. If c is greater than zero and each number in the last row is either positive or zero, "c" is an upper bound for the real zeros of "f".

2. If c is less than zero and each number in the last row are alternately positive and negative, (zero counts as either positive or negative ), "c" is a lower bound for the real zeros of "f".

Therefore -1 is a lower bound and 4 is an upper bound.

Precalculus 1.5 Inverse Functions

2007-09-26T04:03:00.001-10:00

Precalculus 1.5 Inverse Functions:

I. Inverse Functions:
A. Let "f " and "g" be two functions such that
f (g(x)) = x for every "x" in the domain of "g"
and
g( f (x)) = x for every "x" in the domain of "f ".

Under these conditions, the function "g" is the INVERSE of the function "f ":

The function "g" is denoted by f ^-1 (read f-inverse), so:

f ( f ^-1(x)) = x

AND

f ^-1( f (x)) = x

The domain of "f " must be equal to the range of f ^-1
AND
the range of "f " must be equal to the domain of f ^-1

Example:
Given f (x) = {(1,2), (4,5), (6,7)}

then f ^-1 (x) = {(2,1), (5,4), (7,6)}

The graphs of "f " and "f ^-1 are related to each other by the fact the graph of "f ^-1" is a reflection of the graph of "f " in the line y = x.

Example:
If the function f has values: {(1,0), (2,3), (4,7)}

then f ^-1 has values of {(0,1), (3,2), (7,4)}

Example:
Find the inverse of f (x) = 4x + 8

1. let f (x) = y, so by substitution we get

y = 4x + 8

2. Switch the x and y

x = 4y + 8

3. Solve for y

x - 8 = 4y

(1/4) x - 2 = y

4. Now y = f ^-1 so

f ^-1 = (1/4) x - 2

Example: Show that f(x) = (x + 8)/3 and g(x) = 3x - 8 are inverse functions of each other. Recall that f(g(x)) = x and g(f(x)) = x therefore:

f (g (x)) = ( (3x-8)+8)/ 3 = (3x)/3 = x so f(g(x)) = x

g ( f (x)) = 3((x + 8)/3) - 8 = x + 8 - 8 = x so g (f (x)) = x

so these two functions are inverse functions of each other.

Example: Find the inverse function of f (x) = x³+5

1. y = x³ + 5
2. x = y³ + 5
3. x - 5 = y³

(x - 5)^1/3 = y

so

f ^-1(x) = (x - 5)^1/3

Check answer:
f (f ^-1(x)) = x
((x - 5)^1/3)³ + 5 = x - 5 + 5 = x so it checks

II. One - to - one function:
A function is one-to-one if, for "a" and "b" in the domain
f (a) = f (b) implies that a = b.

A function "f " has an inverse function " f " if and only if " f " is one-to-one.

Use the horizontal line test to check if a function is one-to-one. If the equation can have a horizontal line pass through it only once at any value, then the function is one-to-one.

Example: y² = x,
doing the vertical line test (to see if it is a function) you see that when x = 1, y = -1 or y = 1 so
this is not a function
but by using the horizontal line test, the line never hits twice so therefore this equation has a one-to-one relationship.

Example: y = x³
this is a function and it has a one-to-one relationship so therefore
this is a one-to-one function.

Testing for One-to-one Functions:

If the function f (x) = x³ + 7
Show that f (a) = f (b)

a³ + 7 = b³ + 7
a³ = b³
a = b so yes it is.

Example:
g (x) = 5x² + 8

5 (a)² + 8 = 5 (b)² +8

5a² = 5b²

a² = b²

so you could get:
a = b or
-a = b or
a = -b or
-a = -b

so therefore this function is not one-to-one.

Precalculus 1.2b Graphs of Functions

2007-09-26T02:16:00.000-10:00

Precalculus 1.2 b Graphs of functions:

I. to graph a piece-wise function:
Given:

So to find the answer to this, find f(0).
Since 0 is equal to 0, we would use the first function.
so f(0) = √0 + 2 = 2, this means it is a solid dot at (0, 2) so the answer cannot be C.
and it is an open dot at f(0) = -x - 1 = -1 because it does not equal zero here.

Now we look at the graph of each function.
f(x) = -x - 1 has a negative slope so the answer would be A.

A way to check your answer is by using your graphing utility.

In y₁ = (√ x + 2)(x≥0)

In y₂ = (- x - 1)(x is less than 0)

with this you can see that the graph could be A or C. Now using your table, you can see that
y₁ = 2 where y₂ = 0 (this zero means no - the function doesn't exist here) so this means that
y₁ would be a solid dot and y₂ is open.

II. Even Function:

A function f is even if, for each x in the domain of f, f (-x) = f (x)

Example:
f (x) = x² + 4

Test the function:
1st - Is it a function - yes it is
2nd - does f (-x) = f (x)

f (-x) = (-x)² + 4
= x² + 4

Therefore: f (-x) = x² + 4 and
f (x) = x² so
f (-x) = f (x) so this function is even.

III. Odd Function

A function f is odd if, for each x in the domain of f,
f (-x) = - f (x)

Example:
f(x) = x⁵ - x³

f(-x) = (-x)⁵ - (-x)³
= -x⁵ - ^- x³
= - (x⁵ - x³)

-f(x) = - (x⁵ - x³) so...

f (-x) = - f(x) so this function is odd.

IV. Neither an Even Function or Odd Function

A function f is neither if, for each x in the domain of f

a) f (-x) ≠ f (x) AND
b) f (-x) ≠ - f(x)

Example:
f (x) = 5 - 3x

f (-x) = 5 - 3(-x) = 5 + 3x

-f(x) = - (5 - 3x) = -5 + 3x

so since f (x) ≠ f (-x) and
f (-x) ≠ - f(x), this function is neither even or odd

V. Graphically showing even, odd or neither functions:

1. A function whose graph is symmetric with respect to the y-axis is an even function.

2. A function whose graph is symmetric with respect to the origin is an odd function.

3. A function whose graph is not symmetric with respect to the y-axis or the origin is a neither function.

Example:

Can you figure out which of these graphs are even, odd or neither?

1. Graph A is neither because it is not symmetric to the y-axis or the origin.

2. Graph B is an even function because it is symmetric to the y-axis.

3. Graph C is an odd function because it is symmetric to the origin.

VI. The Greatest Interger Function:

is denoted by [[x]] and is defined by

f (x) = [[x]] = the greatest integer less than or equal to x.

The graph of the greatest integer function jumps vertically one unit at each integer and is contant (a horizontal line segment) between each pair of consecutive integers.

Example: f (x) = [[x]]

[[-2]] = -2

[[-1.5]] = -2

[[-0.5]] = -1

[[1.5]] = 1

From this you can see that the value of the x rounded down to the nearest integer.

To put this on the graphing utility:

y₁ = math → num↓#5 int( x)

Example 2:

f (-4) = [[-1]] - 2 = -1 - 2 = -3

f (-3.5) = [[-.875]] - 2 = -1 - 2 = -3

So we can definitely throw C away. So looking at A or B,

and since f (-4) = -3, this means the dot is closed so the answer would be A.

Precalculus 2.2 Polynomial Functions of Higher Degree

2007-09-26T02:14:00.002-10:00

Precalculus 2.2 Polynomial Functions of Higher Degree

You can find extra notes at the following website:

http://scidiv.bcc.ctc.edu/FL/MATH105/sso0202.pdf

I. Graphs of:

Polynomial functions are continuous if -
they have no breaks, holes, or gaps.

Example 1: This graph is continuous

Example 2: Piece-wise functions are not continuous

II. Simple Graphs:
f (x) = xⁿ, where n is greater than zero

1. If n is even, the graph of y = xⁿ touches the axis at the x-intercept.
2. If n is odd, the graph of y = xⁿ crosses the axis at the x-intercept.

III. The Leading Coefficient Test
Given f(x) = a_nxⁿ + ... + a₁x + a₀

1. When "n" is odd:
a. If the Leading Coefficient is positive (a_n is greater than 0), the graph falls to the left and rises to the right.
b. If the Leading Coefficient is negative (a_n is less than 0), the graph rises to the left and falls to the right.

2. When "n" is even:
a. If the Leading Coefficient is positive (a_n is greater than 0), the graph rises to the left and right.
b. If the Leading Coefficient is negative (a_n is less than 0), the graph falls to the left and right.

*This determines ONLY the right and left behavior of the graph!

IV. Zeros of Polynomial Functions:
1. The graph of "f" has at most "n" real zeros.
2. The function has at most n-1 relative extrema (relative minimum or maximums).

Example: Given y = x²
n = 2
so
2 real zeros and (n-1) or 1 minimum

V. Real Zeros:
1. x = a is a zero of the function "f"
2. x = a is a solution of the polynomial equation f (x) = 0
3. (x - a) is a factor of the polynomial f (x)
4. (a, 0) is an x-intercept of the graph of "f".

Example: f (x) = x² - 8x + 15
0 = x² - 8x + 15
0 = (x - 5)(x - 3)
x = 5 and x = 3

therefore the zeros are (5, 0) and (3, 0)

x² so n = 2 so ... there are at least 2 real zeros
graph: a is greater than 0 so this function rises to the left and right.

Example: f (x) = (-3/8) x⁴ - x³ + 2x² + 5
Given the zeros:

1. Leading Coefficient = -3/8
2. Leading Degree is 4 so it is even
Therefore the graph falls to the left and the right.

Note: use the calculator to find the intercepts if algebraically not able to at this stage

When you graph the above function:
Maximum: (-2.914851686, 19.68779879)
Zeros: (-4.141946129, 0) and (1.934035914, 0)

Example: f (x) = x³ - 4x

Leading Coefficient = 1
Leading degree is 3 so it is odd

The graph falls to left and rises to the right

0 = x³ - 4x
0 = x (x² - 4)
0 = x (x + 2)(x - 2)
0 = x, x = -2, and x = 2

Maximum value (-1.15, 3.08)
Minimum value (1.15, -3.08)

VI. Multiplicity for the Factor (x - r)^k

In general, a factor of (x - r)^k yields a repeated zero x = r of multiplicity k.
1. If k is odd, the graph crosses the x-axis at x = r
2. If k is even, the graph touches (but does not cross) the x-axis at x = r.

Example: f (x) = 4x² -6x + 9

Find all zeros:

0 = 4x² - 6x + 9
0 = (2x - 3)(2x - 3)
0 = (2x - 3)²

so we know a zero is (3, 0)

Using #2 above, k = 2 so it is even, therefore the graph touches the x-axis at x = 3

Example 2: Find the x-intercepts and multiplicity of f (x) = 2(x + 2)²(x - 3)
x-intercepts are (-2, 0) and (3, 0) and the mutiplicity of 2.

2.2b Finding a Polynomial with Given Zeros:

I. Finding a Polynomial with given zeros

Example 1: given zeros: -4 and 5
1. For each of the given zeros, form a corresponding factor.
We have: x = -4 and x = 5

f (x) = (x + 4)(x - 5)
= x² - 5x + 4x - 20
= x² - x - 20

Now sketch the graph:
1. Apply the Leading Coefficient test
Leading Coefficient is 1 and 1 is positive and the degree is 2 so it is even
Therefore the graph rises to the left and right.
2. Use zeros given: (-4, 0) and (5, 0)
3. Find the y-intercept by letting x = 0
f (0) = -20
4. Find the vertex
x = -b/(2a) = 1/2
f (1/2) = -20.25

So now you have 4 points to plot so you can sketch the curve.

Example 2: Zeros: 0, 2, and -1/3 so
x = 0, x = 2, and x = -1/3
f (x) = (x)(x - 2)(3x + 1)
= (x²)(3x + 1)
= 3x³ + x² - 6x² - 2x
= 3x³ - 5x² - 2x

Sketch the graph:
1. Leading Coefficient test - odd and positive so falls to the left and rises to the right
2. Use given zeros which is also the y-intercept

Example 3: zeros: 6 + √ 3 and 6 - √3
Therefore:
x = 6 + √3 and x = 6 - √3

f (x) = (x - 6 - √3)(6 - 6 + √3)
= x² - 6x + x√3 - 6x + 36 - 6√3 - x√3 + 6√3 - (√3)²
= x² - 12x + 36 - 3
= x² - 12x + 33

Example 4: (use grouping)
Zeros: 4, 2 + √7, 2 - √7

f (x) = (x - 4)(x - 2 - √7)(x - 2 + √7)
= (x - 4)(x² - 2x + x√7 - 2x + 4 - 2√7 - x√7 + 2√7 - ( √7)²
= (x - 4)(x² - 4x + 4 - 7 )
= (x - 4)(x² - 4x - 3)
= x³ - 4x² - 3x - 4x² + 16x + 12
= x³ - 8x² + 13x + 12

II. The Intermediate Value Theorem:
- This theorem helps locate the real zeros of a polynomial function.

Example 1: Find a value x = a where a polynomial function is positive and another x = b where it is negative, you can conclude that the function has at least one real zero between the two values.

f (x) = x³ - 3x² + 3

1. Using the leading coefficient test, the graph falls left and rises right
2. Finding a couple of points, (0, 3) is one point and (-3, -5) is another point

(0, 3) is above x-axis while (-3, -51) is below the x-axis so there is a real zero between them (the graph crosses the x-axis)

(2, -1) is also below the x-axis so there is a real zero between (0, 3) and (2, -1)

(5, 53) is another point and that is above the x-axis so there is another real zero between (2, -1) and (5, 53).

This tells us that the graph crosses the x-axis in three different places.

Next using the calculator to find the "exact" zeros:

1. In your calculator: y₁ = x³ - 3x² + 3

2. Press 2nd trace, #2 zero
going from the left, go below the x-axis and this will be your lower bound and then above the x-axis, this will be your upper bound, and then guess around the x-axis.

(-.8793852, 0) which was between (-3, 51) and (0, 3)
(1.3472964, 0) which was between (0, 3) and (2, -1)
(2.5320889, 0) which was between (2, -1) and (5, 53)

Example 2: g(x) = (1/8)(x + 1)² (x - 3)³

1. the highest degree is 5 which is odd and the leading coefficient is positive so the graph falls to the left and rises to the right.

2. There could be 5 different x-intercepts so trying some different points we have
(-2, -15.625), (-1, 0), (0, -3.375), (1, -4), (2, -1.125), (3, 0), (4, 3.125)

From our table, we see that there are only 2 x-intercepts.

Check this algebraically:

0 = (1/8)(x + 1)²(x - 3)³
0 = (1/8)(x + 1)²
0 = x + 1
-1 = x

0 = (x - 3)³
0 = x - 3
3 = x

These are the two points that we found graphically.

Precalculus 2.1 Quadratic Functions

2007-09-26T01:51:00.000-10:00

Precalculus 2.1 Quadratic Functions

2.1 Quadratic Functions

You can find extra notes on this website:

http://scidiv.bcc.ctc.edu/FL/MATH105/sso0201.pdf

I. Quadratic Functions
Definition: Let “n” be a non-negative integer and let a_n, a_n-1, …, a₂, a₁, a₀
be real numbers with a_n ≠ 0.

The Function

f (x) = a_nxⁿ + a_n-1x^n-1 + … + a₂x² + a₁x + a₀

is called a POLYNOMIAL FUNCTION of x with degree “n”

Polynomial functions are classified by degree:

f (x) = a is a constant function
f (x) = mx + b is a linear function
f (x) = ax² + bx + c is a quadratic function where a ≠ 0 and {a, b, c } is contained in the set of Real Numbers

Parabola – a graph of a quadratic function is a special type of u-shaped curve.

Since this graph opens upward, given f (x) = ax² + bx + c, "a" is greater than 0.
if "a" is less than 0, then the parabola is opened downward.

Vertex is the turning point and the axis of symmetry is perpendicular through the x-value of the vertex. Can be found by x = -b/(2a)

Key: using f(x) = ax²
if "a" is greater than 1, the graph is a vertical stretch of the graph y= f(x)
if 0 is less than "a" which is less than 1, the graph is a vertical shrink of the graph y = f(x)

Standard Form of a Quadratic Function:

f(x) = a (x - h)² + k, where a ≠ 0,
the axis of symmetry is the vertical line x = h, and the vertex of the function is (h, k)

Example 1: Given the vertex (4, -1) and the point (2, 3), what is the equation of the quadratic function.

y = a (x - h)² + k
3 = a (2 - 4)² + (-1)
4 = a (-2)²
4 = a (4)
1 = a

y = 1 (x - 4)² - 1

Example 2: Given the vertex is (5/2, -3/4) and a point (-2, 4), what is the equation of the quadratic function.

y = a(x - h)² + k
4 = a(-2 - 5/2)² + (-3/4)
4.75 = a(-4.5)²
4.75 = 20.25 a
19/81 = a

y = (19/81)(x - 5/2)² - 3/4

II. Maximum or Minimum values:
To find the maximum or minimum, find the vertex by using
x = -b/(2a)

Example: Given C = 800 - 10x + 0.25 x²
find the minimum cost and the number of fixtures:

x = -b/(2a) = 10/((2)(.25) = 10/.5 = 20

therefore there will be 20 fixtures

C = .25 (20)² - 10(20) + 800
C = 700

So the minimum cost is $700

You can check by graphing, and the minimum point is (20, 700)

III. Identify the vertex and the intercepts Algebraically.

f(x) = x² + 9x + 8

Vertex is x=-b/2a = -9/2 = -4.5

f (-9) = (-4.5)² + 9(-4.5) + 8
= 20.25 - 40.5 + 8
= -12.75

vertex is (-4.5, -12.25)

Intercepts:
Let x = 0
f (0) = 0² + 9(0) + 8 = 8
(0, 8)

Let y = 0
0 = x² + 9x + 8
0 = (x + 8)(x + 1)
x = -8 and x = -1
so
(-8, 0) and (-1, 0)

IV. Find the equation of a quadratic

Given the two x-intercepts (-2, 0) and (10, 0), can you find the equation?

A. if the parabola opens upward, "a" is greater than 0,

y = a (x - p)(x - q)

because there can be lots of answers depending upon what the value of "a", so we will
Let a = 1

f (x) = (x - (-2))(x - 10)
= (x + 2)(x - 10)
= x² - 10x + 2x - 20
= x² - 8x - 20

B. if the parabola opens downward, "a" is less than 0,
y - a (x - p)(x - q)
because again, there can be lots of answers depending upon what the value of "a", so we will
Let a = -1

f(x) = -(x + 2)(x - 10)
= -(x²- 8x - 20)
= -x² + 8x + 20

Precalculus Chapter 1.1b + 1.2a

2007-09-25T02:33:00.000-10:00

Chapter 1.1b Functions + 1.2a Graphs of Functions

I. Application: A right triangle is formed in the first quadrant by the x-axis and y-axis and a line through the point (2, 1). Write the area of the triangle as a function of “x”, and determine the domain of the function.

1. Graph the triangle:
- one side of the triangle is the x-axis and another side is the y-axis. Since we do not know exactly where the points of the third line cross the 2 axes, we use the points (0,y) and (x,0).

A = ½ (base)(height) = ½ xy

Since (0, y) , (2, 1) and (x, 0) all lie on the same line, the slopes between any pairs of points are equal.

M = (1 – y)/(2 – 0) = (1 – 0)/(2 – x)

1 – y = 2/(2 – x)

y = -2/ ( 2 – x) + 1

y = (-2 + (2 – x))/ (2 – x)

y = -x / (2 – x)

y = x / (x – 2)

Now using substitution:

A = ½ x (x / (x – 2)) = x²/(2x – 4)

The domain is (2, ∞ ) since the area has to be greater than zero.

II. Evaluating a Difference Quotient:

If f (x) = 2x, then using the difference quotient below

Using f(x) = 2x, plug into the equation above:

2h/h = 2

Example 2: Given 5x – x²,

-5 – h , where h cannot equal zero

Example 3:

Increasing, Decreasing, and Constant Functions

A function “f” is increasing on an interval if, for any x₁ and x₂ in the interval, when x₁ is less than x₂ implies that f (x₁ ) is less than f(x₂ )
A function “f” is decreasing on an interval if, for any x₁ and x₂ in the interval, when x₁ is less than x₂ implies that f (x₁ ) is greater than f(x₂ )
A function “f” is constant on an interval if, for any x₁ and x₂ in the interval, f (x₁ ) = f(x₂ )

Check out this website:
http://www.mathematicshelpcentral.com/lecture_notes/precalculus_algebra_folder/increasing_and_decreasing_functions.htm

Relative Minimum and Relative Maximum:
1. A function value f (a) is called a relative minimum of “f” if there exists an interval that contains a such that:

So you can see that the function is increasing from (- ∞ , -3) and then again from (2, ∞). The function is decreasing from (-3, 2)

Now lets see an example of relative minimum and relative maximum points:

So you can see that the function's relative minimum point from ( - ∞, 0 ) is
"a" while its relative minimum point from ( 0, ∞) is "c" . The relative maximum point is "b".

Precalculus Chapter 1.1a Functions

2007-09-18T08:06:00.000-10:00

Precalculus Chapter 1.1 Functions and their Graphs

Definition of a Function: A function “f” from a Set A to a Set B is a relation that assigns to each element “x” in the set A exactly one element “y” in the Set B. The Set A is the domain (or set of inputs) of the function “f”, and the Set B contains the range (or set of outputs).

Example:
Set A contains {1, 2, 3, 4, 5} and Set B contains {a, b, c, d, e }

Let's say that Set A maps to Set B in giving the following ordered pairs:
{(1, a), (2, c), (3, c), (4, d), (5, b)}
The mapping would look like:

Since every input there is only one output, therefore this is a function.

Remember, for every input there is only one output.

{(a, 2), (b, 1), (c, 4)}

Where a, b, and c are the input and 1, 2, and 4 are the output

CHARACTERISTICS of a FUNCTION:

1). Each element in Set A must be matched with an element of Set B.
2). Some elements in Set B may no be matched with any elements of Set A.
3). Two or more elements of Set A may be matched with the same element of Set B.

FUNCTIONS are commonly represented in four ways:
1). Verbally – by a sentence that describes how the input variable is related to the output variable.
2). Numerically – by a table or list of ordered pairs
3). Graphically
4). Algebraically by an equation in two variables.
example: y = x² where “y” is a function of “x”

“x” is the independent variable (input or domain)
“y” is the dependent variable (output or range)

TESTING FOR FUNCTION REPRESENTED:
1) Algebraically:
a) 2x + y – 6 = 0 Solve for y.
y = -2x + 6
To each value of “x”, there is only one “y” , therefore this example is a function.

b) 3x² + 2y = 1
2y = -3x² + 1
y = (1/2)(-3x² + 1)

Again, this is a function.

c) 4² + 3y² = 6
3y² = -4x² - 6
y² = (1/3)( -4x² - 6)

This is not a function!

2). Vertical line test – graph the equation. If you can draw a vertical line and it only touches the graph of the equation once, then the equation is a function. If it touches two or more points, then it is not a function.

3) Function Notation:

Input = x
Output was y, it will now be f(x)

Equation example:
Was: y = 2x + 3
Will now be represented by: f (x) = 2x + 3

EVALUATING A FUNCTION:

Example: f (x) = 2x – 3
f ( 1) = 2 (1) – 3 = 2 – 3 = -1
f (-3) = 2 (-3) – 3 = -6 – 3 = - 9
f (x – 1) = 2 (x – 1) – 3 = 2x – 2 – 3 = 2x – 5

Example: g (y) = 7 – 3y
a. g (0) = 7 – 3(0) = 7
b. g (7/3) = 7 – 3 (7/3) = 7 – 7 = 0
c. g ( s + 2) = 7 – 3 (s + 2 ) = 7 – 3s – 6 = 1 – 3s

A PIECEWISE-DEFINED FUNCTION
- this means that the function is defined by two or more equations over a specified domain.

f (x ) =

{ 2x + 1, x ≤ 0

{ 2x + 2, x is greater than 0

Evaluate: f ( -1)

because -1 is less than 0, we use 2x + 1 so

f (-1) = 2 (-1) + 1 = -2 + 1 = -1

Evaluate: f (0 ) because 0 is equal to 0, we use 2x + 2 so

f (0) = 2x + 2 = 2 (0) + 2 = 2 f (2) = 2x + 2 = 2 (2) + 2 = 4 + 2 = 6

Precalculus Prerequisite 5 - Inequalities

2007-09-11T02:36:00.000-10:00

Precalculus Prerequisite Section 5 - Inequalities:

II. Solving Inequalities Algebraically and Graphically:
A. Solve the Inequality:

Precalculus P4.b Completing the square and

2007-09-10T03:21:00.000-10:00

I. Completing the Square:

if x² + bx = c, then

x² + bx + (b/2)² = c + (b/2)²

so:

(x + b/2)² = c + (b²)/4

Example:

x² + 4x - 32 = 0

x² + 4x = 32

therefore b = 4 so:

x² + 4x + (4/2)² = 32 + (4/2)²

(x + 2)² = 32 + 4

(x + 2)² = 36

take the square root of both sides:

x + 2 = ± √36

x + 2 = ±6

x = -2 ±6

so x = 4 and x = -8

Now check your answers:

(-8)² + 4(-8) - 32 = 0

64 - 32 - 32 = 0

0 = 0

(4)² + 4(4) - 32 = 0

16 + 16 - 32 = 0

0 = 0

They both work so our answers are x = 4 and x = -8

Example 2:

x² - 2x - 3 = 0

x² - 2x = 3

x² - 2x + (-2/2)² = 3 + (-2/2)²

x² - 2x + (-1)² = 3 + 1

(x - 1)² = 4

x - 1 = ±√4

x = 1 ±2

x = 1 + 2 = 3

x = 1 - 2 = -1

check answers:

(3)² - 2(3) - 3 = 0

9 - 6 - 3 = 0

0 = 0

(-1)² - 2(-1) - 3 = 0

1 + 2 - 3 = 0

0 = 0

II. Quadratic Formula:

if ax² + bx + c = 0 then

Example:

9x² - 12x - 14 = 0

x = (12 ± √(144 - 4(9)(-18)))/(2(9))

x = (12 ± √648)/18

x = (12 ± √324 √2)/18

x = (12 ± 18 √2)/18

x = (2 ± 3 √2)/3

x = 2/3 ± √2

III. Solving an Equation of Quadratic Type:

x⁴ + 2x³ - 8x - 16 = 0 by grouping

x³ (x + 2) - 8(x + 2) = 0

(x³ - 8 ) ( x + 2) = 0

x³ = 8

x = 2

x + 2 = 0

x = -2

now check the answers:

they both check!

IV. By factoring more difficult equations:

4x²(x - 1)^1/3 + 6x (x - 1)^4/3 = 0

2x [2x (x - 1)^1/3 + 3 (x - 1)^4/3] = 0

remember that (x - 1)^4/3 = (x - 1) ^3/3 (x - 1)^1/3

and 3/3 = 1 so:

2x(x - 1)^1/3 [2x + 3 (x - 1)^3/3 ] = 0

2x(x - 1)^1/3 [ 2x + 3x - 3] = 0

2x(x - 1)^1/3 [5x - 3] = 0

so:

2x = 0

x = 0

(x - 1)^1/3 = 0

x - 1 = 0

x = 1

5x - 3 = 0

5x = 3

x = 3/5

check your answers!

V. Solving an equation involving a radical:

√(-5x + 4) - x = 6

√(-6x + 4) = x + 6

now square both sides:

(√(-6x + 4))² = (x + 6)²

-6x + 4 = x² + 12x + 36

0 = x² + 18 x + 32

0 = (x + 16)(x + 2)

x = -16 and x = -2

check both answers:

√(-6 (-16) + 4) - (-16) = 6

√(96 + 4 ) + 16 = 6

10 + 16 = 6

not true

√(-6(-2) + 4) - (-2) = 6

√(12 + 4) + 2 = 6

√16 + 2 = 6

4 + 2 = 6

this one works! so

x = -2 is the only answer!

VI. Solving an Equation Involving Two radicals:

√x + √(x - 20) = 10

√ x = 10 - √(x - 20)

(√x)² = (10 - √(x - 20))²

x = 100 - 20√(x - 20) + (x - 20)

0 = 80 - 20√(x - 20)

-80 = -20√(x - 20)

4 = √(x - 20)

4² = x - 20

16 = x - 20

36 = x

Now check your answer.

√36 + √(36 - 20) = 10

6 + 4 = 10 so it checks

VII. Solving an Equation with Rational Exponents

Example:

(x - 5)^2/3 = 16

to undo exponents, multiply by the reciprocal

((x - 5)^2/3)^3/2 = 16^3/2

x - 5 = 64

x = 69

Check your answer.

(69 - 5)^2/3 = 16

64^2/3 = 16

16 = 16 so it checks

VIII. Solving an Equation Involving Absolute Value:

recall: |x| = 9

x = 9 and x = -9 so

|3x + 2 | = 7

3x + 2 = 7 and 3x + 2 = -7

3x = 5 and 3x = -9

x = 5/3 and x = -3

Precalculus Prerequisite 4a - Solving Equations Algebraically and Graphically

2007-09-06T10:04:00.000-10:00

Prerequisite 4a - Solving Equations Algebraically and Graphically

I. Vocabulary:

A. An Equation - is a statement that two algebraic expressions are equal.
Example: 2x + 4 = 10

B. To Solve an equation in "x" means to find all values of "x" for which the statement is true. These values are solutions.
Example: 2x + 4 = 10
2x = 6
x = 3

C. An identity equation- an equation that is true for every real number in the domain of the variable.
Example: x² - 6x + 9 = (x - 3)²

D. Conditional Equation - an equation that is true for just some (or even none of the real numbers in the domain of the variable.
Example 1: x² - 6x + 9 = 0
(x - 3)² = 0
x = 3 only

Example 2: 3x + 2 = 4x - 5
x = 7 only

Example 3: x² + 3x + 4 = 4x - 5
no solutions

E. Notations:
1. all real numbers - R
2. no solutions - { } empty set or Æ is the null set

F. Solving an Equation Involving Fractions:
Example 1: x/4 + (2x)/3 = 6

multiply by the LCD
4, 3 are the denominators so the LCD = 12, so multiply each term by 12:

12(x/4) + (12)(2x)/3 = (12)(6)
3x + 8x = 72
11x = 72
x = 72/11

Example 2: 5/x + (3x)/2 = 7

LCD = 2x

(2x)(5/x) + (2x)(3x)/2 = (2x)(7)
10 + 3x² = 14x
3x² - 14 x + 10 = 0

use the quadratic equation

x = (-b ± Ö(b² - 4ac))/(2a)

x = (14 ± Ö(14² - (4)(3)(10)))/((2)(3))

x = (14 ± Ö76)/3

x = (7 + Ö 19)/3 and x = (7 - Ö19)/3

or

x = 3.786299648 and x = .8803670188 (check to make sure they both work - they do!)

You can also check by using your graphing calculator -
put in
y₁ = 5/x + (3x)/2 and
y₂ = 7

calc ® intersect, shows the same answers!

G. Extraneous Solutions - an answer or solution that does not satisfy the original equation.
Example:

6/x - 2/(x+3) = (3(x+5))/(x(x + 3))

LCD = x(x+3)

6(x + 3) - 2x = 3(x + 5)
6x + 18 - 2x = 3x + 15
4x + 18 = 3x + 15
x = -3

substituting x=-3 back into the original equation, we have
-2 - 2/0 = 6/0
This is impossible so there is not a solution

H. To find the x-intercepts (a, 0), let y = 0 and solve for x.
To find the y-intercepts (0, b), let x = 0 and solve for y.

Example:
2x² - 5x + 2 = y

x-intercepts, let y = 0
2x² - 5x + 2 = 0
(2x -1)(x - 2) = 0
2x - 1 = 0 and x - 2 = 0
x = 1/2 and x = 2
(.5, 0) and (2, 0)

y-intercepts, let x = 0
2(0)² - 5(0) + 2 = y
2 = y
(0,2)

I. Finding Solutions Graphically:
24x³ - 36x + 17 = 0

graph this on your calculator and you see there is only one solution:
window:
x-values: -10 to 10
y-values: -5 to 40
You can see it is hard to tell how many times the equation crosses the x-axis
so change the y-values to -3 to 3 and you see it crosses y=0 only once

x = -1.414486, y=0

J. Can find answers more than one way, here are a couple using the calculator:
1. y₁ = 24x³ - 36x + 17
2nd trace ® zero
enter left-bound before x=-2 and then right-bound after x=-1, then guess about -1.5, gives the answer.
2. y₁ = 24x³ - 36x + 17
y₂ = 0
2nd trace, #5 intersect, enter 3 times about where the two equations intersect.

K. Remember with points of intersection, always look for all solutions:
Example: 24x³ - 36x + 17 = 2x + 5 has how many solutions?

(-1.393598, 2.212803)
(.3407855, 5.681571)
(1.052813, 7.105626)

remember there can be no points of intersection (therefore no solutions), one solution or many solutions.

L. Solving Polynomial Equations Algebraically:

1. First degree equation - linear equation -
example: 2x + 4 = 7

2. Second degree equation - quadratic equation -
example: 2x² + 4x + 6 = 0

3. Third degree equation - Cubic equation -
example: 2x³ + 2x² + 5x + 2 = 0

4. Fourth degree equation - Quartic equation -
example: 2x⁴ + 3x³ - 2x² + 3x + 2 = 0

5. Fifth degree equation - Quintic equation -
example: 4x⁵ + 2x⁴ - 3x³+ 2x² + x - 4 = 0

Precalculus - Prerequisite 1 - Graphical Representation of Data, P2 - Graphs of Equations, P3 -Lines in a Plane

2007-08-13T03:17:00.000-10:00

Prerequisite 1 - Graphical Representation of Data

I) Vocabulary -
A) The Cartesian Plane - the rectangular coordinate plane that represents ordered pairs fo real numbers by points in a plane.
1) x - axis: the horizontal real number line
2) y - axis; the vertical real number line
3) origin - the point of intersection of the x-axis and the y-axis
4)Quadrants - the two axes divide the plane into four parts - the upper right quadrant is quadrant I, the upper left quadrant is quadrant II, the lower left quadrant is quadrant III, and the lower right quadrant is quadrant IV. They go counterclockwise from the positive side of the x-axis.
5) Ordered pair - each point in the plane corresponds to an ordered pair of real numbers (x, y)
called the coordinates of the point.
a) x-coordinate: represents the directed distance from the y-axis to the point
b) y-coordinate: represents the directed distance from the x-axis to the point.
B) Representing Data Graphically:
1) Scatterplot
2) bar graph
3) line graph
4) histogram
5) Interpreting a model based upon the data:
a) look at the data and know how to find the best-fit line
C) Distance Formula: d = √((x₁ - x₂)² + (y₁ - y₂)²)
this is derived from the pythagorean theorem, where the difference of the x's is represented by "a" and the difference of the y's is represented by "b" and so
AB = a² + b² = c²
Example: Given points (-2, -5) and (-3, 4), what is the distance between the two points.
√((-2 - ^-3)² + (-5 - 4)²) =
√((1)² + (-9)²) =
√(1+81)=
√82
E) Midpoint Formula: the mean, or average, of the x-coordinates and the y-coordinates.therefore the midpoint formula is: ((x₁+ x₂) /2, (y₁+ y₂) /2)
example: Given A(-1, 7) and B(3, -3) what is their midpoint?((-1 + 3)/2, (7 + -3)/2) = (2/2, 4/2) = (1, 2)
F) Standard equation of a circle with radius r and center (h, k) is:
r² = (x - h)² + (y - k)²
OR
r = √((x - h)² + (y - k)²)

Prerequisite 2: Graphs of Equations
I) Vocabulary:
A) Solution Point - for an equation in variables x and y, a point (a, b) is a solution point if the substitution of x = a and y = b satisfies the equation.
1) Graph of the equation - the set of all solution points of an equation.
B) How to Sketch the Graph of an Equation by Point Plotting
1) If possible, rewrite the equation so that one of the variables is isolated on one side of the equation.
2) Make a table of several solution points (usually 5 - 7)
3) Plot these points in the coordinate plane.
4) Connect the points with a smooth curve.
C) Using a Graphing Utility to Graph an Equation - to graph an equation involving x and y on a graphing utility, use the following procedure.
1) Rewrite the equation so that "y"is isolated on the left side.
2) Enter the equation into a graphing utility.
3) Determine a viewing window that shows all important features of the graph.
4) Graph the equation.
D) Throughout this course, you will learn that there are many ways to approach a problem.
1) a numerical approach: construct and use a table.
2) a graphical approach: draw and use a graph.
3) an algebraic approach: use the rules of algebra.

Prerequisite 3: Lines in the Plane
I) Vocabulary:
A) the slope of a line - represents the number of units a line rises or falls vertically for each unit of horizontal change from left to right.
Slope: (nonvertical line): Ratio of the vertical change (the rise, Δ y) to the horizontal change (the run, Δ x)
Formula: slope = m = (Δy/Δx) = (y₂ - y₁)÷ (x₂ - x₂)

II) Postulate:
A) In the coordinate plane, 2 nonvertical lines are parallel if and only if they have the same slope. Any two vertical or horizontal lines are parallel.
B) Slope- Intercept Form of the Equation of a Line: y = mx + b where m is the slope of the line and b is the y-intercept (where it crosses the y-axis so point (0,b)
C) Point-slope form of an equation of the line that passes through the point (x₁, y₁) and has a slope of m is
y - y₁ = m(x - x₁)
D) Summary of Equations of Lines
1. General form: Ax + By + C = 0
2. Vertical line: x = a
3. Horizontal line: y = b
4. Slope-intercept form: y = mx + b
5. Point-slope form: y - y₁ = m(x - x₁)
E) Parallel Lines - two distinct non-vertical lines are parallel if and only if their slopes are equal.
F) Perpendicular Lines - two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals of each other. If m₁ = a/b then m₂ = -b/a.
The product of the two slopes are -1. (a/b)(-b/a) = -1
m₁ - -1/(m₂)
Example: if m₁ = 2/3 then
m₂ = -3/2

Geometry Chapter 9.5 Trigonometric Ratios

2007-08-08T02:17:00.001-10:00

Chapter 9.5 Trigonometric Ratios:

I) Vocabulary:
A) A Trigonometric Ratio - is a ratio of the lengths of two sides of a right triangle.
1) Sine - abbreviated sin
2) Cosine - abbreviated cos
3) Tangent - abbreviated tan

check out: http://www.tpub.com/math1/20b.htm

Let ΔABC be a right triangle. The sine, the cosine, and the tangent of the acute angle ∠A are defined as follows:

The side opposite angle A = side a
The side opposite angle B = side b
The side opposite angle C = side c = hypotenuse

Sin A = (side opposite angle A)/hypotenuse = a/c

Cos A = (side adjacent angle A)/hypotenuse = b/c

Tan A = (side opposite angle A)/(side adjacent angle A) = a/b

Example: Given ΔABC with AB = 13, AC = 5 and BC = 12 with right ∠C, find the 3 different ratios for angle A and angle B:
Sin A = 12/13
Cos A = 5/13
Tan A = 12/5

Sin B = 5/13
Cos B = 12/13
Tan B = 5/12

Do you notice a pattern?
The Sin A = Cos B and Cos A = Sin B plus the tangents are just reciprocals of each other. This is because the Sine and Cosine are cofunctions of each other.

Example: Find the sine, cosine and tangent of 30°, 45° and 60°.
Sin 30° = 1/2
Cos 30° = √(3)/2
Tan 30° = √(3)/3

Sin 45° = √(2)/2
Cos 45° = √(2)/2
Tan 45° = 1

Sin 60° = √(3)/2
Cos 60° = 1/2
Tan 60° = √3

B)Angle of Elevation - the angle that your line of sight makes with a line drawn horizontally.
Example: The angle of elevation from the base of a water slide to the top of a waterslide is about 13°. The slide extends horizontally about 58.2 meters. Find the vertical height of the slide.

1) Draw a diagram. Place the appropriate values where they belong.
2) Using the trigonometric functions, find the one you need.
a) We need to know the side opposite the angle and we know the side adjacent to the angle so this would be tangent.
b) Tan 13° = h/58.2
h = 58.2 (tan 13°)
h = 13.43652872 meters

Chapter 9.6 Solving Right Triangles:

I) Vocabulary:
A) Solve a right triangle - to determine the measures of all six parts, 3 sides and 3 angles.
Example: Given ΔABC, angle C is the right angle, angle A = 26°, AB = 4.5 inches, find AC, BC and angle B.
measure of angle A = 26°
measure of angle B = ?
measure of angle C = 90°
AB = c = 4.5 inches
BC = a = ?
AC = b = ?

A triangle has 180° so 26°+ 90° = 116°, 180° - 116° = 64° = measure of angle B

From the angle A, we know the hypotenuse or side "c" = 4.5, to find side "a", this is opposite to angle A so we would use sine function.
sin 26° = a/4.5
a = 1.972670161 inches

From the angle A, we know the hypotenuse and need to find side "b", this is adjacent to angle A so we would use the cosine function.
cos 26° = b/4.5
b = 4.044573208 inches

Therefore we have now solved the triangle:
measure of angle A = 26°
measure of angle B = 64°
measure of angle C =90°
a = 1.972670161 inches
b = 4.044573208 inches
c = 4.5 inches

Geometry Chapter 11 Area of Polygons and Circles

2007-08-07T06:27:00.000-10:00

Chapter 11.1 Angle Measures in Polygons:

I) lets look at the sum of polygon angle measures:
Polygon - Number of sides - Number of triangles can make from the polygon - sum of the measures of the interior angles:
A) Triangle - 3 sides - 1 triangle - 180°
B) Quadrilateral - 4 sides - 2 triangles - 360°
C) Pentagon - 5 sides - 3 triangles - 540°
D) Hexagon - 6 sides - 4 triangles - 720°
E) Heptagon - 7 sides - 5 triangles - 900°
F) Octagon - 8 sides - 6 triangles - 1080°
G) Nonagon - 9 sides - 7 triangles - 1260°
H) Decagon - 10 sides - 8 triangles - 1440°
I) Undecagon - 11 sides - 9 triangles - 1620°
J) Dodecagon - 12 sides - 10 triangles - 1800°

Do you see a pattern? The number of sides minus 2 is equal to the number of triangles so...

II) Theorem:
A) Polygon Interior Angles Theorem - the sum of the measures of the interior angles of a convex n-gon is (n - 2) (180°)
Example: What is the sum of the measures of the interior angles of a convex 20-gon?
(20 - 2)(180°) = 3240°
B) The measures of each interior angle of a regular n-gon is:
(1/n)(n-2)(180°)
Example: Given a regular hexagon, what is the measure of each angle?
(1/6)(6-2)(180°) = (1/6)(4)(180°) = 120°
C) Polygon Exterior Angles Theorem - the sum of the measures of the exterior angles of a convex polygon, one angle iat each vertex, is 360°.
Example: If the exterior angles of a polygon is x, 2x, 2x, 4x, and 3x, solve for x.
x + 2x + 2x + 4x + 3x = 360
12 x = 360
x = 30
Example: Given a regular polygon has 20 sides, what is the measure of an exterior angle and an interior angle?
360/20 = 18, so every exterior angle is 18°
180° - 18° = 162° , so every interior angle is 162°

(20 - 2)(180) = (162)(20)
3240 = 3240 so it checks

Example: Given the measrue of each interior angle of a regular n-gon is 120°, how many sides does the polygon have?
(1/n)(n-2)(180) = 120
(n-2)(180) = 120n
180n - 360 = 120n
60n = 360
n = 6

If the interior angle is 120°, then the exterior angle is 60°
360/60 = 6 sides, same answer!

Chapter 11.2 Areas of Regular Polygons:

I) Area of an Equilateral Triangles is:
A = ((√3)/4) s²
Example: Given an equilateral triangle with side length of 4 what is the area?
A = ((√3)/4)(4²)
A = ((√3)/4)(16)
A = 4√3 square units

II) Vocabulary:
A) Center of the polygon - when you circumscribe a circle about a regular polygon, the center point of the circle is the center of the polygon.
B) Radius of the polygon - when you circumscribe a circle about a regular polygon, from the center point of the circle to the vertex of the polygon is the radius of the polygon.
C) Apothem of the polygon - the distance from the center point of the polygon perpendicular to any side of the polygon.
D) A Central angle of a regular polygon - is an angle whose vertex is the center and whose sides contain two consecutive vertices of the polygon. You can divide 360° by the number of sides to find the measure of each central angle of the polygon.

III) Theorem:
A) Area of a Regular Polygon - the area of a regular n-gon with side length "s" is half the product of the apothem "a" and the perimeter "P"
A = (1/2)(a)(P)

Chapter 11.3 Perimeters and Areas of Similar Figures
I) Theorems:
A) Areas of Similar Polygons- if two polygons are similar with the lengths of the corresponding sides in the ratio of a:b, then the ratio of their areas is a²:b²
Example: Given two similar hexagons with corresponding sides of 4 and 5, what is the ratio of their areas?
4²:5²
16:25

Chapter 11.4 Circumference and Arc Length
A) Circumference - of a circle is the distance around the circle.
C = 2π r = πd
B) Arc length corollary:
in a circle, the ratio of the length of a given arc to the circumference is equal to the ratio of the measure of the arc to 360º
Example: recall, the central angle is equal to it's intercepted arc
Given circle P with central angle APB = 30º and radius of 8 radius, what is the measure of the intercepted arc AB?
(measure of the arc AB)/(2π8) = (30/360)
(measure of the arc AB)/ (16π) = (1/6)
measure of the arc AB = 16π/6
measure of the arc AB = 8π/3 inches or about 8.37758041 inches

Chapter 11.5 Areas of Circles and Sectors
I) Vocabulary:
A) Sector of a Circle - is the region bounded by two radii of the circle and their intercepted arec.

II) Theorems:
A) Area of a Circle - the are of a circle is π times the square of the radius
A = πr²
Example: given a circle with radius of 9 cm, what is the area?
A = π(9)² = 81π square cm.
B) Area of a Sector - the ratio of the area A of a sector of a circle to the area of the circle is equal to the ratio of the measure of the intercepted arc to 360º
Example: Given Circle C with points A and B on the circle, the major arc AB is 293º and the radius is 10 cm, what is the area of the sector?
(Area of sector)/(π 10²) = 293º /360º
Area of sector/100π = 293/360
Area of sector = 1465π/18 which is about 255.6907354 square cm.

Chapter 11.6 Geometric Probability
I) Vocabulary
A) Probability is a number from 0 to 1 that represents the chance that an event will occur.
B) Geometric Probability - involves geometric measures such as length or area

II) Theorems:
A) Probability and length - Let line segment AB be a segment that contains the segment CD. If a point K on line segment AB is chosen at random, then the probability that it is on CD is as follows:
P(Point K is on line segment CD) = (length of CD)/(length of AB)
Example: Given AB = 5 and CD = 2, then the probability that point K is on CD is
5/2

B) Probability and Area - Let J be a region that contains region M. If a point K in J is chosen at random, then the probability that it is in region M is as follows:
P(Point K is in region M) = (Area of M)/(Area of J)
Example: You mow a rectangular lawn of 20 feet by 30 feet and it has a garden of 5 feet by 6 feet. What is the geometric probability that if someone tosses a ball that it will land in the garden?
(5)(6) = 30 square feet
(20)(30) = 600 square feet so
30/600 = 1/20

Geometry 10.7 Locus of points

2007-08-07T06:02:00.001-10:00

Chapter 10.7 Locus:

I)Vocabulary:
A) Locus - in a plane is the set of all points in a plane that satisfy a given condition or a set of given conditions. The word locus is derived from the Latin word for location. The plural of locus is loci, pronounced "low-sigh"
- can be described as the path of an object moving in a plane.
B) Finding a locus
1) Draw any figures that are given in the statement of the problem.
2) Locate several points that satisfy the given condition.
3) Continue drawing points until you can recognize a pattern.
4) Draw the locus and describe it in words.

Examples:
1) Given a point C, what is the locus of points 3 inches from C?
When you draw this you will get: the locus of points is a circle with center point C and a radius of 3 inches.

2) Given line AB, what is the locus of points 2 cm from line AB?
the locus of points 2 cm from line AB is 2 lines parallel to AB on either side of line AB 2 cm from AB

3) Given lines AB and CD, what is the locus of points equidistant from lines AB and CD?
the locus of points is a line parallel to both AB and CD that is half way between the lines AB and CD.

4) Given ∠BAC, what is the locus of points equidistant from ray AB and ray AC?
the locus of points is the angle bisect of ∠BAC.

5)Given point A and point B, what is the locus of points equidistant from point A and point B?
the locus of points is the perpendicular bisector of line segment AB.

C) Earthquakes - the epicenter of an earthquake is the point on the Earth's surface that is directly above the earthquake's origin. A seismograph measures ground motion during an earthquake. The seismograph measures the distance to the epicenter, but not he direction to the epicenter. To locate the epicenter, readings from three seismographs in different locations are needed.

1) locating an epicenter:
You are given readings from 3 different seismographs:
1) point A(-2, 2) and distance to the earthquake is 3 miles
2) point B(4, -1) and distance to the earthquake is 6 miles
3) point C(1, -5) and distance to the earthquake is 5 miles

Drawing these three circles, the point of interception is the epicenter. This would be (-2, -1)

Geometry Chapter 10.6 Equations of Circles

2007-08-07T05:27:00.001-10:00

Chapter 10.6 Equations of Circles:

I) Standard equation of a circle with radius r and center (h,k) is:
r² = (x - h)² + (y - k)²
OR
r = √((x - h)² + (y - k)²)

Example: Given circle O with center point (5, 3) and radius of 6, what is the standard equation of circle O?
r² = (x - h)² + (y - k)²
6² = (x - 5)² + (y - 3)²
36 = (x - 5)² + (y - 3)²

Example: Given circle O with center point ( -2, 4) and radius of 9, what is the standard equation of circle O?
9² = (x - ^-2)² + (y - 4)²
81 = (x + 2)² + (y - 4)²

Example: Given the standard equation of circle O: (x + 12)² + (y - 4)² = 25, what is the center point and radius?
center point is (-12, 4) and radius of 5

Geometry Chapter 10.5 Segment Lengths in Circles

2007-08-07T03:34:00.000-10:00

Chapter 10.5 Segment Lengths in Circles:

I) Vocabulary:
A) when two chords intersect in the interior of a circle, each chord is divided into two segments which are called segments of a chord.
B) Tangent Segment - is tangent to a circle at an endpoint.
C) Secant Segment - is a segment that has an endpoint on a circle.
D) External Segment - is part of a secant segment from an external point to the circle. It is does not have any part of the segment in the interior part of the circle.

II) Theorem:
A) If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.
Example: Given circle P with chord AB and chord CD that intersect at point E, AE = 9, ED = 12, CE = 15, EB = x, solve for x.
(9)(x) = (15)(12)
9x = 180
x = 20
B) If two secant segments share the same endpoint outside a circle, then the product of the length of one secant segment and the length of its external segment equals the product of the length of the other secant segment and the length of its external segment.
Example: Given circle A with secant RPQ with PQ being a chord on the circle, and secant RST with ST being a chord on the circle, RP = 9, PQ = 11, RS = 10 and ST = x, solve for x.
(RP)(RQ) = (RS)(RT)
9 (11+9) = 10 (x + 10)
(9)(20) = 10x + 100
180 = 10x + 100
80 = 10x
8 = x
Example: Given circle A with secant RPQ with PQ being a chord on the circle, and secant RST with ST being a chord on the circle, RP = x, PQ = 65, RS = 15, and ST = 35. Solve for x.
(RP)(RQ) = (RS)(RT)
x (x + 65) = (15)(15 + 35)
x² + 65x = 750
x² + 65x - 750 = 0
(x + 75)(x - 10) = 0
x = - 75, x = 10
Since a length cannot be a negative number, the only answer would then be 10 units.
C) If a secant segment and a tangent segment share an endpoint ooutside a circle, then the product of the length of the secant segment and the length of its external segment equals the square of the length of the tangent segment.
Example: Given circle A with secant ECD with CD being a chord on the circle, and tangent EA, EC = x, CD = 12, and EA = 8, solve for x.
(x)(x + 12) = 8²
x² + 12x = 64
x² + 12x - 64 = 0
(x + 16)(x - 4) = 0
x = -16, x = 4
Since a length cannot be a negative number, the only answer would then be 4 units.
Example: Given circle A with secant ECD with CD being a chord on the circle, and tangent EA, EC = 12, CD = 36, and EA = x, solve for x.
(12)(12 + 36) = x²
(12)(48) = x²
576 = x²
24 = x

Geometry Chapter 10.3 Inscribed Angles and 10.4 Other Angle Relationships in Circles

2007-08-01T06:23:00.001-10:00

Chapter 10.3 Inscribed Angles

I) Vocabulary
A) Inscribed Angles - an angle whose vertex is on a circle and whose sides contain chords of the circle.
B) Intercepted arc - the arc that lies in the interior of an inscribed angle and has endpoints on the angle.
C) If all of the vertices of a polygon lie on a circle, the polygon is inscribed in the circle and the circle is circumscribed about the polygon.

II) Theorems:
A) Measure of an inscribed angle - if an angle is inscribed in a circle, then its measure is half the measure of its intercepted arc.
Example: if inscribed angle ABC of circle D is 20º, what is the measure of arc AC?
20º times 2 = 40º so the measure of arc AC is 40º.
B) If two inscribed angles of a circle intercept the same arc, then the angles are congruent.
Example: If angle A and angle B are both inscribed angles and both intercept arc CD, then angle A = angle B
C) If a right triangle is inscribed in a circle, then the hypotenuse is a diameter of the circle. Conversely, if one side of a n inscribed triangle is a diameter of the circle, then the triangle is a right triangle is a right triangle and the angle opposite the diameter is the right angle.
D) A quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary.
Example: Given quadrilateral ABCD is inscribed in circle E, angle A = 115º, angle B = angle D = y, and angle C = x, solve for x and y.
measure of angle A + measure of angle C = 180º
115 º+ x = 180 º
x = 65º
measure of angle B + measure of angle D = 180º
y + y = 180º
2y = 180º
y = 90º
Example: Given quadrilateral ABCD is inscribed in Circle E, angle A = 26y, angle B = 3x, angle C = 2x, and angle D = 21y, solve for both x and y.
measure of angle A + measure of angle C = 180º
26y + 2x = 180
measure of angle B + measure of angle D = 180º
3x + 21y = 180
Now solve both equations for x:
2x = -26y + 180
x = -13y + 90º
3x = -21y + 180
x = -7y + 60
Since x is equal to itself, by substitution:
-13y + 90 = -7y + 60
90 = 6y + 60
30 = 6y
y = 5
Now substitute y = 5,
x = (-7)(5) + 60 = -35 + 60 = 25
check to make sure both answers work!

Chapter 10.4 Other Angle Relationships in Circles
I) Theorems:
A) If a tangent and a chord intersect at a point on a circle, then the measure of each angle formed is one half the measure of its intercepted arc. m∠ 1 = (1/2)measure of arc AB.
Example: Given circle P with angle ACB that intercepts arc AB, the measure of arc AB = 50º, what is the measure of angle ACB?
m∠ACB = (1/2) measure of arc AB
m∠ACB = (1/2) (50)
m∠ACB = 25º
B) If two chords intersect in the interior of a circle, then the measure of each angle in one half the sum of the measures of the arcs intercepted by angle and its vertical angle.
m∠1 = (1/2)(measure arc CD + measure of arc AB)
Example: If Chord AB and chord CD intersect circle P at point E, the measure of arc AD = 25 degrees and the measure of arc CB = 75º what is the m∠AED?
measure of angle AED = (1/2)(25 + 75 )
measure of angle AED = (1/2)(100)
measure of angle AED = 50º
C) If a tangent and a secant, two tangents, or two secants intersect in the exterior of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs.
Example: Given Secant PAB and tangent PC of circle O, the measure of arc BC = 200º and the measure of arc AC = 30º, what is the measure of angle APC?
measure of angle APC = (1/2)(measure of arc BC - measure of arc AC)
measure of angle APC = (1/2)(200 - 30)
measure of angle APC = (1/2)(170)
measure of angle APC = 85º
Example: Given secant PAB and secant PCD of circle O, the measure of arc BD = 105º and measure of arc AC = 51º, what is the measure of angle APC ?
measure of angle APC = (1/2)(105 - 51 )
measure of angle APC = (1/2)(54)
measure of angle APC = 27º
Example: Given tangent PA and tangent PC, the measure of angle APC = 40º, what are the measures of the major arc AC and the minor arc AC?
m∠APC = (1/2)(major arc AC - minor arc AC)
if we let the minor arc AC = x, then the major arc AC = 360 - x
m∠APC = (1/2)((360 -x) - x)
m∠APC = (1/2)(360 - 2x)
40 = 180 - x
x = 220º
so measure of major arc AC = 220º and the measure of minor arc AC = 140º

Geometry Chapter 10.2 Arcs and Chords,

2007-08-01T06:05:00.000-10:00

Chapter 10.2 Arcs and Chords:

I)Vocabulary:
A) Central angle - in a plane, an angle whose vertex is the center of a circle.
B) Minor Arc - if the measure of a central angle APB is less than 180º, then A and B and the points of circle P in the interior of angle APB form a minor arc of the circle.
C) Major Arc - The points A and B and the points of circle P in the exterior of angle APB for a major arc. OR if the measure of a central angle APB is greater than 180º and less then 360º, then A and B and the points of circle P in the interior of ∠APB form a major arc of the circle.
D) Semicircle - if the endpoints of an arc are the endpoints of a diameter.
E) To name a minor arc, use the endpoints of the arc.
Example: given a minor arc on circle C with endpoints A and B, its name is arc AB and notation is the letters AB with a small arc over the two letters.
F) To name a major arc, use the endpoints and a point on the circle between the endpoints.
Example: given a major arc on circle C with endpoints A and B and a point D between A and B, its name is arc ADB.
G) Measure of a minor arc or major arc - is defined to be the measure of its central angle .
Example: measure of minor arc AB = measure of angleACB
II) Theorems:
1) Arc Addition Postulate or Arc Partition Postlate - the measure of an arc formed by two adjacent arcs is the sum of the measures of the two arcs.
Example: Arc ABC = arc AB + arc BC
2) Congruent arcs - two arcs of the same circle or of congruent circles that have the same measure.
3) In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent.
4) If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
5) If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.
6) In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.

Geometry Chapter 10.1 Tangents to Circles

2007-08-01T05:41:00.001-10:00

Chapter 10.1 Tangents to Circles

I) Vocabulary:
A) Circle - is the set of all pints in a plane that are equidistant from a given point, called the center of the circle.
B) Radius - the distance from the center to a point on the circle. Two circles are congruent if they have the same radius.
C) Diameter - the distance across the circle, through its center. The diameter is twice the radius.
D) Chord - is a segment whose endpoints are points on the circle.
E) Diameter - is a chord that passes through the center of the circle.
F) Secant - is a line that intersects a circle in two points.
G) Tangent - is a line in the plane of a circle that intersects the circle in exactly one point.
H) Tangent Circles - coplanar circles that intersect in one point.
I) Concentric - coplanar circles that have a common center.
J) Common tangents - a line or segment that is tangent to two coplanar circles.
1) Common internal tangent - intersects the segment that joins the centers of the two circles.
2) Common external tangent - does not intersect the segment that joins the centers of the two circles.
K) Interior of a circle - consists of the points that are inside the circle.
L) Exterior of a circle - consists of the points that are outside the circle.
M) Point of tangency - the point at which a tangent line intersects the circle to which it is tangent.
N) Theorems:
1) If a line is tangent to a circle, then it is perpendicular to the radius drawn to the point of tangency.
Example: Given line AB is tangent to circle C with point A on circle C, AC is a radius of 5 units, AB is 12 units, what is the length of line segment CB?
Since line AB is a tangent line, we know that AC and AB are perpendicular so angle CAB is a right angle so triangle CAB is a right triangle. Therefore we can use pythagorean theorem to find CB.
5² + 12² = (CB)²
25 + 144 = (CB)²
169 = (CB)²
13 = CB

2) In a plane, if a line is perpendicular to a radius of a circle at its endpoint on the circle, then the line is tangent to the circle.
Example: Given radius DE = 11 units on circle D, there is an external point F, line segment DF = 61 and line segment EF = 60, is line EF a tangent line to circle D?
11² + 60² = 121 + 3600 = 3721
61² = 3721
Therefore, since the two smaller sides squared are equal to the longer side squared, this is the converse of pythagorean theorem so we know that angle DEF is a right angle, so therefore line EF is perpendicular to line segment DE so therefore EF is a tangent line to circle D.

3) If two segments from the same exterior point are tangent to a circle, then they are congruent.
Example: Given tangent SR = 2x + 7 and tangent SB = 5x - 8, where both tangents intersect at point S off of circle C with points R and B are on circle C, solve for x.
2x + 7 = 5x - 8
7 = 3x - 8
15 = 3x
5 = x

Geometry 9.4 Special Right Triangles

2007-07-24T06:20:00.000-10:00

Geometry 9.4 Special Right Triangles:

I) 45°: 45°:90° Triangle Theorem: in a 45°, 45°, 90° triangle,
the hypotenuse is √2 times as long as each leg.
1x: 1x: x√2
Example: if the leg of a 45°: 45°: 90° triangle is 5, what is the other leg length and the hypotenuse length?
Since the legs are equal in length, the other leg is 5.
the hypotenuse would be 5√2

II) 30 :60 : 90 triangle theorem: in a 30º , 60º , 90º triangle,
the hypotenuse is 2 times as long as the shorter leg, and the longer leg is
√3 times as long as the shorter leg.
1x : x√3 : 2x
Example: if the shorter leg of a 30º , 60º , 90º triangle is 7, what is the length of the longer leg and the hypotenuse?
the longer leg is 7√3 and the hypotenuse is (2)(7) = 14

Geometry 9.1 - 9.3 Similar Right Triangles, Pythagorean Theorem

2007-07-24T05:50:00.000-10:00

Geometry 9.1 Similar Right Triangles:

I) Theorems:
A) if the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other.
B) In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments.
The length of the altitude is the geometric mean of the lengths of the two segments.
C) In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.

Geometry 9.2 The Pythagorean Theorem:
I) Theorems:
A) Pythagorean Theorem: in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.
a² + b² = c²
Example: Given the two legs of a right triangle are 7 and 9, what is the hypotenuse?
7² + 9² = c²
49 + 81 = c²
130 = c²
c = √130 ≈ 11.40175425
B) Pythagorean Triple: is a set of three positive intergers, a, b, and c, that satisfy the equation
a² + b² = c²
Examples:
3, 4, 5
5, 12, 13
8, 15, 17
are a few

Geometry 9.3 The Converse of the Pythagorean Theorem:
I) Theorems:
A) Converse of the Pythagorean Theorem: if the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle.
If c² = a² + b², then Δ ABC is a right triangle.
B) If the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides, then the triangle is acute.
if c² is less than a² + b², then ΔABC is acute.
C) If the square of the length of the longest side of a triangle is greater than the sum of the squares of the lengths of the other two sides, then the triangle is obtuse.
if c² is greater than a² + b², then ΔABC is obtuse.
( you can think of this in a doorway, the length of the door does not change nor does the doorway. As you open the door, the angle formed between the door and the doorway gets larger, therefore the distance from the doorway to the end of the door not on the hinge (in the room) is getting longer. So if the angle formed from the intersection of the doorway and door makes a right angle, the triangle is right. If the angle formed is smaller than the right angle, the triangle is acute. If the angle formed is larger than the right angle, the triangle is obtuse.)
Examples:
1) Given a triangle has lengths of 2, 10, and 11 is it acute, right or obtuse?
2² + 10² = 4 + 100 = 104
11² = 121
therefore the c² is longer so the triangle is obtuse.
2) Given a triangle has lengths of 6, 5 and 7, is it acute, right or obtuse?
5² + 6² = 25 + 36 = 61
7² = 49
therefore the c² is smaller so the triangle is acute.
3) Given a triangle has lengths of 6, 8 and 10, is it acute, right or obtuse?
6² + 8² = 36 + 64 = 100
10² = 100
therefore the c² is equal so the triangle is right.
4) Given a triangle has lenghts of 5, 6, and 11, is it acute, right or obtuse?
Recall that a triangle has to have the sum of two sides greater than the third side and since 5 + 6 = 11, these side lenghts cannot form a triangle so it is none of them.

Geometry Chapter 6.2 - 6.6 Properties of Quadrilaterals

2007-07-24T04:05:00.000-10:00

I) 6.2 Properties of Parallelograms
A) Parallelogram: A quadrilateral with both pairs of opposite sides parallel.
B) theorems for Parallelograms:
If a quadrilateral is a parallelogram, then
1. both pairs of opposite sides are congruent.
2. both pairs of opposite angles are congruent.
3. both pairs of consecutive angles are supplementary.
4. the diagonals bisect each other.
5. a diagonal bisects the parallelogram into 2 congruent triangles
6. if one pair of opposite sides are parallel and congruent.

II) 6.3 Proving Quadrilaterals are Parallelograms:
1. Show that both pairs of opposite sides are parallel.
2. Show that both pairs of opposite sides are congruent.
3. Show that both pairs of opposite angles are congruent.
4. Show that one angle is supplementary to both consecutive angles.
5. Show that the diagonals bisect each other.
6. Show that one pair of opposite sides are congruent and parallel.

III) 6.4 Rhombuses, Rectangle, and Squares
A) Rhombus: A parallelogram with 4 congruent sides
1) the diagonals are perpendicular
2) the diagonals bisect opposite angles
B) Rectangle: A parallelogram with 4 congruent right angles.
1) the diagonals are congruent
C) Square: a parallelogram with 4 congruent sides and 4 congruent right angles.

IV) 6.5 Trapezoids and Kites
A) Trapezoid: A quadrilateral with exactly one pair of parallel sides called the bases. It has 2 pairs of base angles. The nonparallel sides are called the legs.
B) Isosceles Trapezoid: A trapezoid that has:
1) nonparalel sides are congruent
2) base angles are congruent
3) diagonals are congruent.
C) Kite: a quadrilateral with 2 pairs of consecutive congruent sides but the opposite sides are not congruent.
1) the diagonals are perpendicular
2) it has exactly one pair of opposite angles congruent. (the pair of angles that is between the noncongruent consecutive sides)
D) Midsegment of a trapezoid: the midsegment of a trapezoid connects the midpoints of the legs and is parallel to both bases. Its length is the average of the 2 bases.
Example: given trapezoid ABCD with BC ll AD and midsegment MN where M is the midpoint of AB and N is the midpoint of CD, then we know:
MN ll AD, MN ll BC, MN = .5(AD + BC)

V) 6.6 Special Quadrilaterals:
Summarizing Properties of Quadrilaterals

A) Quadrilateral - 4 sided polygon

1) Kite
2) Parallelogram
a) Rhombus
b) Rectangle
i) Square
3) Trapezoid
a) Isosceles Trapezoid

VI) Areas of Triangles and Quadrilaterals:
A) Rectangles, Square,Parallelogram
A = (base)(height) where the base and height must be perpendicular
B) Triangle
A = (1/2)(base)(height) where the base and height must be perpendicular
C) Trapezoid
A = (1/2)h (B₁ + B₂)
D) Rhombus and Kite
A =(1/2)( d₁ )(d₂ )

Geometry Chapter 6.1 Polygons

2007-07-24T03:38:00.000-10:00

6.1 Polygons

check out this website:

http://www.math.com/tables/geometry/polygons.htm

A) Polygon - is a plane figure with 3 or more sides, with each side intersecting with exactly 2 other sides, and that meets the following conditions:
1. If is formed by three or more segments called sides, such that no two sides with a common endpoint are collinear.
2. Each side intersects exactly two other sides, one at each endpoint.
3. each endpoint of a side is a vertex of the polygon.

B) Polygons are named by the numbe of sides they have.
3 sides = triangle
4 sides = quadrilateral
5 sides = pentagon
6 sides = hexagon
7 sides = heptagon
8 sides = octagon
9 sides = nonagon
10 sides = decagon
12 sides = dodecagon
n sides = n-gon
C) A polygon is convex if no line that contains a side of the polygon contains a point in the interior of the polygon.
D) a polygon that that is not convex is called nonconvex or concave.
E) Equilateral - a polygon with all of its side are congruent.
F) Regular - a polygon that is equilateral and equiangular.
G) Diagonal of a Polygon: A segment that joins 2 nonconsecutive vertices.
H) Theorem:
1) Interior Angles of a quadrilateral - the sum of the measures of the interior angles of a quadrilateral is 360 degrees.

Geometry 5.6 Indirect Proof and Inequalities in Two Triangles

2007-07-19T02:55:00.005-10:00

Geometry 5.6 Indirect Proof and Inequalities in Two Triangles

I) Theorems:
a) Hinge Theorem - if two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second, then the third side of the first is longer than the third side of the second.
Example: Given ΔRST and ΔVWX, RS = VW and ST = WX, angle S = 100 degrees and angle W = 80 degrees, we can conclude that RT is greater than VX.
B) Converse of the Hinge theorem - if two sides of one triangle are congruent to two sides of another triangle, and the third side of the first is longer than the third side of the second, then the included angle of the first is larger than the included angle of the second.

II) Using Indirect Proofs:
A) Indirect Proof - is a proof in which you prove that a statement is true by first assuming that its opposite is true. If this assumption leasds to an impossibility, then you have proved that the original statement is true.
1) Guidelines for writing an indirect proof:
a) identify the statement that you want to prove is true.
b) begin by assuming the statement is false; assum its opposite is true.
c) obtain statements that logically follow from your assumption.
d) if you obtain a contradiction, then the original statement must be true.

Example: Prove a triangle cannot have 2 right angles.

1) Given ΔABC.
2) Assume angle A and angle B are both right angles is true by one of two possibilities (it is either true or false so we assume it is true).
3) measure of angle A = 90 degrees and measure of angle B = 90 degrees by definition of right angles.
4) measure of angle A + measure of angle B + measure of angle C = 180 degrees by the sum of the angles of a triangle is 180 degrees.
5) 90 + 90 + measure of angle C = 180 by substitution.
6) measure of angle C = 0 degrees by subtraction postulate
7) angle A and angle B are both right angles is false by contradiction (an angle of a triangle cannot equal zero degrees)
8) A triangle cannot have 2 right angles by elimination (we showed since that if they were both right angles, the third angle would be zero degrees and this is a contridiction so therefore our assumption was false ).

Here is another direct proof in t-table form:

Example 2:

This is a direct proof.

Given: LM = MN

Prove: line segment LM is congruent to line segment MN.

This is example 2 as an indirect proof:

Example 3:
Given: LM = MN

Prove: line segment LM is congruent to line segment MN.

Example 4: Given angle PQR is a straight angle, prove the measure of the angle is equal to 180 degrees indirectly.

Example 5: Prove the following indirectly.

As you may have noticed, each indirect proof has 4 steps that have the same concepts:

1. given

2. Assume the opposite of what you want to prove is true.

2nd to last step, your assumption was false by contradiction.

last step, what you wanted to prove is proven by elimination.

Geometry Chapter 5.5 Inequalities in One Triangle

2007-07-18T05:46:00.000-10:00

Geometry chapter 5.5 Inequalities in One Triangle

I) Theorems:
A) If one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side.
B) If one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle.
C) Exterior Angle Inequality - the measure of an exterior angle of a triangle is greater than the measure of either of the two nonadjacent interior angles.
Example: Given ΔABC with side BC extended to point D forming exterior angle ACD,
measure of angle ACD > measure of angle A and
measure of angle ACD > measure of angle B
D) Triangle Inequality - the sum of the lengths of any two sides of a triangle is great than the length of the third side.
Example: Given the triangle has lengths of 5 and 15, what is the other length x?
5 + 15 = 20 and 15 - 5 = 10, therefore the third length is
10 < x < 20

Geometry Chapter 5.4 Midsegment Theorem

2007-07-18T05:28:00.000-10:00

Geometry Chapter 5.4 Midsegment Theorem

I) Vocabulary:
A) Midsegment of a triangle - is a segment that connects the midpoints of two sides of a triangle.
B) Theorem: Midsegment Theorem - the segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half as long.
Example: given ΔABC with point D the midpoint of AC and point E the midpoint of BC and point F is the midpoint of AB, we can conclude
1) DE ll AB
2) DE = 1/2 AB
3) EF ll AC
4) EF = 1/2 AC
5) FD ll BC
6) FD = 1/2 BC
Therefore, you end up with 4 triangles that are congruent.
Check out websites dealing with Fractals- a fractal is created with midsegments. Beginning with any triangle, shade the triangle formed by the three midsegments. Continue this process for each unshaded triangle. Here is one:
http://mathforum.org/alejandre/applet.mandlebrot.html

Geometry Chapter 5.3 Medians and Altitudes of a Triangle

2007-07-18T05:03:00.000-10:00

Geometry Chapter 5.3 Medians and Altitudes of a Triangle

I) Vocabulary:
A) Median of a triangle - is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side.
B) Centroid of the triangle - the point of concurrency of the three medians of a triangle .
C) Altitude of the triangle - is the perpendicular segment from a vertex to the opposite side or to the line that contains the opposite side. An altitude can lie inside, on, or outside the triangle.
D) Orthocenter of the triangle - the point of concurrency of the three altitudes of a triangle.

II) Theorems:
A) Concurrency of Medians of a Triangle - the medians of a triangle intersect at a point that is called the centroid and that is two thirds of the distance from each vertex to the midpoint of the opposite side.
Example: If point P is the centroid of ΔABC, then AP = 2/3 AD, BP = 2/3 BF, and CP = 2/3 CE.
B) Concurrency of Altitudes of a Triangle - the lines containing the altitudes of a triangle are concurrent at the orthocenter.
Example: If AE, BF, and CD are the altitudes of ΔABC, then the lines AE, BF and CD intersect at the orthocenter point H.

Geometry Chapter 5.2 Bisectors of a Triangle

2007-07-18T04:10:00.001-10:00

Geometry Chapter 5.2 Bisectors of a Triangle

I) Vocabulary:
A) A Perpendicular bisector of a triangle - is a line (or ray or segment) that is perpendicular to a side of the triangle at the midpoint of the side.
B) Concurrent lines - when three or more lines (or rays or segments) intersect in the same point.
C) Point of concurrency - the point of intersection of the concurrent lines.
D) Circumcenter of the triangle - the point of concurrency of the perpendicular bisectors of a triangle.
E) Angle bisector of a triangle - is a bisector of an angle of the triangle.
F) Incenter of the triangle - the point of concurrency of the angle bisectors of a triangle.
II) Theorems:
A) concurrency of Perpendicular Bisectors of a Triangle - the perpendicular bisectors of a triangle intersect at a point that is equidistant from the vertices of the triangle.
Example: Given ΔABC with point P being the circumcenter of the triangle, we can conclude that
PA = PB = PC.
Using point P as the center of the circle and PA, PB, or PC as a radius, the circleis circumscribed about the triangle.
B) Concurrency of Angle Bisectors of a Triangle - the angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle.
Example: Given ΔABC with point P being the incenter of the triangle, we can conclude that
PD = PE = PF and PD ll BC, PE ll AC and PF ll AB.
Using point P as the center of the circle and PD, PE, or PF as a radius, the circle is inscribed within the triangle.

Geometry Chapter 5.1 Perpendiculars and Bisectors

2007-07-18T02:02:00.000-10:00

Geometry Chapter 5.1 Perpendiculars and Bisectors:

I) Vocabulary:
A) Perpendicular Bisectors - a segment, ray, line or plane that is perpendicular to a segment at its midpoint.
B) Equidistant - a point is equidistant from two points if its distance from each point is the same.
C) Distance from a point to a line - is defined as the length of the perpendicular segment from the point to the line.
D) Equidistant from the two lines - when a point is the same distance from one line as it is from another line.
II) Theorems:
A) Perpendicular Bisector Theorem - if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
Example: If line CP is the perpendicular bisector of line segment AB, then CA = CB.
B) Converse of the Perpendicular Bisector Theorem - if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.
Example: If DA = DB, then point D lies on the perpendicular bisector of line segment AB.
C) Angle Bisector Theorem - if a point is on the bisector of an angle, then it is equidistant from the two sides of the angle.
Example: if measure of angle BAD = measure of angle CAD, then BD = DC.
D) Converse of the Angle Bisector theorem - if a point is in the interior of an angle and is equidistant from the sides of the angle, then it lies on the bisector of the angle.
Example: If DB = DC, then measure of angle BAD = measure of angle CAD

Geometry Chapter 4.7 Triangles and Coordinate Proof

2007-07-17T06:25:00.000-10:00

Geometry Chapter 4.7 Triangles and Coordinate Proof

A coordinate proof involves placing geometric figures in a coordinate plane. Then you will use:

1) Distance Formula - to show segments are congruent
2) Midpoint formula - to show that the segment is bisected
3) Slope formula -
a) if 2 lines have the same slope, then the lines are parallel
b) if 2 lines have slopes that are negative reciprocals of each other, then the lines are perpendicular

Example: To show that ΔABC is an isosceles right triangle given A(0,0), B(0,6) and C(6,0).
1. slope of AB = (0-6)/(0-0) = undefined
2. slope of BC = (6 - 0)/(0 - 6) = 6/(-6) = -1
3. slope of AC = (0 - 0)/(0 - 6) = 0 / 6 = 0

Therefore AB is perpendicular to AC because horizontal and vertical lines are perpendicular to each other. therefore angle A is a right angle because perpendicular lines form right angles.

4. AB = 6, AC = 6, and BC = 6√2
so we can conclude that AB = AC because they have the same measure.

Therefore since ΔABC has one right angle and 2 congruent sides, it is an isosceles right triangle.

Geometry Chapter 4.6 Isosceles, Equilateral, and Right Triangles

2007-07-17T06:18:00.000-10:00

Geometry Chapter 4.6 Isosceles, Equilateral, and Right Triangles

I) Vocabulary:
A) Base angles - the two angles adjacent to the base of an isosceles triangle.
B) Vertex angles - the angle opposite the base of an isosceles triangle.
II) Theorems:
A) Base Angles Theorem: if two sides of a triangle are congruent, then the angles opposite them are congruent.
B) Converse of the Base Angles Theorem: if two angles of a triangle are congruent, the the sides opposite them are congruent.
C) Hypotenuse - Leg Congruence Theorem (HL = HL): if the hypotenuse and a leg of a right triangle are congruent to the hypotenuse and leg of a second right triangle, then the two triangles are congruent.
Example: Given Δ ABC and ΔDEF are both right triangles, BC = EF and AC = DF,
then ΔABC = ΔDEF.
III) Corollaries:
A) If a triangle is equilateral, then it is equiangular.
B) If a triangle is equiangular, then it is equilateral.

Geometry Chapter 4.3 and 4.4 Triangle are Congruent: SSS, SAS, ASA, and AAS

2007-07-17T06:04:00.000-10:00

Geometry Chapter 4.3 Triangle are Congruent by SSS and SAS

I) Postulates:
SSS = SSS Congruence Postulate - if three sides of one triangle are congruent to three sides of a second triangle, then the two triangles are congruent.
Example: Given ΔABC and ΔDEF,
if AB = DE, BC = EF and AC = DF, then ΔABC = ΔDEF.

SAS = SAS Congruence Postulate - if two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the two triangles are congruent.
Example: Given ΔABC and ΔDEF,
if AB = DE, BC = EF and ∠B = ∠E, then ΔABC = ΔDEF.

Geometry Chapter 4.4 Triangles are Congruent by ASA and AAS

II) Postulates:
ASA = ASA Congruence Postulate - if two angles and the included side of one triangle are congruent to two angles and the included side of a second triangle, then the two triangles are congruent.
Example: Given ΔABC and ΔDEF,
if angle A = angle D, AB = DE, and angle B = angle E, then Δ ABC = Δ DEF.

AAS = AAS Congruence Postulate - if two angles and a nonincluded side of one triangle are congruent to two angles and the corresponding nonincluded side of a second triangle, then the two triangles are congruent.
Example: Given Δ ABC and Δ DEF,
if angle A = angle D, angle C = angle F, and BC = EF , then Δ ABC = Δ DEF.

Geometry chapter 4.2 Congruence and Triangles

2007-07-17T05:46:00.000-10:00

Geometry Chapter 4.2 Congruence and Triangles

I) Vocabulary:
A) When two figures are congruent, there is a correspondence between their angles and sides such that corresponding angles are congruent and corresponding sides are congruent.

Example: Given Δ ABC is congruent to Δ PQR then we know
1) angle A = angle P, angle B = angle Q, and angle C = angle R
2) AB = PQ, BC = QR, and AC = PR

Make sure that you list the corresponding angles in the same order with the triangle congruence.
Example: ΔABC = ΔDEF is not the same as ΔABC = ΔEFD.

B) Third Angles Theorem - if two angles of one triangle are congruent to two angles of another triangle, then the third angles are also congruent.
C) Reflexive Postulate - Every triangle is congruent to itself
D) Symmetric Postulate - If ΔABC = ΔDEF, then ΔDEF = ΔABC
E) Transitive Postulate - If ΔABC = ΔDEF and ΔDEF = ΔJKL, then ΔABC = ΔJKL

Geometry Chapter 4.1 Triangles and Angles

2007-07-17T03:50:00.000-10:00

Geometry Chapter 4.1 Triangles and Angles

I) Vocabulary:
A) Triangle - is a figure formed by three segments joining three noncollinear points.
B) Classification of triangles:
1) By sides:
a) Equilateral Triangle - has 3 congruent sides
b) Isosceles Triangle - has at least 2 congruent sides
c) Scalene Triangle - has no congruent sides
2) By angles:
a) Acute Triangle - has 3 acute angles
b) Equiangular Triangle - has 3 congruent angles that measure 60 degrees each
c) Right Triangle - has one right angle and 2 acute angles
d) Obtuse Triangle - has one obtuse angle and 2 acute angles
C) Vertex - each of the three points joining the sides of a triangle (plural - vertices)
D) Adjacent Sides - in a triangle, two sides sharing a common vertex.
E) Right triangles have 2 sides that form the right angle called the legs. The side opposite the right angle is the hypotenuse of the triangle.
F) Interior angles - when the sides of a triangle are extended, other angles are formed. the three original angles are the interior angles.
G) Exterior angles - the angles that are adjacent to the interior angles.
H) Theorem:
1) Triangle sum theorem - the sum of the measures of the interior angles of a triangle is 180 degrees.
2) Exterior Angle Theorem - the measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles.
I) Corollary to a theorem - is a statement that can be proved easily using the theorem.
Example: The acute angles of a right triangle are complementary.

Example: given Triangle ABC with side BC extended through point D, if angle A = 65 degrees and angle ACD = 2x + 10 and angle B = x, solve for x.

angle A + angle B = angle ACD
65 + x = 2x + 10
55 = x

Geometry Chapter 3.7 Perpendicular lines in the Coordinate Plane

2007-07-16T07:24:00.001-10:00

Geometry chapter 3.7 Perpendicular lines in the Coordinate Plane:

I) Postulate:
A) In a coordinate plane, 2 non-vertical lines are perpendicular if and only if they product of their slopes is equal to (-1). The two slopes are negative reciprocals of each other.

all vertical and horizontal lines are perpendicular to each other.

Given the slope of the first line is a/b,
then the slope of the perpendicular line is (-b/a).

(a/b)(-b/a) = (-1)

B) To show that two lines are perpendicular lines:
1) Find the slopes of each line
2) multiply the slopes
3) Perpendicular if they = -1

Example: Given the first line has points (-1,2) and (5, -1) and the second line has points (4,2) and (1, -4), find out if the two lines are perpendicular:

M₁ = (2- (-1))/(-1 - 5) = 3/(-6) = -1/2
M₂ = (2 - (-4))/(4 - 1) = 6/3 = 2

when you multiply these together (-1/2)(2) = -1
so these lines are perpendicular to each other.

Example 2:
Given the slope of the first line is 3/4, what is the slope of a line perpendicular to this line and what is the equation of the line perpendicular through the point (9, -2)

y = mx + b
the slope was 3/4 so the perpendicular slope is (-4/3)

-2 = (-4/3)(9) + b
-2 = -12 + b
10 = b

therefore the equation of the line perpendicular to the first line through the point (6, -2) is
y = (-4/3) x + 10

Geometry Chapter 3.6 Parallel lines in the Coordinate Plane

2007-07-16T07:17:00.001-10:00

Geometry Chapter 3.6 Parallel lines in the Coordinate Plane

I) Vocabulary:
A) Slope: (nonvertical line): Ratio of the vertical change (the rise, Δ y) to the horizontal change (the run, Δ x)

Formula: slope = m = (Δy/Δx) = (y₂ - y₁)÷ (x₂ - x₂)

II) Postulate:
A) In the coordinate plane, 2 nonvertical lines are parallel if and only if they have the same slope. Any two vertical or horizontal lines are parallel.
B) Equation of a line:
y = mx + b
where m is the slope of the line and b is the y-intersept (where it crosses the y-axis)

Geometry 3.3 Parallel Lines and Transversals and 3.4 Proving lines are parallel and 3.5 Using Properties of Parallel lines

2007-07-16T07:08:00.002-10:00

Geometry 3.3 Parallel Lines and Transversals

I) Vocabulary:
A) Transversal: A line that intersects 2 lines at different places.
II) Postulate:
A) If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.
B) If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles are supplementary.
C) If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent.
D) If two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent.
E) If a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other.

II) Converse of above:
A) If 2 lines are cut by a transversal such that the corresponding angles are congruent, then the lines are parallel.
B) if 2 lines are cut by a transversal such that the pairs of consecutive interior angles are supplementary, then the lines are parallel.
C) if 2 lines are cut by a transversal such that the pairs of alternate exterior angles are congruent, then the lines are parallel.
D) if 2 lines are cut by a transversal such that the pairs of alternate interior angles are congruent, then the lines are parallel.

III) Using Properties of Parallel lines:
A) if two lines are parallel to the same line, then they are parallel to each other.
B) In a plane, if two lines are perpendicular to the same line, then they are parallel to each other.

Geometry 3.2 Proof and Perpendicular lines

2007-07-16T07:06:00.000-10:00

Geometry 3.2 Proof and Perpendicular Lines:

I) Theorems:
1. If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular.
2. If two sides of two adjacent acute angles are perpendicular, then the angles are complementary.
3. If two lines are perpendicular, then they intersect to form four right angles.

Geometry Chapter 3.1 Lines and angles

2007-07-16T06:56:00.000-10:00

Geometry Chapter 3.1 Lines and Angles:

I) Vocabulary:
A) Parallel Lines (ll): two lines that are coplanar and do not intersect.
B) Skew lines: two lines that doe not intersect and are not coplanar.
C) Parallel Planes: Planes that do not intersect.
II) Postulates:
A) If there is a line and a point not on the line, then there is exactly one line through the point parallel to the given line.
B) If there is a line and a point not on the line, then there is exactly one line through the point perpendicular to the given line.
III) Identifying Angels formed by transversals:
A) Transversal: is a line that intersects two or more coplanar lines at different points.
B) Corresponding angles: if you have two lines that are intersected by a transversal, the the two angles that occupy the same position - (eg. to the left of the transversal and on top of the two lines)
C) Alternate interior angles: if you have two lines that are intersected by a transversal, then the two angles that lie between the two lines and on opposite sides of the transversal.
D) Alternate exterior angles: if you have two lines that are intersected by a transversal, then the two angles that lie outside the two lines on opposite sides of the transversal.
E) Consecutive Interior Angles: two angles if they lie between the two lines on the same side of the transversal. Also known as same side interior angles.

Geometry 2.5 Proving Statements about Segments and 2.5 Proving Statements about Angles

2007-07-16T06:45:00.000-10:00

Geometry 2.5 Proving Statements about Segments:

I) Vocabulary:
A) Reflexive: any segment or angle is congruent to itself.
B) Symmetric: If AB = CD, then CD = AB or angle A = angle B, then angle B = angle A.
C) Transitivity: If AB = CD and CD = EF, then AB = EF or if angle A = angle B and Angle B = angle C, then angle A = angle C.
D) Theorem: a true statement that follows as a result of other true statements.
All theorems must be proven true for all cases. Here are a few ways of doing theorems.

1. Two-Column Proof: has numbered statements and reasons that show the logivcal order of an argument.
2. Paragraph proof: a proof can be written in paragraph form.
3. Flow proof: a chart that has arrows going from one statement to the next with the reasons written underneath the statement.

E) Theorems:
1. All right angles are congruent.
2. If two angles are supplementary to the same angle (or to congruent angles), then they are congruent.
3. If two angles are complementary to the same angle (or to congruent angles), then they are congruent.
4. If two angles form a linear pair, then they are supplementary.
5. Vertical angles are congruent.
6. Corresponding parts of congruent triangles are congruent (CPCTC)
7. If two lines form congruent adjacent angles, then the lines are perpendicular.
8. The supplement of a right angle is a right angle.

Geometry Chapter 2.3 Deductive Reasoning and 2.4 Reasoning with Properties from Algebra

2007-07-16T06:26:00.000-10:00

Geometry Chapter 2.3 Deductive Reasoning

I) Using Symbolic Notation:
1) " → " means implies
2) "~" means not
3) "↔" means if and only if

A) conditional statement has a hypothesis (symbolically "p") and a conclusion (symbolically "q") so: if p then q or p → q
B) Converse: q → p
C) Inverse: ~p → ~ q
D) Contrapositive: ~q → ~ p
E) Biconditional: p ↔ q

F) Deductive Reasoning: uses facts, definitions, and accepted properties in a logical order to write a logical argument.
1) Law of detachment: if "p → q" is a true statement and "p" is a true statement, then we can conclude that "q" is true.
Example: If Mark gets time off of work then he is going to Hawaii on vacation. (this is true)
Mark gets time off of work. (this is true).
Therefore we can conclude that Mark is going to Hawaii on vacation.

Symbolically:
Let "p" be Mark gets time off of work.
Let "q" be Mark is going to Hawaii on vacation.

p → q is true
p is true.
Therefore q is true.

2) Law of Syllogism:
If p → q AND q → r are both true, then we can conclude that p → r .
This is like the transitivity postulate.

Example: If Carol buys a new swimsuit, then she is going to the Lake.
If Carol goes to the Lake, then she will go swimming.
Therefore we can conclude that
If Carol buys a new swimsuit, then she will go swimming.

2.4 Reasoning with Properties from Algebra:

G) Postulates:
1. Reflexive Postulate: a = a
2. Symmetric Postulate: if a = b then b = a
3. Transitivity Postulate: if a = b and b = c, then a = c
4. Addition Postulate: if a = b and c is not equal to 0, then a + c = b + c
5. Subtraction Postulate: if a = b and c is not equal to 0, then a - c = b - c
6. Multiplication Postulate: if a = b and c is not equal to 0, then ac = bc
7. Division Postulate: if a = b and c is not equal to 0, then a/c = b/c
8. Distribution Postulate: if a(b + c) then ab + ac

H) Definitions:
1. Perpendicular lines - form right angles
2. Right angle - an angle that measures 90 degrees.

Geometry Chapter 2.2 Definitions and Biconditional Statements

2007-07-16T06:18:00.001-10:00

Geometry Chapter 2.2 Definitions and Biconditional Statements

I) Vocabulary:
A) Perpendicular: Two lines that intersect to form right angles.
B) Line Perpendicular to a Plane: A line that intersects a plane in a point and is perpendicular to every line that includes that point in the plane that intersects it.
C) Biconditional Statement: A statement that contains the words "if and only if" (iff) and is equivalent to writing a statement combining a conditional statement and its converse. For the truth value of a biconditional statement to be true, both the conditional statemene and its converse have to have the same truth value.

Example: Conditional Statement: If two sides of a triangle are congruent, then the angles opposite them are congruent.
Converse Statement: If two angles of a triangle are congruent, then the sides opposite them are congruent.

Biconditional Statement: Two sides of a triangle are congruent if and only if the two angles of the triangle are congruent.

Geometry Chapter 2.1 Conditional Statements

2007-07-16T03:53:00.000-10:00

Geometry chapter 2.1 Conditional Statements:

I) Vocabulary:

A. Conditional Statements - (if-then) - has two parts:

1. Hypothesis - (if)

2. Conclusion - (then)

Example: If it is noon in Georgia, then it is 9 am in California.

B) Converse Statement: Switch the hypothesis and conclusion of a Conditional Statement.

Example: If it is 9 am in California, then it is noon in Georgia.

C) Inverse Statement: negate both the hypothesis and conclusion of a Conditional Statement.

Example: If it is not noon in Georgia, then it is not 9 am in California.

1) Negation - write the negative of the statement.
Example: Statement: Angle A is acute.
Negation: Angle A is not acute.
or
It is not true that angle A is acute.
D) Contrapositive Statement: Switch and negate both the Hypothesis and Conclusion of a Conditional Statement.

Example: If it is not 9 am in California, then it is not noon in Georgia.

E) Equivalent Statements: 2 statements that are both true or both false, they have the same truth value. Contrapositive statements is always equivalent to its Conditional statement. The converse statement is always equivalent to its inverse statement.

F) Counterexample: An example that shows that a conditional statement is false.

Example: If x² = 25, then x = 5
A counterexample is x = (-5) because (-5)² = 25 but 5 is not equal to (-5)

II) Postulates:

#5) Postulate 5: Through any 2 points there exists one line.
#6) Postulate 6: A line contains at least two points.
#7) Postulate 7: If two lines intersect, then their intersection is exactly one point.
#8) Postulate 8: Through any three noncollinear points there exists exactly one plane.
#9) Postulate 9: A plane contains at least three noncollinear points.
#10) Postulate 10: If two points lie in a plane, then the line containing them lies in the plane.
#11) Postulate 11: If two planes intersect, then their intersection is a line.

Geometry Chapter 1.7 Introduction to Perimeter, Circumference, and Area

2007-07-12T04:45:00.000-10:00

Geometry Chapter 1.7 Introduction to Perimeter, Circumference, and Area

I) Vocabulary and Formulas:

Notation:
A = Area
P = Perimeter

A) Triangle:
A = (1/2)(base)(height)
P = (side a) + (side b) + (side c)

Example: Given Triangle ABC with side a = 10, b = 17 and c = 21 and the height of the triangle is 8. Find the area and the perimeter.

Area = (1/2)(21)(8) = 84 square units.
Perimeter = 10 + 17 + 21 = 48 units

B) Square: has all the sides equal so the base = height so we will call them all sides.
A = (side)(side) = s²
P = side + side + side + side = 4s

Example: let one side of a square is equal to 5 inches. what is the area and perimeter?
A = (5)² = 25 square inches
P = 4 (5) = 20 inches

C) Rectangle:
A = (base)(height)
P = 2b + 2h

Example: let the base = 3 and the height = 4
A = (3)(4) = 12 square units
P = (2)(3) + (2)(4) = 6 + 8 = 14 units

D) Circle: 2 radius = diameter
A = π r²
P = Circumference = 2 π r

Example: If a circles diameter is 20 cm, what is the area and circumference?
A = π (10)² = 100π square cm.
C = (2) π (10) = 20 π cm.

Geometry Chapter 1.6 Angle Pair Relationships

2007-07-12T04:36:00.000-10:00

Geometry Chapter 1.6 Angle Pair Relationships

I) Vocabulary

A) Vertical Angles: is when the sides of 2 angles form 2 pairs of opposite rays.
1. Vertical angles are congruent.

B) Linear Pair: 2 adjacent angles are a linear pair if their noncommon sides are opposite rays.
1. If two angles form a linear pair, then they are supplementary.
2. If two angles form a linear pair, then their sum is 180 degrees.

C) Complementary Angles: 2 angles whose measures total 90 degrees.
1. If two angles have the same complement, then the 2 angles are congruent.
2. If two angles are complementary to congruent angles, then the two angles are congruent.

D) Supplementary Angles: 2 angles whose measures total 180 degrees.
1. If two angles have the same supplement, then the 2 angles are supplementary.
2. If two angles are supplementary to congruent angles, then the two angles are supplementary.

Geometry Chapter 1.5 Segment and Angle Bisectors

2007-07-12T04:18:00.000-10:00

I) Vocabulary:

A) Midpoint: A midpoint of a segment is the point that divides, or bisects, the segment into two congruent segments.

B) Segment Bisector: is a segment, ray, line, or plane that intersects a segment at its midpoint.

Example: Given segment AB bisects segment CD at point E. If CD = 10, what does AE and EB equal?
10/2 = 5 so they both equal 5 units.

C) Angle Bisector: is a ray that divides an angle into two adjacent angles that are congruent.

Example: Given angle PQR is bisected by ray QS, and measure of angle SQR = 22 degrees, what is the measure of angle PQR and angle PQS?

22 x 2 = 44 so measure of angle PQR = 44 degrees and when an angle is bisected, the two angles formed are congruent so angle SQR = angle PQS = 22 degrees.

D) Midpoint Formula: the mean, or average, of the x-coordinates and the y-coordinates.

therefore the midpoint formula is: ((x₁+ x₂) /2, (y₁+ y₂) /2)
example: Given A(-1, 7) and B(3, -3) what is their midpoint?

((-1 + 3)/2, (7 + -3)/2) = (2/2, 4/2) = (1, 2)

Geometry chapter 1.4 Angles and their measures

2007-07-12T02:45:00.000-10:00

Geometry Chapter 1.4 Angles and their Measures:
I) Vocabulary and Postulates:

A) Angle - consists of two different rays that have the same initial point.
1) Vertex - the initial point of the angle.
2) Rays of an angle - the sides of the angle.

B) Congruent Angles - angles that have the same measure.
C) Angle Addition Postulate - If P is on the interior of angle RST, then measure of angle RSP + measure of angle PSP = measure of angle RST.

D) Acute Angle - an angle that's measure is between zero and ninety degrees.

E) Right Angle - an angle that's measure is 90 degrees.

F) Obtuse Angle - an angle that's measure is between 90 degrees and 180 degrees.

G) Straight Angle - an angle that's measure is 180 degrees.

H) Adjacent Angles - two angles are adjacent angles if they share a common vertex AND a common side, but do not have any common interior points.

Geometry Chapter 1.3 Segments and Their Measures

2007-07-10T05:50:00.000-10:00

Geometry Chapter 1.3 Segments and Their Measures

I) Vocabulary & Postulates:

A) theorems: Rules that have been proven to be true.
Example: Triangle Sum Theorem: The sum of the measures of the interior angles of a triangle is 180 degrees.

B) Postulates or Axioms: Rules that we accept as true without proof.

C) Between - When 3 points are one a line, you may say that 1 point is between the others.

D) Segment Addition Postulate or Partition Postulate: If B is between point A and point C, then
AB + BC = AC

E) Distance formula: AB = √((x₁ - x₂)² + (y₁ - y₂)²)

this is derived from the pythagorean theorem, where the difference of the x's is represented by "a" and the difference of the y's is represented by "b" and AB = c so

a² + b² = c²

F) Congruent Segments - Segments that have the same measure

Written:

Lengths are equal in length so: AB = CD

and

Segments are congruent so: line segment AB is congruent to line segment CD.

Geometry Chapter 1.2 Points, Lines and Planes

2007-07-10T04:01:00.000-10:00

Chapter 1.2 Points, Lines, and Planes

I) Undefined Terms:

A) Point: a point has no dimension, usually represented by a small dot.

B) Line: A line extends in one dimension. It is usally represented by a straight line with two arrowheads to indicate that the line extends without end in two directions.

C) Plane: a plane extends in two dimensions, goes on forever.

D) Collinear Points: 2 or more points on the same line.

E) Coplanar Points: Points that lie on the same plane.

F) Line Segment: a segment of a line; consists of 2 endpoints and all points in between.

G) Ray: An initial point and all points one side of that point.

H) Opposite Rays: 2 rays that form a straight line.

I) Intersect: Two or more geometric figures intersect if they have one or more points in common.

Geometry - Chapter 1.1 Patterns and Inductive Reasoning

2007-07-10T03:45:00.000-10:00

Chapter 1.1 - Patterns and Inductive Reasoning

A. Using Inductive Reasoning
1. Look for a pattern. - look at several examples. Use diagrams and tables to help discover a pattern.
2. Make a Conjecture. - a conjecture is an unproven statement that is based on observation.
3. Verify the Conjecture - use logical reasoning to verify that the conjecture is true in all cases.

Example - the sum of the first n odd positive integers is...
first odd positive integer is 1 = 1
sum of first two odd positive integers is 1 + 3 = 4
sum of first three odd positive integers is 1 + 3 + 5 = 9
sum of first four odd positive integers is 1 + 3 +5 + 7 = 16

looking at these answers: 1, 4, 9, 16 we see they are perfect squares so
1², 2², 3², 4²
Therefore we can make a conjecture that the sum of the first n odd positive integers is n² .

B. Counterexample - is an example that shows a conjecture is false.
Example: All prime numbers are odd.
Since 2 is a prime number but 2 is even this would be a counterexample.

Precalculus 12.5b Area of a Plane Region

2007-05-07T12:17:00.008-10:00

12.5b Area of a Plane Region

Area of a Region bounded by:

x = a, x = b, y = 0 and y = f (x) where f(x) is less than or equal to 0.

The boundary of the area is from x₀ = a to x_n = b.

What is the area under the curve?

George Riemanns - German who came up with technique to find the orange area.

1. Subdivide the interval from a to b into smaller intervals. This is called a partition of [a,b] specifically:

a = x₀ ≤ x₁ ≤ x₂ ≤ x₃ ... ≤ x_{n - 1} ≤ x_n = b

2. Choose a point w_i in interval [x_{i - 1}, x_i] , i = 1 ... n

so...

when i = 1, w₁ is contained in [x₀, x₁]

when i = 2, w₂ is contained in [x₁, x₂]

3. Let Δ x_i = x_i - x_i-1 for i = 1 ... n

i= 1; Δ x₁ = x₁ - x₀

i = 2; Δ x₂ = x₂ - x₁

Δ x_i = length of the i^th subinterval.

4. Form f(w_i)(Δx_i) for i = 1 ... n

5. The area of the region:

A = f (w₁)(Δ x₁) + f(w₂)(Δ x₂) + ... + f (w_n)(Δx_n)

This is called a Riemann's Sum.

So by increasing the number of rectangles that you make, you can obtain a closer and closer approximation

Therefore to find the Area of a Plane Region:

Let "f" be continuous function and nonnegative on the interval [a, b]. The Area A of the region bounded by the graph of "f", the x-axis (y = 0), and the vertical lines x = a and x = b is

Area of a rectangle = height times width

check out this website: http://tutorial.math.lamar.edu/classes/calcI/areaproblem.aspx

Example 1:
Find the Area of the Region: f(x) = 2 - x², -1 £ x £ 1

Let's let n = 4

so the width = (1 - ^-1)/4 = 1/2

height =
[-1, -1/2] = 1.75
[-1/2, 0] = 2
[0, 1/2] = 1.75
[1/2, 1] = 1

so the Area when f(x) = 2 - x² and is divided up into 4 rectangles =
(1/2)(1.75) + (1/2)(2) + (1/2)(1.75) + (1/2)(1) = 3.25

but to find the area with more rectangles, let's use the above formulas:

Therefore to find the area, use the summation formulas:

To check this on the graphing calculator:

put in y₁ = 2 - x²

Now on your homescreen, math arrow down to #9 fnInt (Y₁, x, -1, 1) enter

should give you 10/3.

In the calculator (function , variable, lower bound, upper bound) = area under the curve.

Precalculus 12.5 The Area Problem

2007-05-05T07:52:00.002-10:00

12.5 The Area Problem

I. Limits of Summation

S = a₁ + a₁r + a₁r² + ... =

_¥
å (a₁(r^i-1)) =
^{i = 1}

a₁/(1-r) provided that the absolute value of r < 1

Using Limit Notation, this sum can be written as:

S =
¥
lim å (a₁ r^i-1)
^{n®¥ i = 1}

= lim a₁ (1 - rⁿ)/(1-r)
^n®¥
Recall that since r < 1 thus making it a fraction, when you take the limit or rⁿ as x approaches infinity, this would equal zero. So:

= (a₁ (1 - 0))/ (1 - r)

= a₁/(1 - r) which you have already seen in this coursework.

Summation Formulas and Properties:

Example 1: Evaluating a Summation:

What is the sum of
i² when i = 1 to i = 30?

=(30)(30 + 1)(2(30) + 1)/6 = (30)(31)(61)/6 = 56730/6 = 9455

Example 2: Evaluating a Summation:

What is the sum of:
(2i + 1) when i = 1 to i = 50?

You have to break this down into parts:

2( å i ) from i = 1 to i = 50 AND
å(1) from i = 1 to i = 50 is equal to

2(50(50+1)/2) + 1(50) = (50)(51) + 50 = 2550 + 50 = 2600

II. Finding the Limit of a Summation

_n
å (2i + 3)/(n²)
^{i = 1}

= (1/n²)(2(n(n+1))/2) + 3n) = (1/n²)(n(n+1) + 3n)
= (1/n)(n + 1 + 3) = (n + 4)/n

which in essence is infinity over infinity so
Therefore the Summation of
lim S_n =
^n®¥

lim (n + 4)/n = 1/1 = 1 so
^n®¥

lim S_n = 1
^n®¥

You can check this by using a table of values:

(n, S(n)) = {(1, 5), (10, 1.4), (100, 1.04), (1000, 1.004), (10000, 1.0004)...}
So you can see the values of S(n) approach 1.

Example 3: Finding the Limit of a Summation:

_n
å (3/n³)(1 + i²)=
^{i = 1}

= (3/n³)(n + (n(n + 1)(2n + 1)/6) =
(3/n² + (3(2n² + 3n + 1)(6n²)
= 3/n² + (6n² + 9n + 3)/(6n²)

lim 3/n² = 0 and
^{n ®¥}

lim (6n² + 9n + 3)/(6n²) = 6/6 = 1
^{n ®¥}

So:

lim S_n = 1
^{n ®¥}

Again, you can check this using a table of values:
(n, S(n)) = {(1, 6), (10, 1.185), (100, 1.0154), (1000, 1.0015), (10000, 1.0002)...}

Precalculus 12.4 Limits at Infinity and Limits of Sequences

2007-05-02T05:23:00.001-10:00

12.4 Limits at Infinity and Limits of Sequences

Definition of Limits at Infinity
If "f" is a function and L₁ and L₂ are real numbers, the statements

lim f(x) = L₁
^x®-¥

AND

lim f(x) = L₁
^x®+¥

denote the limits at infinity. The first is read "the limit of f(x) as x approaches negative infinity is L₁," and the second is read "the limit of f(x) as x approaches infinity is L₂"

To help evaluate limits at infinity, you can use the following:

Limits at Infinity
If "r" is a positive real number, then

lim (1/(x^r)) = 0 Limit toward the right.
^x®+¥

Furthermore, if x^r is defined when x is less then 0, then

lim (1/(x^r) = 0 the limit toward the left.
^{x®- ¥}

Thought: -1/10 = -.1, -1/100 = -.01, -1/1000 = -.001, -1/10000 = -.0001
so as you can see, in the denominator as x approaches negative infinity, f(x) approaches zero.

Example 1: Evaluating a Limit at Infinity:

lim (5/(2x)) = 0
^x®¥

As you can see when you graph this function, as x gets larger, f(x) gets closer to zero so that is how we can conclude the the limit equals zero.

Example 2:
lim (1 - 3/(x²)) =
^x®¥

We can separate each part of the function to:

lim (1 )= 1
^x®¥

and

lim (3/(x²)) = 0
^x®¥

so
lim (1 - 3/(x²)) = 1 - 0 = 1
^x®¥

Compare the following limits:
Find the limit as x approaches infinity:

a). f(x) = (-5x + 2)/(4x² - 1)

b). g(x) = (4x² + 1)/(3x² - 1)

c). j(x) = (-3x⁴ + 1)/(2x³-1)

When you take the limit of each of these functions as x approaches infinity, you get:

1st, look in the denominator and find the largest exponent and use this term by dividing all the terms of the function by that variable raised to the largest exponent.

a).
lim (-5x + 2)/(4x² - 1)=
^x®¥

so divide each term by x²

lim [(-5x)/(x²) + 2/(x²)]/[(4x²)/(x²) - 1/(x²)]
^x®¥

= (-0 + 0)/(4 - 0) = 0

b).
lim (4x² + 1)/(3x² - 1)=
^x®¥

so divide each term by x²

lim [4 + 1/(x²)]/[3 - 1/(x²)]=
^x®¥

(4 + 0)/(3 - 0) = 4/3

c).
lim (-3x⁴ + 1)/(2x³-1)
^x®¥

and so divide each of the terms by x³

lim (-3x + 1/x³)/(2 - 1/x³)=
^x®¥

-3x/2 so by putting infinity in for x, the limit does not exist because the numerator decreases without bound as the denominator approaches 2.

Therefore we can conclude:

Limits at Infinity for Rational Functions

for the rational function f(x) = N(x)/D(x) when

N(x) = a_nxⁿ + ... + a₀

And

D(x) = b_mx^m + ... + b₀

the limit as x approaches positive or negative infinity is as follows:

lim f(x) = 0, when n is less than m
^x®¥

lim f(x) = a_n/b_m, when n = m
^x®¥

And if n is greater than m, the limit does not exist.

Limit of a Sequence:
Let "f" be a function of a real variable, such that
lim f(x) = L
^x®¥

If {a_n} is a sequence such that f(n) = a_n
for every positive integer "n", then the
lim a_n = L
^n®¥

Example:
Given: 1/3, 1/9, 1/27, 1/81, ...

As n increases without bound, the terms of the sequence
a_n = 1/3ⁿ is said to

CONVERGE to zero.

lim 1/3ⁿ = 0
^n®¥

___________________________________________

A sequence that does not converge is said to DIVERGE!

Example of this is: the sequence 1, -1, 1, -1, ... this diverges.

Find the limit of the Sequence:

Example 1:
lim (4x - 1)/(x + 3)=
^x®¥
{3/4, 7/5, 11/6, 15/7, ...} = 4

Example 2:
lim (3x + 2)/(2x² - 1)=
^x®¥
{5/1, 8/7, 11/17, 14/31, ...} = 0

Example 3:
lim (7x² - 1)/(8x²)=
^x®¥
{6/8, 27/32, 62/72, 111/128, ...} = 7/8

Use the site: http://www.calculus-help.com/funstuff/phobe.html

Chapter 1, lesson 4 to help with this lesson.

Precalculus 12.3 The Tangent Line Problem

2007-04-30T07:50:00.001-10:00

12.3 The Tangent Line Problem

The Slope of the graph of a function can be used to analyze rates of change at particular points on the graph.

The Slope of a line indicates the rate at which a line rises or falls.
- For a straight line, this rate (or slope) is the same at every point on the line.
- For other graphs than lines, the rate at which the graph rises or falls changes from point to point.

Recall: slope formula = (the change in y's)/(the change in x's) = Δy/Δx

Review: With circles, you were taught about tangent lines. A tangent line is a line that intersects a circle in exactly one point, called the point of tangency. Using this process, to determine the rate at which a graph rises or falls at a single point, you can find the slope of the tangent line at that point.

The tangent line to a graph of a function "f" at a point P(x₁, y₁) is the line that best approximates the slope of the graph at that point.

Example1: f(x) = x² - 2x + 1

find the tangent line at the point (1, f(x))

f(1) = 1² -2(1) + 1 = 0
so the tangent line at (1,0) is a straight line and the equation of the tangent line at (1, f(x)) is y=0.

A more precise method then "eyeballing" the tangent line
is making use of the secant line through the point of tangency
and a second point on the line where (x, f(x)) is the first point
and (x + h, f(x + h)) is a second point on the graph of "f", the
slope of the secant line through these two points is:

M_secant = (f(x + h) - f(x))/h

Using the limit process, you can find the
exact slope of the tangent line at (x, f(x)).

Definition of the Slope of a Graph:

The slope m of the graph of "f" at the point (x, f(x)) is equal to the slope
of its tangent line at (x, f(x)) and is given by:

M =

lim m_sec =
^h®0

lim (f(x+h) - f(x))/h
^h®0

provided the limit exists!

Example 2: given f(x) = 2x + 5, find the slope of the tangent line at (-1, -3)

M_sec = (f(x+h) - f(x))/h

= (f(-1 + h) - f(-1))/h
= (2(-1 + h) + 5 - (2(-1) + 5))/h
= (-2 + 2h + 5 +2 - 5)/h
= (2h)/h
= 2
The graph has a slope of 2 at the point (-1, -3)
To find the equation of the tangent line:
y = mx + b
(-3) = 2(-1) + b
-1 = b

so the equation of the tangent line at point (-1, -3) is
y = 2x -1

Example 3: given f(x) = 10x - 2x² at point (3, 12)

M_sec = [10(x+h) - 2(x + h)² - (10x - 2x²)]/h

M_sec = [10x + 10h -2(x² +2xh + h²) -10x + 2x²]/h

M_sec = [10h - 2x² - 4xh - 2h² + 2x²]/h

M_sec = [10h - 4xh - 2h²]/h

M_sec = 10 - 4x - 2h

Slope of the tangent line = M =
lim M_sec =
^h®0

lim 10 - 4x - 2h = 10 - 4x so at point (3, 12), M = 10 - 4(3) = -2
^h®0

Therefore the slope of the tangent line at the point (3, 12) is -2.

To find the equation of the tangent line, use the point and the slope:

y = mx + b
12 = (-2)(3) + b
12 = -6 + b
18 = b

y = 2x + 18 is the equation of the tangent line at the point (3, 12)

Finding a Formula for the slope of a Graph:
Example4: g(x) = x³ at points (1, 1) and (-2, -8)

M_sec = [(x + h)³ - (x³)]/h

M_sec = [x³ + 3x²h + 3xh² + h³ - x³]/h

M_sec = [3x²h + 3xh² + h³]/h

M_sec = 3x² + 3xh + h² , where h¹0

Next take the limit of M_sec as h approaches 0.

The slope of the tangent line M =

lim 3x² + 3xh + h² = 3x²
^h®0

Now use this M = 3x² for the slope at (1,1)

M = 3(1)² = 3

now y = mx + b
1 = 3(1) + b
b = -2

so the equation of the tangent line at point (1,1) is y = 3x - 2

Again now at point (-2, -8)

Now use this M = 3x² for the slope at (-2,-8)

M = 3(-2)² = 3(4) = 12

now y = mx + b
-8 = 12(-2) + b
-8 = -24 + b
b = 16

so the equation of the tangent line at point (-2,-8) is y = 12x +16

The formula that you derived from the function f(x) = x³ and used the limit process to get M = 3x² represents the slope of the graph of "f" at the point (x, f(x)).

The Derived function is called the derivative of f at x. It is donoted by f ' (x), which is reas as "f prime of x".

Definition of the Derivative:

The Derivative of "f" at "x" is

f ' (x) = lim_h®0 (f(x+h) - f(x))/h

provided the limit exists!

Example 5: f(x) = x² - 3x + 4

f ' (x) = lim_h®0 [(x + h)² - 3(x + h) + 4 - (x² - 3x + 4)]/h

f ' (x) = lim_h®0 [x² + 2xh + h² - 3x -3h + 4 - x² + 3x -4]/h

f ' (x) = lim_h®0 [2xh + h² - 3h]/h

f ' (x) = lim_h®0 2x + h - 3 = 2x - 3

f '(x) = 2x - 3

So therefore the derivative of f(x) = x² - 3x + 4 is f ' (x) = 2x - 3

NOTE that in addition to f ' (x) , other notations can be used to denoted the derivatives of y = f(x). The most common are:

(dy)/(dx)

y'

(d/(dx))[f(x)]

and D_x[y]

Using the derivative of f(x) = x³ - x, find the slope of the tangent line at point (2,6).

f'(x) = lim_h®0 [(x + h)³ - (x + h ) - (x³ - x)]'h

f'(x) = lim_h®0 [x³ + 3x²h + 3xh² + h³ - x - h - x³ - x]/h

f'(x) = lim_h®0 [3x²h + 3xh² + h³ - h]/h

f'(x) = lim_h®0 (3x² + 3xh + h² - 1)

f ' (x) = 3x² - 1

So at the point (2, 6), the slope is
f ' (2) = 3(2)² - 1 = 3(4) - 1 = 12 - 1 = 11

So the equation of the tangent line at point (2,6) is
6 = 11(2) + b
6 = 22 + b
b = -16

so the equation of the tangent line at point (2,6) is y = 11x - 16

Example 6: Find the derivative of f. Use the derivative to determine any points on the graph of f where the tangent line is horizontal.

f(x) = x³ - 3x

f ' (x) = lim_h®0 [(x + h)³ - 3(x + h) - (x³ - 3x)]/h

f ' (x) = lim_h®0 [x³ + 3x²h + 3xh² + h³ - 3x -3h - x³ + 3x]/h

f ' (x) = lim_h®0 [3x²h+ 3xh² + h³ -3h]/h

f ' (x) = lim_h®0 [3x² + 3xh+ h² -3]

f ' (x) = 3x² - 3

Recall a horizontal line has a slope of zero so

f ' (x) = 0 so
0 = 3x² - 3
3 = 3x²
1 = x²
x = ±1

putting these values in for x:

So "f" has horizontal tangents at (1, 0) adn (-1, 0)

If you need extra help, watch the Chapter 2 lesson 1 tutorial at
http://www.calculus-help.com/funstuff/phobe.html

12.2b Precalculus - One-Sided Limits

2007-04-26T02:06:00.001-10:00

12.2b One-Sided Limits

One-Sided Limits - when a limit approaches one value from the left side of "c" and a different value from the right side of "c"

lim f(x) = L
^{x®c^-}

and

lim f(x) = K
^x®c⁺

where if L = K, then the limit exists but if L ¹K, then the limit does not exist for the value of x.

Example 1:

lim f(x)
^x®1

Given f(x) = 4 - x² when x£1 or f(x) = 3 - x when x is greater than 1

therefore:

lim f(x) = 4 - (1)² = 4 - 1 = 3
^{x®1^-}

and

lim f(x) = 3 - 1 = 2
^x®1⁺

so since 3 ¹ 2, then
lim f(x) does not exist since the function approaches different values
^x®1
from the left and from the right.

I. Existence of a Limit:

If "f" is a function and "c" and "L" are real numbers, then

lim f(x) = L
^x®c

if and only if BOTH the left and the right limits exist and are equal to L.

Example 2:

f(x) = 3 - x when x is less than 1 or 3x - x² when x is greater than 1

lim f(x) = 2
^{x ®1^-}

lim f(x) = 2
^{x ®1⁺}

Because the one-sided limits both exist and are equal to the same number 2, it follows that

lim f(x) = 2 so the limit does exist at x = 1.
^x®1

II. A limit from Calculus:

Evaluating a Limit from Calculus - the limit of a difference quotient:

lim (f(x+h) - f(x))/h where h¹0
^h®0

Direct substitution into the difference quotient ALWAYS produces the indeterminate
form 0/0.

Example 3: find the difference quotient of f(x) = 5 - 6x

lim [5 - 6(x + h) - (5 - 6x)]/h
^h®0

lim [5 - 6x - 6h - 5 + 6x]/h
^h®0

lim -6h/h = -6
^h®0

therefore:
lim [5 - 6(x + h) - (5 - 6x)]/h = -6
^h®0

Example 4: find the difference quotient of f(x)= 4 - 2x - x²

lim ([4 - 2(x + h) - (x+h)²] - [4 - 2x - x²])/h =
^h®0

lim ([4 - 2x -2h -(x² + 2xh + h²] - 4 + 2x + x²)/h =
^h®0

lim (4 - 2x -2h -x² -2xh - h² - 4 + 2x + x²)/h =
^h®0

lim (- 2h - 2xh - h²)/h =
^h®0

lim (- 2 -2x -h) = -2 -2x
^h®0

Precalculus - 12.2a Techniques for Evaluating Limits

2007-04-25T03:19:00.001-10:00

12.2 Techniques for Evaluating Limits

I. Limits of Polynomial and Rational Functions

1. If p is a polynomial function and "c" is a real number, then

lim p(x) = p(c)
^x®c

2. If "r" is a rational function given by r(x) = p(x)/q(x) and "c" is a real number such that q(x) ¹0.

Example 1: Evaluating Limits by Direct Substitution

lim (.5x³ - 5x) = (.5)(.2)² - 5(2) = 6
^x®-2

Example 2:
lim (x² + 1)/(x) = (3² + 1)/ 3 = (9 + 1)/ 3 = 10/3
^x®3

Example 3:
lim (x²- 1)^1/3 = (3² - 1)^1/3 = (9 - 1)^1/3 = 8^1/3 = 2
^x®3

II. Factoring: Dividing out Technique

Example 4:
lim (x² - 3x)/x = (x)(x-3)/(x)
^x®0

lim x - 3 = -3
^x®0

Example 5:
lim (x² - 1)/(x + 1) =
^x®-1

lim(x + 1)(x - 1)/(x + 1) =
^x®-1

lim (x - 1) = -2
^x®-1

This technique should be applied only when direct substitution produces zero in both the numerator and the denominator. The resulting fraction 0/0, has no meaning as a real number.

It is called an indeterminate form because you cannot, from the form alone, determine the limit.

III. Rationalizing Technique

First Rationalize the numerator of the function.

Example 6:
lim (3 - Ö (x))/(x - 9)
^x®9

multiply the numerator and the denominator by the conjugate of the numerator:

lim[ (3 - Ö (x))(3 + Ö(x))] /[(x - 9) (3 + Ö(x))] =
^x®9

lim (9 - x) / [(x - 9 )(3 + Ö(x))] =
^x®9

As you can see in the numerator, you have (9 - x) and in the denominator you have (x - 9) so if we multiply the numerator by (-1) you would have:

lim (-1)(- 9 + x) / [(x - 9 )(3 + Ö(x))] =
^x®9

so now x - 9 divides out of the numerator and denominator

lim -1 / (3 + Ö(9)) = -1/ (3 + 3 ) = -1/6
^x®9

IV. Using Technology:

lim (1 + 2x)^1/x
^x®0

Since you cannot use substitution, use a table:
(x, f(x)) = {(-.1, 9.313), (-.01, 7.540), (-.001, 7.404), (-.0001, 7.391), (-.00001, 7.3892), (-.000001, 7.38907), (.000001, 7.38904), (.00001, 7.3898), etc...}

so you can see that it is estimated about

lim (1 + 2x)^1/x »7.3890561 » e²
^x®0

Algebraically:
Choose 2 points that are equidistant from 0 and close to 0 and find their average:

(7.38907 + 7.38904)/2 »7.389055 » e²

Example 7:

lim (1/(2+x) - 1/2)/x =
^x®0

again using the calculator when x = -.0000000001, f(x) = -1/4
and when x = .0000000001, f(x) = -1/4

so

lim (1/(2+x) - 1/2)/x = -1/4
^x®0

Algebraically:

lim (1/(2+x) - 1/2)/x =
^x®0

simplifying the fraction:

[(2 - (2 + x))/(2 + x)(2)]/x =

(- x)/ [(2 + x)(2)(x)] =

-1/(4 + 2x) so:

lim (1/(2+x) - 1/2)/x = -1/(4 + 2(0)) = -1/4 same answer!!
^x®0

12.1 Introduction to Limits

2007-04-19T08:06:00.002-10:00

12.1 Introduction to Limits

Watch the tutorials on lessons 1 - 3 on the following website.

http://www.calculus-help.com/funstuff/phobe.html

Lesson 1. What is a limit
Lesson 2. When does a limit exist.
Lesson 3. How do you evaluate a limit?

______________________________________________________

The Limit Concept - this notion of a limit is a fundamental concept of calculus.

Example 1: You are given 32 feet of fence and are asked to form a rectangular pen whose area is as large as possible. What is the largest area the pen would cover and what dimensions should the pen be?

If we let w be the width of the pen and l be the length of the pen.
Therefore we know that 2w + 2l = 32 for the perimeter.

Therefore if we solve for the length, l = 16 - w

then the area would be

A = lw
A = (16 - w)(w)
A = 16w - w²

Using this equation, we can experiment with different values of w to see how to obtain the maximum area.

(w, A) = {(1, 15), (2, 28), (3, 39), (4, 48), (5, 55), (6, 60), (7, 63), (8, 64), (9, 63), (10, 60), (11, 55), (12, 48), (13, 39) ...}

so as you can see it appears that (8, 64) is a maximum point so therefore when w = 8, the maximum area would be 64. The dimensions of the pen would be an 8 by 8 pen.

In limit terminology, you can say that "the limit of A as w approaches 8 is 64"
This is written as

lim A = 64
^w®8

Definition of Limit:

If f(x) becomes arbitrarily close to a unique number L as x approaches c from either side, the limit of f(x) as x approaches c is L. This is written as

lim f(x) = L.
^x®c

Example 2: Estimating a Limit Numerically

lim (4 - 3x)
^x®3

let's plug in some values and see what happens when x approaches 3.
(x,f(x)) = {(2.9, -4.7), (2.99, -4.97), (2.999, -4.997), (3, ?), (3.001, -5.003), (3.01, -5.03), (3.1, -5.3)}

So we can conclude that as x approaches 3, f(x) approaches -5.

So lim (4 - 3x) = -5
^x®3

this graph is continuous. For graphs that are not continuous, finding a limit can be more difficult.

Example 3: Estimating a Limit Numerically:

lim (x+1)/(x²-x-2)
^x®-1

Again, let's look at some values for x around -1.

(x, f(x)) = {(-1.1, -.3226), (-1.01, -.3322), (-1.001, -.3332), (-1, error),
(-.999, -.3334), (-.99, -.3344), (-0.9, -.3448)}

So as you can see in the table when x = -1, f(x) = error because if you place -1 in for x in the equation, f(x) = 0/0 which is impossible to have because you cannot divide by 0.

But if you place values close to -1, f(x) ® -1/3 so

lim (x+1)/(x²-x-2) = -1/3
^x®-1

Now try:

lim (x + 1)/(x²-x-2)
^x®2

again f(2) = 3/0 which you cannot do so:

when we look at the value of f(x) at values of x around 2, you have two different answers.

1. as x approaches 2 from the left side:

lim (x + 1)/(x²-x - 2)
^{x®2^-}

you can see the limit approaches -¥

2. as x approaches 2 from the right side:

lim (x + 1)/(x²-x - 2)
^x®2⁺

you can see the limit approaches +¥

so since f(x) is not bound as x approaches 2, you can conclude that the limit does not exist at x = 2.

Example 4:

lim (3x²-12)/(x-2)
^x®2

Again plugging x = 2, f(2) = 0/0 which is impossible to have so

(x, f(x)) = {(0, 6), (1, 9), (2, error), (3, 15), (4, 18), (5, 21), ...} so you can see that it is suppose to be 12. we can do this if we use factoring:

(3x² - 12)/(x - 2) =

(3(x² - 4))/(x - 2) =

( 3 (x + 2)(x - 2))/ (x - 2) =

3 (x + 2) because the (x - 2) divide out so we can simplify the limit to look like:

lim 3(x + 2)
^x®2

= 3 ( 2 + 2) = 12

Example 5:

lim (cos (x) - 1)/ x
^x®0

again when x =0, f(0) = error but when we look at the graph, as x approaches 0 from the left side and the right side, f(x) = 0 so we can conclude

lim (cos (x) - 1)/ x = 0
^x®0

Example 6:

lim (cos (1/x))
^x®0

Try using different windows:
let x be between -2p and +2p and y between -1 and +2

Now zoom in alittle: let x be between -1 and +1 by .1

Now zoom in more: let x be between -.1 and +.1 by .01

Continue this, what do you notice?

The smaller the window, so the closer to x = 0 you get, the graph does not approach any particular number (it goes between +1 and -1).

Therefore this limit does not exist because no matter how close you are to zero, it is possible to choose values of x₁ and x₂ such that cos(1/x₁) and cos(1/x₂) that do not equal.

Example 7: Finding Limits by Direct Substitution:

lim (x + 4)/(x - 3)
^x®1

Do the numerator separately from the denominator:

lim (x + 4) = 5
^x®1

lim (x - 3) = -2 so
^x®1

lim (x + 4)/(x - 3) = -5/2
^x®1

____________________________________________________________

I. Conditions Under Which Limits Do not exist:

The Limit of f(x) as x ® c does not exist if any of the following conditions is true.

1. f(x) approaches a different number from the right side of "c" than from the left side of "c".

2. f(x) increases or decreases without bound as x approaches c.

3. f(x) oscillates between two fixed values as x approaches c.

II. Properties of Limits:

Let "b" and "c" be real numbers and let "n" be a positive integer.

1. lim b = b
^x®c

2. lim x = c
^x®c

3. lim xⁿ = cⁿ
^x®c

4. lim ⁿ Ö(x) = ⁿÖ(c) for n even and c greater than zero.
^x®c

III. Operations with Limits:

check out: http://homepage.usask.ca/~mha040/Limit%20Formulas.pdf

Given: lim f(x) = L
^x®c
and

lim g(x) = K
^x®c

1. Scalar multiple:
lim [ bf(x)] = bL
^x®c

2a. Sum of limits:
lim [ f(x) + g(x)] = L + K
^x®c

2b. Difference of limits:
lim [ f(x) - g(x)] = L - K
^x®c

3. Product:
lim [ f(x) ´ g(x)] = L ´ K
^x®c

4. Quotient:
lim [ f(x) ¸ g(x)] = L ¸ K
^x®c

5. Power:
lim [ f(x)]ⁿ = Lⁿ
^x®c

Example 8:

lim f(x) = 3x
^x®c

and

lim g(x) = -2x + 1
^x®c

Find:
lim [ f(x) + g(x)] =3(3) + (-2)(3) + 1 = 9 - 6 + 1 = 4
^x®3

Find:
lim [ f(x) g(x)] = [(3)(4)][-2(4) + 1] = (12)(-8 + 1) = (12)(-7) = -84
^x®4

Example 9:
lim f(x) = 3
^x®c

and

lim g(x) = -2
^x®c

Find:
lim [ f(x) + g(x)]² = (3 + -2)² = 1
^x®c

Find:
lim [ 6f(x)g(x)] = 6(3)(-2) = -36
^x®c

Precalculus 10.7 Graphs of Polar Equations

2007-04-04T05:35:00.001-10:00

10.7 Graphs of Polar Equations

A. Graphing a Polar Equation by Point Plotting by:

1. Convert to Rectangular Form. Multiply both sides of the Polar Equation by "r" and convert the result to rectangular form.

Example 1:
r = 3 cos q
r² = r 3 cos q
x² + y² = 3x
x² -3x + y² = 0

Using complete the square:

x² - 3x + (-3/2)² + y² = 0

(x - 3/2)² + y² = 9/4

This is the graph of a circle with center point (3/2, 0) and r = 3/2

2. Use a Polar Coordinate Mode

Put your calculator in polar mode and place the equation in the r =

r = 3 cos q
Use 0 £ q £ 2p , -6 £ x £ 6, and -4 £ y £ 4

This produces the same graph.

3. Use the Parametric Mode:

The graph of the Polar Equation r = f (q ) can be written in parametric form, using t as a parameter, as follows:

x = f (t) cos t and y = f (t) sin t

r = 3 cos q

x = 3 cos t cos t

y = 3 cos t sin t

Again, this will produce the same graph.

B. Tests for Symmetry:

The graph of a polar equation is symmetric with respect to the following if the given substitution yields an equivalent equation.

1. The line q = p/2
Replace (r, q ) by (r, p-q ) or (-r, -q )

2. The polar axis:
Replace (r, q) by (r, -q) or (-r, p-q)

3. The Pole:
Replace (r, q) by (r, p+ q) or (-r, q)

Example 2: Use symmetry to sketch the graph r = 16 cos 3q
1. The line q = p /2 replace (r, q) by (r, p - q) or (-r, -q)

-r = 16 cos (3(-q))
-r = 16 cos (-3q )
-r = 16 cos 3q
Þthis is not the same equation so it is not equivalent

2. The polar axis replace (r, q) by (r, -q)
r = 16 cos 3(-q)
r = 16 cos 3q
Þthis is the same equation so it is equivalent

3. The pole replace (r, q) by (r, p + q) or (-r, q)
-r = 16 cos 3q
Þthis is not equivalent

Therefore r = 16 cos 3q is symmetric with respect to polar axis.

Plotting the points in the table and using the polar axis symmetry, you obtain the graph which is called limacon. (the c has a ' attached to the bottom of it)

The points are (r, q) = (0, 16), (p/6, 0), (p/3, -16), (p/2, 0), (2p/3, 16), (5p/6, 0), (p, -16)

Example 3: r = 4 - sin q

1. line q = p/2
r = 4 - sin (p- q )
r = 4 - (sin p cos q - cosp sin q)
r = 4 - ((0) cos q - (-1) sinq)
r = 4 - sin q
Þsame equation

2. Polar axis
r = 4 - sin (-q)
r = 4 + sin q
Þnot same equation

or -r = 4 - sin ( p- q)
-r = 4 - (sinp cos q - cos p sin q)
-4 = 4 - sin q
Þnot equivalent

3. Pole
-r = 4 - sin q not equivalent
or
r = 4 - sin ( p+ q)
r = 4 - (sinp cos q + cosp sin q)
r = 4 - (-sin q)
r = 4 + sin q
Þnot equivalent

Therefore r = 4 - sin q is symmetric with respect to q = p/2

Example 4: r = 2 csc q cos q
r = 2(1/sinq)(cos q) = 2 cot q

1. q = p /2
-4 = 2 cot (-q)
-r = -2 cot q
r = 2 cot q
Þtherefore it is the same equation or equivalent

2. Polar axis:
-r = 2 cot ( p- q)
-r = 2 cot (-q)
-r = -2 cot q
r = 2 cot q
ÞEquivalent

3. Pole:
r = 2 cot ( p+ q)
r = 2 cot q
Þ Equivalent

ÞTherefore r = 2 csc q cos q is symmetric with respect to q = /2, polar axis and pole.

B. Zeros and Maximum r - values

½ r ½ is the maximum value for r and knowing the q - values for which r = 0 are two helpful ways to sketch the graph.

Example 4: Find the maximum value of r = 6 + 12 cos q

½r ½ = ½6 + 12 cosq ½

½ r ½ = ½ 6 + 12 cos q ½ £ ½6 ½ + ½ 12 cos q½

½ r ½ = 6 + 12½ cos q½£ 18 (from the thought that 6 + 12 = 18)

so if cos q = 1 then q = 0

So when q = 0, ½ r ½ = 18

Zeros:
Let r = 0

0 = 6 + 12 cos q
-6 = 12 cos q
-1/2 = cos q

Therefore q = 2p/3, 4p/3

Now for the table:
(r , q) goes with (x, y)
(18, 0°) and (18, 0)
(6, 90°) and (0, 6)
(0 , 120°) and (0 , 0)
(-6, 180 °) and (6, 0)
(0, 240°) and (0, 0)
(6, 270°) and (0, -6)
(18, 360°) and (18, 0)

Put your calculator in polar mode and you can get the polar coordinate. Graph the equation and go to 2nd trace, value and place the angle measure in, and the x - values and y - values will be displayed.

Example 5: r = 5 sin (2q)
½ r ½ = ½ 5 sin 2q ½

½ r ½ = 5

If we let sin b = 1 so b = 90° and b = 2q so therefore q = 45° or p/4

so when q = p/4, 3p/4 , 5p /4, 7p/4

So again to get the table of values:

(r, q) goes with (x, y)
(0, 0°) and (0, 0)
(5, 45°) and (3.5, 3.5)
(0, 90°) and (0, 0)
(-5, 135°) and (3.5, 3.5)
(0, 180°) and (0, 0)
(5, 225°) and (-3.5, -3.5)
(0, 270°) and (0, 0)
(-5, 315 °) and (-3.5, 3.5)
(0, 360°) and (0, 0)
This is a rose curve.

check out this video at
http://sam.ntpi.spcollege.edu/spjc/view/eventListing.jhtml?eventid=5323&c=13169

Precalculus 10.6 Polar Coordinates

2007-04-03T04:59:00.002-10:00

10.6 Polar Coordinates:

So far, you have been representing graphs of equations as collections of points (x, y) on the rectangular coordinate system, where x and y represent the directed distances from teh coordinate axes to the point (x, y). In this section, you will study a second system - the Polar Coordinate System.

I. To form the polar coordinate system in the plane, fix a point O, called the pole (or origin) and construct from O an initial ray called the polar axis. Then each point P in the plane can be assigned polar coodinates (r,q ) as follows:

1. r = directed distance from O to P.
2. q = directed angle, counterclockwise from polar axis to segment OP.

To find other points in the same place, do the following:

(r, q ) = (r, q ± 2np )

or

(r, q) = (-r, q ±(2n + 1)p )

Example: Given the polar coordinate (3, p/4)
= (3, p/4 + 2p) = (3, 9p/4)

OR

(-3, p/4 + (2 (0) +1)p) = (-3, 5p/4)

(-3, p/4 - p) = (-3, -3p/4)

II. Coordinate Conversion:
Because (x, y) lies on a circle of radius r, it follows that r² = x² + y². Moreover, for r> 0, the definitions of the trigonometric functions imply:

tan q = y / x

cos q = x/ r

sin q = y/r

The Polar Coordinates (r, q) are related to the rectangular coordinates (x, y) as follows:

x = r cos q

y = r sin q

tan q = y/x

r² = x² + y²

III. Polar to Rectangular Conversion

Example 2: Convert each point to rectangular coordinates
a). (-2, 7p/6) means r = -2 and q = 7p/6

x = (-2)(cos 7p/6) = (-2)(- Ö(3)/2) = Ö3

y = (-2)(sin 7p/6) = (-2)(-1/2) = 1

Therefore the rectangular coordinate is (Ö3 , 1)

b). (8.25, 3.5) means r = 8.25 and q = 3.5 radians

x = (8.25)(cos 3.5) = -7.72576767
y = (8.25)(sin 3.5) = -2.89391628

Therefore the rectangular coordinate is (-7.72576767, -2.89391628)

IV. Rectangle to Polar Conversion

Example 3: Convert each point to polar coordinates

a) Let 0£ q £ 2p and using the rectangular coordinate (0, -5)

tan q = -5/0 = undefined, therefore q = 270° based on where the point is.

r = Ö (x² + y²)
= Ö(0² + (-5)²)
= 5

Therefore a polar coordinate is (5, 270°) or (5, 3p/2)

and the negative would be (-5, -90°) or (-5, p/2)

b) (3, -1)

Tan q = -1/3

q = -18.43494882° or -.3217505544 radians

so q = 341.5650512° or 5.961434753 radians

r = Ö(3² + (-1)²) = Ö(9 + 1) = Ö10

Therefore the polar coordinate is (Ö10, 5.961434753 radians)

To find another coordinate for the same location: 5.961434753-p = 2.819842099
so (-Ö10 , 2.819842099 radians)

V. Equation Convertion

Recall x = r cos q and y = r sin q and x² + y² = r²

Given: x² + y² - 6x = 0 substitute in the above equations for x and y

r² - 6(r cos q) = 0

r ( r - 6 cos q) = 0

r = 0 and r - 6 cos q = 0
r = 6 cos q

so the polar equation is r = 6 cos q

VI. Convert the polar equation to rectangular form:

Example: r = 4 cos q

r² = 4r cos q

x² + y² = 4x

x² -4x + y² = 0 would be the rectangular equation

Example: q = 5p/3

Tan q = tan 5p/3 = -Ö3

y/x = -Ö3/1
y = -Ö3 x
y = xÖ3 = 0

Example: r = 6/(2cosq - 3 sin q)

r = 6 / (2(x/r) - 3(y/r))

r = 6r/(2x - 3y)

2x - 3y = 6r/r

2x - 3y = 6