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) = anxn + ... + a1x + a0
1. When "n" is odd:
a. If the Leading Coefficient is positive (an is greater than 0), the graph falls to the left and rises to the right.
b. If the Leading Coefficient is negative (an 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 (an is greater than 0), the graph rises to the left and right.
b. If the Leading Coefficient is negative (an 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.
