**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) = x

^{n}, where n is greater than zero

1. If

*n *is even, the graph of

*y* = x

^{n} touches the axis at the x-intercept.

2. If

*n* is odd, the graph of

*y* = x

^{n} crosses the axis at the x-intercept.

**III. The Leading Coefficient Test**

Given f(x) = a

_{n}x

^{n} + ... + a

_{1}x + a

_{0}**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 = x

^{2}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) = x

^{2} - 8x + 15

0 = x

^{2} - 8x + 15

0 = (x - 5)(x - 3)

x = 5 and x = 3

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

x

^{2} 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) x

^{4} - x

^{3} + 2x

^{2} + 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) = x

^{3} - 4x

Leading Coefficient = 1

Leading degree is 3 so it is odd

The graph falls to left and rises to the right

0 = x

^{3} - 4x

0 = x (x

^{2} - 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) = 4x

^{2} -6x + 9

Find all zeros:

0 = 4x

^{2} - 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)

= x

^{2} - 5x + 4x - 20

= x

^{2} - x - 20

**Now sketch the graph:**

1. Apply the Leading Coefficient testLeading 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 vertexx = -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)

= (x

^{2})(3x + 1)

= 3x

^{3} + x

^{2} - 6x

^{2} - 2x

= 3x

^{3} - 5x

^{2} - 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)

= x

^{2} - 6x

+ x√3 - 6x + 36

- 6√3 - x√3 + 6√3 - (√3)

^{2}= x

^{2} - 12x + 36 - 3

= x

^{2} - 12x + 33

**Example 4: (use grouping)**

Zeros: 4, 2 + √7, 2 - √7

f (x) = (x - 4)(x - 2 - √7)(x - 2 + √7)

= (x - 4)(x

^{2} - 2x

+ x√7 - 2x + 4

- 2√7 - x√7 + 2√7 - ( √7)

^{2}= (x - 4)(x

^{2} - 4x + 4 - 7 )

= (x - 4)(x

^{2} - 4x - 3)

= x

^{3} - 4x

^{2} - 3x - 4x

^{2} + 16x + 12

= x

^{3} - 8x

^{2} + 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) = x

^{3} - 3x

^{2} + 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: y

_{1} = x

^{3} - 3x

^{2} + 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.