math 31 lessons chapter 4: max / min chapter 5: sketching 3. sketching polynomials

MATH 31 LESSONS

Chapter 4: Max / Min

Chapter 5: Sketching

3. Sketching Polynomials

Steps for Sketching Polynomial Functions

Step 1. Degree

State the degree of f (x) and identify the shape.

When the degree is even, then:

if a > 1, it opens up

Deg = 2

Deg = 4

When the degree is even, then:

if a > 1, it opens up

if a < 1, it opens down

Deg = 2

Deg = 4

Deg = 2

Deg = 4

When the degree is odd, then:

if a > 1, it rises to the right

Deg = 3Deg = 5

When the degree is odd, then:

if a > 1, it rises to the right

if a < 1, it falls to the right

Deg = 3Deg = 5

Step 2. Intercepts

y-intercept(s)

Let x = 0, and then solve for y

x-intercept(s)

Let y = 0, and then solve for x

Note:

To find the x-intercept:

Factor completely and find the zeros

- for complex functions, you may need to use

the factor theorem and long division

To find the x-intercept:

For second degree factors of the form Ax2 + Bx + C =

0,

you can use the quadratic formula to solve.

i.e.

A

ACBBx

2

42

To find the x-intercept:

If you can’t factor or use the quadratic formula,

then use Newton’s method for finding roots.

i.e.

Take a first guess x1.

Then, to find x2:

1

112 xf

xfxx

Step 3. First Derivative Test

Differentiate and state the critical values

i.e. When f (x) = 0

Differentiate and state the critical values

i.e. When f (x) = 0

Use the interval test to show where the function is

increasing and decreasing

Differentiate and state the critical values

i.e. When f (x) = 0

Use the interval test to show where the function is

increasing and decreasing

Identify local (and absolute) max / mins

- substitute into the original function to get the y-values

Step 4. Sketch the Function

Place all intercepts and critical values on the grid

Using your knowledge of where the function is increasing

and decreasing, connect the dots

Extend the arms on either side to infinity

- recall that polynomial functions are continuous and

have a domain x

Ex. 1 Sketch the following function:

Try this example on your own first.Then, check out the solution.

xxy 483

Degree:

This function is of degree 3 and a > 0.

Thus, this function will rise to the right.

xxy 483

Intercepts:

y-intercept: (x = 0)

So, there is a y-intercept at (0, 0).

00480 3 y

xxy 483

x-intercepts: (y = 0)

xxy 483

0483 xx

0482 xx

0x 482 x

3448 x

So, there are x-intercepts at (0, 0) and (6.93, 0)

First Derivative Test:

xxxf 483

483 2 xxf

163 2 x

443 xx

xxxf 483

483 2 xxf

163 2 x

443 xx

0443when0 xxxf

4x Find CV’s

Interval test:

4f

-4

Sketch a number line, using the CV’s as boundaries

443 xxxf

x < -4:

Since f > 0, it is increasing.

e.g. x = -5

454535 f

0913

443 xxxf

4f

-4

() ()

-4 < x < 4:

Since f < 0, it is increasing.

e.g. x = 0

404030 f

0443

4f

-4

() () (+) () 443 xxxf

x > 4:

Since f > 0, it is increasing.

e.g. x = 5

454535 f

0193

4f

-4

() () (+) () (+) (+)

443 xxxf

Find the y-values by subbing them into the original function:

Local Min

Local Max

4f

-4

() () (+) () (+) (+)

xxxf 483

12844844atmaxLocal 3 f

12844844atminLocal 3 f

Sketch:

-6.93 6.93

128

-128

y

x

First, put the intercepts and the CV’s on the graph

-4 4

y

Then, use the interval test to connect the dots.

4f

-4

-6.93 6.93

128

-128

x-4 4

Ex. 2 Sketch the following function:

Try this example on your own first.Then, check out the solution.

45 24 xxy

Degree:

This function is of degree 4 and a > 0.

Thus, this function will open up.

45 24 xxy

Intercepts:

y-intercept: (x = 0)

So, there is a y-intercept at (0, 4).

45 24 xxy

44050 24 y

x-intercepts: (y = 0)

So, there are x-intercepts at (1, 0) and (2, 0)

45 24 xxy

045 24 xx

041 22 xx

02211 xxxx

2,1 x

First Derivative Test:

45 24 xxxf

xxxf 104 3

522 2 xx

0522when0 2 xxxf

0522when0 2 xxxf

02 x

0x

052 2 x

2

52 x

58.12

5x

Interval test:

Sketch a number line, using the CV’s as boundaries

522 2 xxxf

0f

-1.58 1.58

x < -1.58:

Since f < 0, it is decreasing.

e.g. x = -3

532323 2 f

01332

522 2 xxxf

0f

-1.58

() (+)

1.58

-1.58 < x < 0:

Since f > 0, it is increasing.

e.g. x = -1

512121 2 f

0312

522 2 xxxf

0f

-1.58

() (+) () ()

1.58

0 < x < 1.58:

Since f < 0, it is decreasing.

e.g. x = 1

512121 2 f

0312

522 2 xxxf

0f

-1.58

() (+) () () (+) ()

1.58

x > 1.58:

Since f > 0, it is increasing.

e.g. x = 3

532323 2 f

01332

522 2 xxxf

0f

-1.58

() (+) () () (+) (+) (+) ()

1.58

Find the y-values by subbing them into the original function:

25.2458.1558.158.1atminsLocal 24 f

Local Min

Local Max

0f

-1.58

() (+) () () (+) (+) (+) ()

1.58Local Min

45 24 xxy

440500atmaxLocal 24 f

Sketch:

-3

4

-4

y

x

First, put the intercepts and the CV’s on the graph

-1-2 1 32

(-1.58, -2.25) (1.58, 2.25)

Then, use the interval test to connect the dots.

0f

-1.58 1.58

-3

4

-4

y

x-1-2 1 32

(-1.58, -2.25) (1.58, 2.25)