M
Math Coach Amy
Amy Ferguson Moncure

H Algebra 2/Trig · Unit 3

Sketch a Polynomial

Use the factored equation to name the degree, the intercepts, the end behavior, and the turns you can see without a calculator.

Jump to section:

Add the exponents for the degree

Even multiplicity bounces

Odd multiplicity crosses

What to Remember

Build the sketch from the factors

Degree: add the exponents.

Each factor x − r gives an intercept. The exponent says whether it crosses, bounces, or flattens.

Then use the end behavior to aim the arms.

Turning points

A bounce on the x-axis is a max or a min. If the curve is below the axis there, the bounce is a maximum. If it is above, the bounce is a minimum.

There is also one turn in a gap between intercepts when the curve has to change direction to reach the next intercept.

Level 1

Easy

Three intercepts, and the curve crosses at each one.

Example
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (x + 2)(x − 1)(x − 4)

a. Add the exponents.

Degree: 1 + 1 + 1 = 3

b. Set each factor equal to 0.

x = −2, multiplicity 1. The graph crosses the x-axis.

x = 1, multiplicity 1. The graph crosses the x-axis.

x = 4, multiplicity 1. The graph crosses the x-axis.

c. Odd degree and a positive leading coefficient: the left end goes down and the right end goes up.

As x → −∞, y → −∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

There is a local maximum between x = −2 and x = 1.

There is a local minimum between x = 1 and x = 4.

A sketch of that shape:

−4−3−2−1123456−16−14−12−10−8−6−4−2246810121416

Answer: Degree 3
x = −2, multiplicity 1; x = 1, multiplicity 1; x = 4, multiplicity 1
As x → −∞, y → −∞
As x → ∞, y → ∞
There is a local maximum between x = −2 and x = 1
There is a local minimum between x = 1 and x = 4

1
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (x + 4)(x + 1)(x − 2)
Show Answer

a. Add the exponents.

Degree: 1 + 1 + 1 = 3

b. Set each factor equal to 0.

x = −4, multiplicity 1. The graph crosses the x-axis.

x = −1, multiplicity 1. The graph crosses the x-axis.

x = 2, multiplicity 1. The graph crosses the x-axis.

c. Odd degree and a positive leading coefficient: the left end goes down and the right end goes up.

As x → −∞, y → −∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

There is a local maximum between x = −4 and x = −1.

There is a local minimum between x = −1 and x = 2.

A sketch of that shape:

−6−5−4−3−2−11234−16−14−12−10−8−6−4−2246810121416

Answer: Degree 3
x = −4, multiplicity 1; x = −1, multiplicity 1; x = 2, multiplicity 1
As x → −∞, y → −∞
As x → ∞, y → ∞
There is a local maximum between x = −4 and x = −1
There is a local minimum between x = −1 and x = 2

2
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (13)(x + 3)(x − 2)(x − 5)
Show Answer

a. Add the exponents.

Degree: 1 + 1 + 1 = 3

b. Set each factor equal to 0.

x = −3, multiplicity 1. The graph crosses the x-axis.

x = 2, multiplicity 1. The graph crosses the x-axis.

x = 5, multiplicity 1. The graph crosses the x-axis.

c. Odd degree and a positive leading coefficient: the left end goes down and the right end goes up.

As x → −∞, y → −∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

There is a local maximum between x = −3 and x = 2.

There is a local minimum between x = 2 and x = 5.

A sketch of that shape:

−5−4−3−2−11234567−10−8−6−4−2246810121416

Answer: Degree 3
x = −3, multiplicity 1; x = 2, multiplicity 1; x = 5, multiplicity 1
As x → −∞, y → −∞
As x → ∞, y → ∞
There is a local maximum between x = −3 and x = 2
There is a local minimum between x = 2 and x = 5

3
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = −(x + 5)(x + 2)(x − 1)
Show Answer

a. Add the exponents.

Degree: 1 + 1 + 1 = 3

b. Set each factor equal to 0.

x = −5, multiplicity 1. The graph crosses the x-axis.

x = −2, multiplicity 1. The graph crosses the x-axis.

x = 1, multiplicity 1. The graph crosses the x-axis.

c. Odd degree and a negative leading coefficient: the left end goes up and the right end goes down.

As x → −∞, y → ∞

As x → ∞, y → −∞

d. Even multiplicities put a max or a min on the x-axis.

There is a local minimum between x = −5 and x = −2.

There is a local maximum between x = −2 and x = 1.

A sketch of that shape:

−7−6−5−4−3−2−1123−16−14−12−10−8−6−4−2246810121416

Answer: Degree 3
x = −5, multiplicity 1; x = −2, multiplicity 1; x = 1, multiplicity 1
As x → −∞, y → ∞
As x → ∞, y → −∞
There is a local minimum between x = −5 and x = −2
There is a local maximum between x = −2 and x = 1

4
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (12)(x + 1)(x − 2)(x − 6)
Show Answer

a. Add the exponents.

Degree: 1 + 1 + 1 = 3

b. Set each factor equal to 0.

x = −1, multiplicity 1. The graph crosses the x-axis.

x = 2, multiplicity 1. The graph crosses the x-axis.

x = 6, multiplicity 1. The graph crosses the x-axis.

c. Odd degree and a positive leading coefficient: the left end goes down and the right end goes up.

As x → −∞, y → −∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

There is a local maximum between x = −1 and x = 2.

There is a local minimum between x = 2 and x = 6.

A sketch of that shape:

−3−2−112345678−14−12−10−8−6−4−2246810

Answer: Degree 3
x = −1, multiplicity 1; x = 2, multiplicity 1; x = 6, multiplicity 1
As x → −∞, y → −∞
As x → ∞, y → ∞
There is a local maximum between x = −1 and x = 2
There is a local minimum between x = 2 and x = 6

5
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = −(12)(x + 3)(x − 1)(x − 4)
Show Answer

a. Add the exponents.

Degree: 1 + 1 + 1 = 3

b. Set each factor equal to 0.

x = −3, multiplicity 1. The graph crosses the x-axis.

x = 1, multiplicity 1. The graph crosses the x-axis.

x = 4, multiplicity 1. The graph crosses the x-axis.

c. Odd degree and a negative leading coefficient: the left end goes up and the right end goes down.

As x → −∞, y → ∞

As x → ∞, y → −∞

d. Even multiplicities put a max or a min on the x-axis.

There is a local minimum between x = −3 and x = 1.

There is a local maximum between x = 1 and x = 4.

A sketch of that shape:

−5−4−3−2−1123456−14−12−10−8−6−4−2246810

Answer: Degree 3
x = −3, multiplicity 1; x = 1, multiplicity 1; x = 4, multiplicity 1
As x → −∞, y → ∞
As x → ∞, y → −∞
There is a local minimum between x = −3 and x = 1
There is a local maximum between x = 1 and x = 4

Level 2

Medium

One intercept bounces.

Example
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (17)(x + 4)²(x − 1)(x − 3)

a. Add the exponents.

Degree: 2 + 1 + 1 = 4

b. Set each factor equal to 0.

x = −4, multiplicity 2. The graph bounces off the x-axis.

x = 1, multiplicity 1. The graph crosses the x-axis.

x = 3, multiplicity 1. The graph crosses the x-axis.

c. Even degree and a positive leading coefficient: both ends go up.

As x → −∞, y → ∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

At x = −4 the graph is above the axis on both sides, so that bounce is a local minimum.

There is a local maximum between x = −4 and x = 1.

There is a local minimum between x = 1 and x = 3.

A sketch of that shape:

−6−5−4−3−2−112345−10−8−6−4−22468101214

Answer: Degree 4
x = −4, multiplicity 2; x = 1, multiplicity 1; x = 3, multiplicity 1
As x → −∞, y → ∞
As x → ∞, y → ∞
Local minimum at (−4, 0)
There is a local maximum between x = −4 and x = 1
There is a local minimum between x = 1 and x = 3

1
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (110)(x + 2)(x − 3)²(x − 6)
Show Answer

a. Add the exponents.

Degree: 1 + 2 + 1 = 4

b. Set each factor equal to 0.

x = −2, multiplicity 1. The graph crosses the x-axis.

x = 3, multiplicity 2. The graph bounces off the x-axis.

x = 6, multiplicity 1. The graph crosses the x-axis.

c. Even degree and a positive leading coefficient: both ends go up.

As x → −∞, y → ∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

At x = 3 the graph is below the axis on both sides, so that bounce is a local maximum.

There is a local minimum between x = −2 and x = 3.

There is a local minimum between x = 3 and x = 6.

A sketch of that shape:

−4−3−2−112345678−14−12−10−8−6−4−22

Answer: Degree 4
x = −2, multiplicity 1; x = 3, multiplicity 2; x = 6, multiplicity 1
As x → −∞, y → ∞
As x → ∞, y → ∞
Local maximum at (3, 0)
There is a local minimum between x = −2 and x = 3
There is a local minimum between x = 3 and x = 6

2
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = −(17)(x + 5)²(x + 1)(x − 2)
Show Answer

a. Add the exponents.

Degree: 2 + 1 + 1 = 4

b. Set each factor equal to 0.

x = −5, multiplicity 2. The graph bounces off the x-axis.

x = −1, multiplicity 1. The graph crosses the x-axis.

x = 2, multiplicity 1. The graph crosses the x-axis.

c. Even degree and a negative leading coefficient: both ends go down.

As x → −∞, y → −∞

As x → ∞, y → −∞

d. Even multiplicities put a max or a min on the x-axis.

At x = −5 the graph is below the axis on both sides, so that bounce is a local maximum.

There is a local minimum between x = −5 and x = −1.

There is a local maximum between x = −1 and x = 2.

A sketch of that shape:

−7−6−5−4−3−2−11234−10−8−6−4−22468101214

Answer: Degree 4
x = −5, multiplicity 2; x = −1, multiplicity 1; x = 2, multiplicity 1
As x → −∞, y → −∞
As x → ∞, y → −∞
Local maximum at (−5, 0)
There is a local minimum between x = −5 and x = −1
There is a local maximum between x = −1 and x = 2

3
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (17)(x + 3)²(x − 2)(x − 4)
Show Answer

a. Add the exponents.

Degree: 2 + 1 + 1 = 4

b. Set each factor equal to 0.

x = −3, multiplicity 2. The graph bounces off the x-axis.

x = 2, multiplicity 1. The graph crosses the x-axis.

x = 4, multiplicity 1. The graph crosses the x-axis.

c. Even degree and a positive leading coefficient: both ends go up.

As x → −∞, y → ∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

At x = −3 the graph is above the axis on both sides, so that bounce is a local minimum.

There is a local maximum between x = −3 and x = 2.

There is a local minimum between x = 2 and x = 4.

A sketch of that shape:

−5−4−3−2−1123456−10−8−6−4−22468101214

Answer: Degree 4
x = −3, multiplicity 2; x = 2, multiplicity 1; x = 4, multiplicity 1
As x → −∞, y → ∞
As x → ∞, y → ∞
Local minimum at (−3, 0)
There is a local maximum between x = −3 and x = 2
There is a local minimum between x = 2 and x = 4

4
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = −(12)(x + 1)(x − 2)²(x − 5)
Show Answer

a. Add the exponents.

Degree: 1 + 2 + 1 = 4

b. Set each factor equal to 0.

x = −1, multiplicity 1. The graph crosses the x-axis.

x = 2, multiplicity 2. The graph bounces off the x-axis.

x = 5, multiplicity 1. The graph crosses the x-axis.

c. Even degree and a negative leading coefficient: both ends go down.

As x → −∞, y → −∞

As x → ∞, y → −∞

d. Even multiplicities put a max or a min on the x-axis.

At x = 2 the graph is above the axis on both sides, so that bounce is a local minimum.

There is a local maximum between x = −1 and x = 2.

There is a local maximum between x = 2 and x = 5.

A sketch of that shape:

−3−2−11234567−224681012

Answer: Degree 4
x = −1, multiplicity 1; x = 2, multiplicity 2; x = 5, multiplicity 1
As x → −∞, y → −∞
As x → ∞, y → −∞
Local minimum at (2, 0)
There is a local maximum between x = −1 and x = 2
There is a local maximum between x = 2 and x = 5

5
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (17)(x + 6)²(x + 2)(x − 1)
Show Answer

a. Add the exponents.

Degree: 2 + 1 + 1 = 4

b. Set each factor equal to 0.

x = −6, multiplicity 2. The graph bounces off the x-axis.

x = −2, multiplicity 1. The graph crosses the x-axis.

x = 1, multiplicity 1. The graph crosses the x-axis.

c. Even degree and a positive leading coefficient: both ends go up.

As x → −∞, y → ∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

At x = −6 the graph is above the axis on both sides, so that bounce is a local minimum.

There is a local maximum between x = −6 and x = −2.

There is a local minimum between x = −2 and x = 1.

A sketch of that shape:

−8−7−6−5−4−3−2−1123−14−12−10−8−6−4−2246810

Answer: Degree 4
x = −6, multiplicity 2; x = −2, multiplicity 1; x = 1, multiplicity 1
As x → −∞, y → ∞
As x → ∞, y → ∞
Local minimum at (−6, 0)
There is a local maximum between x = −6 and x = −2
There is a local minimum between x = −2 and x = 1

Level 3

Hard

Two bounces, or a fraction in front.

Example
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (1113)(x + 4)²(x − 1)²(x − 6)

a. Add the exponents.

Degree: 2 + 2 + 1 = 5

b. Set each factor equal to 0.

x = −4, multiplicity 2. The graph bounces off the x-axis.

x = 1, multiplicity 2. The graph bounces off the x-axis.

x = 6, multiplicity 1. The graph crosses the x-axis.

c. Odd degree and a positive leading coefficient: the left end goes down and the right end goes up.

As x → −∞, y → −∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

At x = −4 the graph is below the axis on both sides, so that bounce is a local maximum.

At x = 1 the graph is below the axis on both sides, so that bounce is a local maximum.

There is a local minimum between x = −4 and x = 1.

There is a local minimum between x = 1 and x = 6.

A sketch of that shape:

−6−4−22468−14−12−10−8−6−4−22

Answer: Degree 5
x = −4, multiplicity 2; x = 1, multiplicity 2; x = 6, multiplicity 1
As x → −∞, y → −∞
As x → ∞, y → ∞
Local maximum at (−4, 0)
Local maximum at (1, 0)
There is a local minimum between x = −4 and x = 1
There is a local minimum between x = 1 and x = 6

1
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (172)(x + 5)²(x − 2)²(x − 4)
Show Answer

a. Add the exponents.

Degree: 2 + 2 + 1 = 5

b. Set each factor equal to 0.

x = −5, multiplicity 2. The graph bounces off the x-axis.

x = 2, multiplicity 2. The graph bounces off the x-axis.

x = 4, multiplicity 1. The graph crosses the x-axis.

c. Odd degree and a positive leading coefficient: the left end goes down and the right end goes up.

As x → −∞, y → −∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

At x = −5 the graph is below the axis on both sides, so that bounce is a local maximum.

At x = 2 the graph is below the axis on both sides, so that bounce is a local maximum.

There is a local minimum between x = −5 and x = 2.

There is a local minimum between x = 2 and x = 4.

A sketch of that shape:

−6−4−2246−14−12−10−8−6−4−22

Answer: Degree 5
x = −5, multiplicity 2; x = 2, multiplicity 2; x = 4, multiplicity 1
As x → −∞, y → −∞
As x → ∞, y → ∞
Local maximum at (−5, 0)
Local maximum at (2, 0)
There is a local minimum between x = −5 and x = 2
There is a local minimum between x = 2 and x = 4

2
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = −(114)(x + 3)²(x − 1)(x − 5)
Show Answer

a. Add the exponents.

Degree: 2 + 1 + 1 = 4

b. Set each factor equal to 0.

x = −3, multiplicity 2. The graph bounces off the x-axis.

x = 1, multiplicity 1. The graph crosses the x-axis.

x = 5, multiplicity 1. The graph crosses the x-axis.

c. Even degree and a negative leading coefficient: both ends go down.

As x → −∞, y → −∞

As x → ∞, y → −∞

d. Even multiplicities put a max or a min on the x-axis.

At x = −3 the graph is below the axis on both sides, so that bounce is a local maximum.

There is a local minimum between x = −3 and x = 1.

There is a local maximum between x = 1 and x = 5.

A sketch of that shape:

−5−4−3−2−11234567−8−6−4−2246810121416

Answer: Degree 4
x = −3, multiplicity 2; x = 1, multiplicity 1; x = 5, multiplicity 1
As x → −∞, y → −∞
As x → ∞, y → −∞
Local maximum at (−3, 0)
There is a local minimum between x = −3 and x = 1
There is a local maximum between x = 1 and x = 5

3
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (14)(x + 2)²(x − 3)²
Show Answer

a. Add the exponents.

Degree: 2 + 2 = 4

b. Set each factor equal to 0.

x = −2, multiplicity 2. The graph bounces off the x-axis.

x = 3, multiplicity 2. The graph bounces off the x-axis.

c. Even degree and a positive leading coefficient: both ends go up.

As x → −∞, y → ∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

At x = −2 the graph is above the axis on both sides, so that bounce is a local minimum.

At x = 3 the graph is above the axis on both sides, so that bounce is a local minimum.

There is a local maximum between x = −2 and x = 3.

A sketch of that shape:

−4−3−2−112345−224681012

Answer: Degree 4
x = −2, multiplicity 2; x = 3, multiplicity 2
As x → −∞, y → ∞
As x → ∞, y → ∞
Local minimum at (−2, 0)
Local minimum at (3, 0)
There is a local maximum between x = −2 and x = 3

4
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = −(119)(x + 6)²(x + 1)²(x − 2)
Show Answer

a. Add the exponents.

Degree: 2 + 2 + 1 = 5

b. Set each factor equal to 0.

x = −6, multiplicity 2. The graph bounces off the x-axis.

x = −1, multiplicity 2. The graph bounces off the x-axis.

x = 2, multiplicity 1. The graph crosses the x-axis.

c. Odd degree and a negative leading coefficient: the left end goes up and the right end goes down.

As x → −∞, y → ∞

As x → ∞, y → −∞

d. Even multiplicities put a max or a min on the x-axis.

At x = −6 the graph is above the axis on both sides, so that bounce is a local minimum.

At x = −1 the graph is above the axis on both sides, so that bounce is a local minimum.

There is a local maximum between x = −6 and x = −1.

There is a local maximum between x = −1 and x = 2.

A sketch of that shape:

−8−7−6−5−4−3−2−11234−22468101214

Answer: Degree 5
x = −6, multiplicity 2; x = −1, multiplicity 2; x = 2, multiplicity 1
As x → −∞, y → ∞
As x → ∞, y → −∞
Local minimum at (−6, 0)
Local minimum at (−1, 0)
There is a local maximum between x = −6 and x = −1
There is a local maximum between x = −1 and x = 2

5
Sketch the graph. Give the degree, the x-intercepts and their multiplicities, the end behavior, and name each maximum or minimum you can locate.
y = (16)(x + 3)³(x − 2)
Show Answer

a. Add the exponents.

Degree: 3 + 1 = 4

b. Set each factor equal to 0.

x = −3, multiplicity 3. The graph flattens and still crosses the x-axis.

x = 2, multiplicity 1. The graph crosses the x-axis.

c. Even degree and a positive leading coefficient: both ends go up.

As x → −∞, y → ∞

As x → ∞, y → ∞

d. Even multiplicities put a max or a min on the x-axis.

There is a local minimum between x = −3 and x = 2.

A sketch of that shape:

−5−4−3−2−11234−14−12−10−8−6−4−22

Answer: Degree 4
x = −3, multiplicity 3; x = 2, multiplicity 1
As x → −∞, y → ∞
As x → ∞, y → ∞
There is a local minimum between x = −3 and x = 2

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