Algebra 3 Quiz: Polynomial Graph Behavior

12 questions · exam conditions

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Polynomial Graph BehaviorQuestion 1 of 12

The only real zeros of $p$ are $-3,\ 0,$ and $4$ , with multiplicities $1,\ 2,$ and $1$ , respectively. If $p(1)>0$ , which statement about the graph is true?

The leading coefficient is positive, so

p(x)\to\infty

x\to-\infty

and as

x\to\infty

The leading coefficient is negative, so

p(x)\to\infty

x\to-\infty

and

p(x)\to-\infty

x\to\infty

The leading coefficient is positive, so

p(x)\to-\infty

x\to-\infty

and

p(x)\to\infty

x\to\infty

The leading coefficient is negative, so

p(x)\to-\infty

x\to-\infty

and as

x\to\infty

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Algebra 3 Quiz

Algebra 3 Quiz: Polynomial Graph Behavior

Practice Polynomial Graph Behavior in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Polynomial Graph Behavior, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The only real zeros of $p$ are $-3,\ 0,$ and $4$ , with multiplicities $1,\ 2,$ and $1$ , respectively. If $p(1)>0$ , which statement about the graph is true?

The leading coefficient is positive, so $p(x)\to\infty$ as $x\to-\infty$ and as $x\to\infty$ .
The leading coefficient is negative, so $p(x)\to\infty$ as $x\to-\infty$ and $p(x)\to-\infty$ as $x\to\infty$ .
The leading coefficient is positive, so $p(x)\to-\infty$ as $x\to-\infty$ and $p(x)\to\infty$ as $x\to\infty$ .
The leading coefficient is negative, so $p(x)\to-\infty$ as $x\to-\infty$ and as $x\to\infty$ . (correct answer)

Explanation: Whenever you see zeros, multiplicities, and a test point like

p(1)>0

, your first move is to write the polynomial in factored form:

p(x)=a(x+3)x^2(x-4)

The multiplicities become the exponents, and the degree is

1+2+1=4

, which is even. Now use the given sign condition:

p(1)=a(4)(1^2)(-3)=-12a>0

a<0

. That means the leading coefficient is negative. Since the degree is even, a negative leading coefficient sends both ends of the graph downward:

p(x)\to-\infty

x\to-\infty

and as

x\to\infty

.
This matches the statement that the leading coefficient is negative and both ends go to

-\infty

. The choices claiming the leading coefficient is positive contradict

p(1)>0

, so they cannot be true. The choices that describe the two ends going in opposite directions — whether positive leading coefficient with left end going down and right end going up, or negative leading coefficient with left end going up and right end going down — are actually describing odd-degree end behavior. This polynomial has degree

4

, so both ends must go the same direction. On this exam, remember the pairing: even degree means same-direction ends, odd degree means opposite ends. Then use a test point to decide whether both ends go up or down.

Question 2

A polynomial graph touches the x-axis at $x=-1$ , crosses at $x=2$ and $x=4$ , and has $f(x)\to-\infty$ as $x\to-\infty$ and as $x\to\infty$ . What is the minimum possible degree, and what is the sign of the leading coefficient?

Degree $3$ ; leading coefficient negative.
Degree $4$ ; leading coefficient negative. (correct answer)
Degree $5$ ; leading coefficient negative.
Degree $4$ ; leading coefficient positive.

Explanation: When you see a question about x-intercepts and end behavior, translate the graph into multiplicities. A graph that touches the x-axis comes from a factor with even multiplicity; a graph that crosses comes from a factor with odd multiplicity. Here

x=-1

touches, so that factor contributes at least multiplicity

2

. The two crossings at

x=2

and

x=4

each contribute at least multiplicity

1

. Adding the minimum multiplicities gives degree

2+1+1=4

. For the leading coefficient, look at end behavior: since

f(x)\to-\infty

x\to-\infty

and as

x\to\infty

, the graph falls on both ends. That requires an even-degree polynomial with a negative leading coefficient. So the correct combination is degree

4

with leading coefficient negative. The choice saying degree

3

, leading coefficient negative fails because an odd-degree polynomial with a negative leading coefficient rises on the left and falls on the right—it cannot go down on both sides. It also misses the even multiplicity from the touch. The choice saying degree

5

, leading coefficient negative has the same end-behavior problem: odd degree means the two ends go in opposite directions, and it is not minimal. The choice saying degree

4

, leading coefficient positive gets the degree right but the sign wrong; positive leading coefficient would make both ends go up to

+\infty

, not

-\infty

.When you see a question about x-intercepts and end behavior, translate the graph into multiplicities. A graph that touches the x-axis comes from a factor with even multiplicity; a graph that crosses comes from a factor with odd multiplicity. Here

x=-1

touches, so that factor contributes at least multiplicity

2

. The two crossings at

x=2

and

x=4

each contribute at least multiplicity

1

. Adding the minimum multiplicities gives degree

2+1+1=4

. For the leading coefficient, look at end behavior: since

f(x)\to-\infty

x\to-\infty

and as

x\to\infty

, the graph falls on both ends. That requires an even-degree polynomial with a negative leading coefficient. So the correct combination is degree

4

with leading coefficient negative. The choice saying degree

3

5

, leading coefficient negative has the same end-behavior problem: odd degree means the two ends go in opposite directions,and it is not minimal. The choice saying degree

4

, leading coefficient positive gets the degree right but the sign wrong; positive leading coefficient would make both ends go up to

+\infty

, not

-\infty

Question 3

Let $g(x)=-x(x-4)^3(x+1)^4$ . Which statement correctly describes the graph?

The graph rises to the left, falls to the right, crosses at $x=-1$ and $x=0$ , touches at $x=4$ , and is positive on $(-\infty,0)\cup(4,\infty)$ .
The graph falls to the left and right, crosses at $x=-1$ , touches at $x=0$ and $x=4$ , and is negative on $(0,4)$ .
The graph falls to the left and right, touches at $x=-1$ , crosses at $x=0$ and $x=4$ , and is positive on $(0,4)$ . (correct answer)
The graph falls to the left and right, touches at $x=-1$ and $x=4$ , crosses at $x=0$ , and is negative on $(0,4)$ .

Explanation: Whenever you see a polynomial already factored, start by finding the degree, the leading coefficient, and the multiplicity of each zero. Here

g(x)=-x(x-4)^3(x+1)^4

has degree

1+3+4=8

and leading coefficient

-1

, so the graph falls to the left and falls to the right. Next, look at each zero:

x=0

has multiplicity

1

(odd), so it crosses;

x=4

has multiplicity

3

(odd), so it also crosses;

x=-1

has multiplicity

4

(even), so it only touches. That already identifies the correct description: falls both ways, touches at

-1

, crosses at

0

and

4

. To confirm sign on

(0,4)

, test

x=1

-1(1-4)^3(2)^4

is positive—so yes. In fact,

(0,4)

is the only positive interval. The distractor saying "rises to the left" has end behavior backwards, treats

x=4

as a touch instead of a cross, and gives the wrong sign interval. The distractor saying "crosses at

x=-1

" and "touches at

0

and

4

" swaps the odd/even multiplicities, making

x=4

touch and sign negative. The distractor saying "touches at

-1

and

4

" again mistakes the odd multiplicity at

4

, and it also claims the graph is negative on

(0,4)

, which fails the test value. Study takeaway: for each zero, ask "odd multiplicity?" If yes, cross;if even, touch. Then determine end behavior from degree and leading coefficient. For intervals, plug in one test point per region—don't rely on memory.

Question 4

Let $m(x)=(x-2)^2(x+3)(x-5)$ . Which statement about the graph of $m$ is true?

The graph falls to the left, rises to the right, and is below the x-axis on $(-\infty,-3)\cup(5,\infty)$ .
The graph falls to the left and right, and is below the x-axis on $(-\infty,-3)\cup(5,\infty)$ .
The graph rises to the left and right, and is above the x-axis on $(-3,2)\cup(5,\infty)$ .
The graph rises to the left and right, and is below the x-axis on $(-3,2)\cup(2,5)$ . (correct answer)

Explanation: When you see a polynomial graph question, focus on two things: the leading term for end behavior, and each factor's multiplicity for how the graph behaves at its x-intercepts. The leading term of

m(x)

is essentially

x^2 \cdot x \cdot x = x^4

, a positive even-degree term, so the graph rises on both the left and the right. The roots are

x=-3

x=2

, and

x=5

. The factor

(x-2)^2

has even multiplicity, so the graph touches the x-axis at

2

and does not change sign there. Testing intervals: on

(-3,2)

, choose

0

m(0)=(4)(3)(-5)<0

. On

(2,5)

, choose

3

m(3)=(1)(6)(-2)<0

. So the graph is below the x-axis on

(-3,2)\cup(2,5)

, and rises to the left and right. That makes the statement "rises to the left and right, and is below the x-axis on

(-3,2)\cup(2,5)

" correct. The choice saying "falls to the left, rises to the right" treats the degree as odd, and its intervals

(-\infty,-3)\cup(5,\infty)

are actually where the graph is above, not below. The choice saying "falls to the left and right" treats the leading coefficient as negative; its interval choice is also above the axis. The choice saying "rises to the left and right, and is above the x-axis on

(-3,2)\cup(5,\infty)

" gets end behavior right but misreads the sign on

(-3,2)

and forgets the graph is below there. A quick strategy: always test one number in each interval between roots, and remember that even-multiplicity roots cause a bounce, not a sign change.

Question 5

A polynomial's graph falls to the left, rises to the right, crosses the x-axis at $x=-2$ , touches the x-axis at $x=5$ , and has y-intercept $50$ . Which equation could define this polynomial?

$p(x)=(x+2)^2(x-5)$
$p(x)=-(x+2)(x-5)^2$
$p(x)=2(x+2)(x-5)^2$
$p(x)=(x+2)(x-5)^2$ (correct answer)

Explanation: When you see a polynomial's end behavior, crossing, and touching, think about multiplicity: an odd exponent means the graph crosses, an even exponent means it touches. Here, falling left and rising right means an odd degree with a positive leading coefficient. Crossing at

x=-2

requires a single factor

(x+2)

, not squared. Touching at

x=5

requires

(x-5)^2

. So the skeleton is

a(x+2)(x-5)^2

, with

a>0

. The y-intercept is

p(0)=a(2)(25)=50a

, and since the intercept is

50

a=1

. Thus the polynomial is

(x+2)(x-5)^2

. The choice

(x+2)^2(x-5)

reverses the multiplicity roles: it would touch at

-2

and cross at

5

, opposite to the description, and its y-intercept is

-20

. The choice

-(x+2)(x-5)^2

has the correct crossing and touching pattern, but its negative leading coefficient makes it fall right and rise left, and its y-intercept is

-50

. The choice

2(x+2)(x-5)^2

also has correct crossing/touching and positive end behavior, but its y-intercept is

2\cdot 50=100

, not

50

. Your takeaway: when graphing polynomials, convert "crosses" to an odd-powered factor and "touches" to an even-powered factor, then use the y-intercept to solve for the leading coefficient. This prevents all three common traps: wrong multiplicity, wrong sign, and wrong stretch factor.

Question 6

Let $p(x)=(2x-1)^3(3x-x^2)^2(x^3+2)$ . Which statement about the graph of $p$ is true?

The degree is $10$ , and $p(x)\to\infty$ as $x\to-\infty$ and as $x\to\infty$ . (correct answer)
The degree is $8$ , and $p(x)\to-\infty$ as $x\to-\infty$ and as $x\to\infty$ .
The degree is $10$ , and $p(x)\to\infty$ as $x\to-\infty$ but $p(x)\to-\infty$ as $x\to\infty$ .
The degree is $8$ , and $p(x)\to-\infty$ as $x\to-\infty$ but $p(x)\to\infty$ as $x\to\infty$ .

Explanation: Whenever you see a polynomial in factored form, find the degree by adding the degrees of each factor, and determine end behavior using the degree's parity and the sign of the leading coefficient. Here,

(2x-1)^3

contributes degree

3

(3x-x^2)^2

contributes degree

4

because the inside has degree

2

and is squared, and

(x^3+2)

contributes degree

3

. So the total degree is

3+4+3=10

. The leading coefficient is positive: from

(2x-1)^3

it's

2^3=8

, from

(3x-x^2)^2

the leading term is

x^4

, and from

(x^3+2)

it's

x^3

. Since the degree is even and the leading coefficient is positive,

p(x)\to\infty

x\to-\infty

and as

x\to\infty

. That matches the statement: degree

10

, both ends going to

+\infty

. The choices saying degree

8

come from miscounting — perhaps treating

(3x-x^2)^2

as degree

2

or forgetting the

(x^3+2)

factor entirely. The choice saying degree

10

but

p(x)\to-\infty

x\to\infty

confuses the sign: the squared factor is always nonnegative, and its leading term is positive, so the overall leading coefficient is not negative. End behavior is controlled only by the highest-degree term, not by the lower-degree details. Study tip: for end behavior, always compute leading coefficient and degree from the factored form first — then apply: even degree with positive leading coefficient means both ends rise.

Question 7

Let $r(x)=(2-x)(x+1)^2(x-4)^2$ . Which description matches the graph of $r$ ?

As $x\to-\infty,\ r(x)\to-\infty$ ; as $x\to\infty,\ r(x)\to\infty$ ; the graph touches at $-1$ and $4$ and crosses at $2$ .
Because the degree is odd, $r(x)\to-\infty$ at both ends; the graph crosses at $2$ .
As $x\to-\infty,\ r(x)\to\infty$ ; as $x\to\infty,\ r(x)\to-\infty$ ; the graph touches at $-1$ and $4$ and crosses at $2$ . (correct answer)
As $x\to-\infty,\ r(x)\to\infty$ ; as $x\to\infty,\ r(x)\to-\infty$ ; the graph crosses at $-1$ and $4$ and touches at $2$ .

Explanation: Whenever you see a factored polynomial, your first moves should be to identify the degree, the sign of the leading coefficient, and each zero's multiplicity. Those three facts determine end behavior and whether the graph crosses or touches. Here,

r(x)=(2-x)(x+1)^2(x-4)^2

. The factor

(2-x)

is degree 1 with leading coefficient

-1

, so the overall leading term is

-x^5

. That means the degree is odd and the leading coefficient is negative: as

x\to-\infty

r(x)\to+\infty

, and as

x\to\infty

r(x)\to-\infty

. For the zeros,

x=-1

and

x=4

each come from squared factors, so their multiplicity is even and the graph touches the x-axis there. The zero at

x=2

comes from the factor

(2-x)

, multiplicity 1, so the graph crosses. Thus the correct description is: as

x\to-\infty,\ r(x)\to\infty

; as

x\to\infty,\ r(x)\to-\infty

; the graph touches at

-1

and

4

and crosses at

2

. The choice that claims the opposite end behavior, with negative left and positive right, misses the negative leading coefficient. The choice saying an odd degree means both ends go to

-\infty

confuses "odd degree" with the sign of the leading coefficient; odd degree gives opposite ends, not identical ends. The choice with correct end behavior but "crosses at

-1

and

4

and touches at

2

" reverses the multiplicity rule: even multiplicity touches, odd multiplicity crosses. On exam day, write the sign of the leading term and list each zero with its multiplicity before reading the answer choices. That single step prevents most graph-matching mistakes.

Question 8

A 6th-degree polynomial $p$ has a negative leading coefficient and real zeros at $-2,\ 0,\ 3,$ and $5$ . Which statement must be true?

$p(x)\to-\infty$ as $x\to-\infty$ and as $x\to\infty$ . (correct answer)
The graph has exactly five turning points.
The graph crosses the x-axis at all four given zeros.
$p(x)<0$ on the interval $(0,3)$ .

Explanation: When you see a polynomial with a specified degree and leading coefficient, first isolate what those two facts alone determine: end behavior. A 6th-degree polynomial is even-degree, so both ends go in the same direction. With a negative leading coefficient, both ends must point downward:

p(x)\to -\infty

x\to\pm\infty

. That is the statement that must be true. The given zeros matter for

x

-intercepts, but they do not control the ends. The "exactly five turning points" choice is a trap: a degree 6 polynomial can have at most

6-1=5

turning points, but multiplicities or non-real zeros can make fewer. Similarly, "crosses the

x

-axis at all four given zeros" is not guaranteed, because a zero with even multiplicity makes the graph touch the axis and bounce, not cross. Finally,

p(x)<0

(0,3)

assumes a particular sign pattern; but if the zeros at 0 or 3 have even multiplicity, the sign on that interval can flip. For example,

p(x)=-(x+2)x^2(x-3)^2(x-5)

is degree 6 with the stated zeros and a negative leading coefficient, yet it is positive on

(0,3)

. Study tip: for end-behavior questions, ask only two things — is the degree even or odd, and is the leading coefficient positive or negative? Those determine the ends; zeros and multiplicities shape the middle.

Question 9

Let $h(x)=-(x-2)(x+3)^2(x+1)(x-5)^2$ . Which statement about the graph of $h$ is true?

The graph falls to the left and right; it crosses at $x=-1$ and $x=2$ and touches at $x=-3$ and $x=5$ . (correct answer)
The graph rises to the left and falls to the right; it crosses at $x=-3$ and $x=5$ and touches at $x=-1$ and $x=2$ .
The graph falls to the left and right; it crosses at $x=-3$ and $x=5$ and touches at $x=-1$ and $x=2$ .
The graph rises to the left and right; it crosses at $x=-1$ and $x=2$ and touches at $x=-3$ and $x=5$ .

Explanation: Whenever you analyze a polynomial graph, let two ideas guide you: the sign of the leading coefficient and degree determine end behavior, and the multiplicity of each factor determines whether the graph crosses or touches at its zero. Here

h(x)

has total degree

1+2+1+2=6

, and the leading coefficient is negative because of the minus sign in front. An even-degree polynomial with a negative leading coefficient falls on both ends. Now inspect zeros:

x=-1

comes from

(x+1)

with multiplicity 1, so the graph crosses;

x=2

comes from

(x-2)

with multiplicity 1, so it crosses. But

x=-3

comes from

(x+3)^2

and

x=5

comes from

(x-5)^2

, both multiplicity 2, so the graph touches at those points instead of crossing. Thus the true statement is: the graph falls to the left and right; it crosses at

-1

and

2

and touches at

-3

and

5

. Each wrong choice misses one piece. The choice that says rises left and falls right mistakes the end behavior as if the leading coefficient were negative with odd degree, and it also swaps which zeros cross and touch. The choice that says falls left/right but crosses at

-3,5

and touches at

-1,2

gets end behavior right but assigns the wrong behavior to each zero. The choice that says rises to the left and right correctly identifies the crossing/touching zeros but ignores the negative leading coefficient, which forces both ends to fall. By checking degree, leading sign, and multiplicity, you can eliminate all of these traps.

Question 10

Let $p(x)=(x+1)^2(2x-3)(x^2+4)$ . Which statement about the graph of $p$ is true?

The graph crosses the x-axis at $x=-1$ , and $p(x)>0$ for $x>\frac{3}{2}$ .
The graph touches but does not cross the x-axis at $x=-1$ , and $p(x)>0$ for $x>\frac{3}{2}$ . (correct answer)
The graph has real zeros at $x=-1$ and $x=\frac{3}{2}$ , and the y-intercept is $12$ .
The graph has y-intercept $-12$ , and it touches the x-axis at $x=\frac{3}{2}$ .

Explanation: When you see a factored polynomial, your first moves should be: identify each real zero, note its multiplicity, and test the sign on one side of each zero. Multiplicity tells you the graph's behavior: odd multiplicity means the graph crosses the x-axis; even multiplicity means it touches and turns around. Here,

x=-1

comes from

(x+1)^2

, so it has multiplicity 2 — the graph touches but does not cross at

x=-1

. The factor

2x-3

gives

x=\frac{3}{2}

with multiplicity 1, so the graph crosses there. Meanwhile,

x^2+4

has no real zeros, and it is always positive. For

x>\frac{3}{2}

, both

(x+1)^2

and

x^2+4

are positive, and

2x-3>0

, so

p(x)>0

. That confirms the correct statement: touches at

x=-1

and positive to the right of

\frac{3}{2}

. Now the traps. The choice saying the graph crosses at

x=-1

gets the multiplicity wrong — crossing requires an odd multiplicity. The choice claiming real zeros at

-1

and

\frac{3}{2}

with y-intercept

12

misses the y-intercept calculation:

p(0)=(1)^2(-3)(4)=-12

, not

12

. Finally, the choice with y-intercept

-12

says it touches at

x=\frac{3}{2}

, but that zero has multiplicity 1, so the graph crosses there, not touches. Study tip: for every factored polynomial, list each factor, its zero, and its multiplicity before answering anything about the graph. Then sign-test one interval to anchor the whole picture.

Question 11

Suppose a polynomial $p$ has only the real zeros $-4,\ 2,\ 7$ . If $p(x)>0$ on $(-\infty,-4)$ , $p(x)<0$ on $(-4,2)$ , $p(x)>0$ on $(2,7)$ , and $p(x)<0$ on $(7,\infty)$ , which conclusion follows?

$p$ has odd degree and a positive leading coefficient, and the zero at $2$ has even multiplicity.
$p$ has even degree and a positive leading coefficient, and all three zeros have odd multiplicity.
$p$ has even degree and a negative leading coefficient, and the zeros at $-4$ and $7$ have even multiplicity.
$p$ has odd degree and a negative leading coefficient, and all three zeros have odd multiplicity. (correct answer)

Explanation: Whenever you see a sign chart for a polynomial, translate it into two things: crossings and end behavior. A sign change at a zero means the factor has odd multiplicity; no sign change would mean even multiplicity. Here the signs go

+,-,+,-

across

-4,2,7

, so every zero is a crossing: all three zeros have odd multiplicity. Next look far left and far right. The chart gives

p(x)>0

x\to-\infty

and

p(x)<0

x\to\infty

. Since the right tail is negative, the leading coefficient must be negative. If the degree were even, anegative leading coefficient would make both ends negative; but the left tail is positive. If the degree were odd, anegative leading coefficient makes left positive and right negative, exactly what is given. Thus degree odd, leading coefficient negative, all three zeros odd multiplicity— that matches the choice stating odd degree, negative leading coefficient, and all three zeros odd multiplicity. The other choices fail: "odd degree and a positive leading coefficient" would make the right tail positive, not negative, and at

2

the sign changes from negative to positive, so that zero cannot have even multiplicity."Even degree and a positive leading coefficient" would make both ends positive, contradicted by the negative right tail."Even degree and a negative leading coefficient" would make both ends negative, contradicted by the positive left tail; also the zeros at

-4

and

7

change sign, so they must have odd, not even, multiplicity. Remember the pattern: sign change = odd multiplicity; no sign change = even multiplicity. For end behavior, opposite end signs mean odd degree, and the right tail sign tells you the leading coefficient sign.

Question 12

Let $f(x)=-2(x+3)(x-1)^2(x-5)$ . Which statement correctly describes the graph?

The graph rises to the left, falls to the right, and is positive on $(-3,1)\cup(1,5)$ .
The graph rises to the left and right, and is negative on $(-\infty,-3)\cup(5,\infty)$ .
The graph falls to the left and right, and is positive on $(-3,1)\cup(1,5)$ . (correct answer)
The graph falls to the left and right, and is negative on $(-3,1)\cup(1,5)$ .

Explanation: Whenever you see a polynomial like this, start by identifying the degree and leading coefficient: they control end behavior, while zeros control sign intervals. Here, the degree is

4

because the factors contribute

1+2+1

, and the leading coefficient is

-2

. A negative leading coefficient with even degree means the graph falls to the left and falls to the right, so any description saying it rises on either side is out. The zeros are

-3

1

, and

5

, with

1

coming from the squared factor

(x-1)^2

. Testing intervals,

f(x)

is negative on

(-\infty,-3)

, positive on

(-3,1)

, positive on

(1,5)

, and negative on

(5,\infty)

. The squared factor never changes sign, so the graph touches the

x

-axis at

x=1

and stays positive across that point. Thus the correct description is: falls to the left and right, and is positive on

(-3,1)\cup(1,5)

. The "rises to the left, falls to the right" option confuses a negative leading coefficient with an odd-degree graph. The "rises to the left and right" option uses the correct even degree but ignores that the leading coefficient is negative. The "falls to the left and right, and is negative on

(-3,1)\cup(1,5)

" option gets end behavior right but flips the sign: test

x=0

, giving

f(0)=-2(3)(1)(-5)=30>0

, so that interval is positive. Study tip: always test one point in each interval, and check multiplicity to know whether the sign actually changes at a zero.