Algebra 3 Quiz: Graphing Rational Functions

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Graphing Rational FunctionsQuestion 1 of 12

Let $f(x)=\frac{x^2-x-6}{x-2}$ . Which statement correctly describes the graph's asymptotes and end behavior?

There is a vertical asymptote at

x=2

, and as

x\to\pm\infty

the graph approaches the horizontal line

y=1

There is a vertical asymptote at

x=2

, and as

x\to\pm\infty

the graph approaches the line

y=x+1

There is a vertical asymptote at

x=2

, and as

x\to\pm\infty

the graph approaches the line

y=x-1

There are vertical asymptotes at

x=-2

and

x=3

, and as

x\to\pm\infty

the graph approaches the line

y=x+1

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Algebra 3 Quiz

Algebra 3 Quiz: Graphing Rational Functions

Practice Graphing Rational Functions 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 Graphing Rational Functions, 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

Let $f(x)=\frac{x^2-x-6}{x-2}$ . Which statement correctly describes the graph's asymptotes and end behavior?

There is a vertical asymptote at $x=2$ , and as $x\to\pm\infty$ the graph approaches the horizontal line $y=1$ .
There is a vertical asymptote at $x=2$ , and as $x\to\pm\infty$ the graph approaches the line $y=x+1$ . (correct answer)
There is a vertical asymptote at $x=2$ , and as $x\to\pm\infty$ the graph approaches the line $y=x-1$ .
There are vertical asymptotes at $x=-2$ and $x=3$ , and as $x\to\pm\infty$ the graph approaches the line $y=x+1$ .

Explanation: When you see a rational function, your first move is to factor. Here,

f(x)=\frac{x^2-x-6}{x-2}=\frac{(x-3)(x+2)}{x-2}

. The denominator is zero at

x=2

, and the numerator is not zero there, so there is a vertical asymptote at

x=2

. Now compare degrees: the numerator has degree 2 and the denominator degree 1, so there is no horizontal asymptote; instead, there is a slant asymptote. Divide to find it:

x^2-x-6 \div (x-2) = x+1

with remainder

-4

. Thus

f(x)=x+1-\frac{4}{x-2}

. As

x\to\pm\infty

, the fraction vanishes, so the graph approaches the line

y=x+1

. That matches the statement with a vertical asymptote at

x=2

and end behavior toward

y=x+1

. What about the other choices? The one claiming a horizontal line

y=1

confuses this with the equal-degree case, where you'd use leading coefficients; here the numerator's degree is higher by one. The choice

y=x-1

is a division mistake — the quotient is

x+1

, not

x-1

. And the choice listing vertical asymptotes at

x=-2

and

x=3

confuses zeros of the numerator with zeros of the denominator: those are

x

-intercepts, not vertical asymptotes. For your exam, remember the procedure: factor, check denominator zeros for vertical asymptotes, then compare degrees to decide between horizontal and slant asymptotes. This will keep you from falling for those classic traps.

Question 2

Which statement correctly lists the x-intercepts and end behavior of $f(x)=\frac{2x^2-3x-5}{x^2+1}$ ?

x-intercepts at $x=1$ and $x=-\frac{5}{2}$ ; as $x\to\pm\infty$ , $f(x)\to2$
x-intercepts at $x=-1$ and $x=\frac{5}{2}$ ; as $x\to\pm\infty$ , $f(x)\to2$ (correct answer)
x-intercepts at $x=-1$ and $x=\frac{5}{2}$ ; as $x\to\pm\infty$ , $f(x)\to0$
x-intercepts at $x=-1$ and $x=\frac{5}{2}$ ; as $x\to\pm\infty$ , $f(x)\to1$

Explanation: Whenever you're asked about a rational function's x-intercepts and end behavior, think two separate pieces: x-intercepts come from the numerator, and end behavior comes from the highest-degree terms. For

f(x)=\frac{2x^2-3x-5}{x^2+1}

, set the numerator equal to zero:

2x^2-3x-5=0

Factoring gives

(2x-5)(x+1)=0

, so [ x=\frac{5}{2}\quad\text{or}\quad x=-1. ] These are the only x-intercepts; the denominator

x^2+1

is always positive, so there are no vertical asymptotes to worry about. Now for end behavior, as

x\to\pm\infty

, the highest-power terms dominate: numerator behaves like

2x^2

, denominator like

x^2

, so [ f(x)\to \frac{2x^2}{x^2}=2. ] Thus the correct statement is the one listing x-intercepts at

x=-1

and

x=\frac{5}{2}

, with

f(x)\to2

x\to\pm\infty

. The choice with x-intercepts at

x=1

and

x=-\frac{5}{2}

is incorrect because neither value makes the numerator zero — plugging in gives nonzero results; that comes from a sign error in factoring. The choices with the correct intercepts but end behavior

f(x)\to0

f(x)\to1

are traps: they ignore the fact that the degrees of numerator and denominator are equal, or they misread the leading coefficients. Since the leading coefficient of the numerator is 2 and the denominator's is 1, the horizontal asymptote is

y=2

, not 0 or 1. Remember: for rational functions, x-intercepts solve numerator

=0

;end behavior compares degrees — when they are equal, the horizontal asymptote is the ratio of leading coefficients.

Question 3

Which statement about the graph of $f(x)=\frac{2x-4}{x^2-1}$ is NOT true?

The graph has vertical asymptotes at $x=-1$ and $x=1$ .
The graph has a horizontal asymptote at $y=2$ . (correct answer)
The graph has an x-intercept at $x=2$ .
The graph has a y-intercept at $y=4$ .

Explanation: When you see a rational function like this, you should think about the four classic graph features: vertical asymptotes, horizontal asymptotes, x-intercepts, and y-intercepts. Each comes from a different piece of algebra, so it helps to check them separately. The statement that is not true is the claim that the graph has a horizontal asymptote at

y=2

. For a rational function, the horizontal asymptote is determined by comparing the degrees of the numerator and denominator. Here the numerator

2x-4

has degree 1, while the denominator

x^2-1

has degree 2. Since the denominator's degree is larger, the horizontal asymptote is

y=0

, not

y=2

. A horizontal asymptote at

y=2

would occur if the degrees were equal and the leading coefficients had ratio

2/1

, but that is not the case. The other statements are true. The graph has vertical asymptotes at

x=-1

and

x=1

because those make the denominator zero while the numerator is nonzero there. It has an x-intercept at

x=2

because setting the numerator

2x-4=0

gives

x=2

. It has a y-intercept at

y=4

because

f(0)=\frac{-4}{-1}=4

. Study tip: for rational functions, remember the source of each feature — denominator zeros give vertical asymptotes, numerator zeros give x-intercepts,

f(0)

gives the y-intercept, and degree comparison gives the horizontal asymptote.

Question 4

Let $f(x)=\frac{x+3}{x-2}$ . Which statement correctly describes the graph's end behavior relative to its horizontal asymptote?

As $x\to-\infty$ , the graph approaches $y=1$ from above; as $x\to+\infty$ , it approaches $y=1$ from below.
As $x\to-\infty$ and as $x\to+\infty$ , the graph approaches $y=0$ from above.
As $x\to-\infty$ , the graph approaches $y=1$ from below; as $x\to+\infty$ , it approaches $y=1$ from above. (correct answer)
As $x\to-\infty$ and as $x\to+\infty$ , the graph approaches $y=1$ from above.

Explanation: Whenever you analyze end behavior of a rational function, first locate the horizontal asymptote. Here the numerator and denominator have equal degree, so the horizontal asymptote is

y=1

, the ratio of leading coefficients. To see whether the graph is above or below this asymptote, compare

f(x)

with

1

by subtracting:

f(x)-1=\frac{x+3}{x-2}-1=\frac{5}{x-2}.

x\to+\infty

, the denominator

x-2

is positive, so

f(x)-1>0

, meaning

f(x)>1

. The graph approaches

y=1

from above. As

x\to-\infty

, the denominator is negative, so

f(x)-1<0

, meaning

f(x)<1

. The graph approaches

y=1

from below. The reversed description — saying it approaches from above on the left and from below on the right — has the two sides swapped, because it ignores the sign of

x-2

. The answer that uses

y=0

confuses the horizontal asymptote with a zero of the function; the actual asymptote is

y=1

, not

y=0

. The answer that says both tails approach from above is only half right: it correctly handles the right side but misses that for

x<2

, the denominator is negative and flips the sign of the difference. A quick check: compute

f(x)-\text{horizontal asymptote}

. If the difference is positive, the graph is above the asymptote; if negative, it is below. Equal-degree rational functions always have horizontal asymptote equal to the ratio of leading coefficients.

Question 5

Let $f(x)=\frac{x^2+x-2}{x^2+1}$ . Which statement about the graph is true?

The graph has horizontal asymptote $y=1$ but never intersects it.
The graph has horizontal asymptote $y=0$ and x-intercepts at $x=-2$ and $x=1$ .
The graph has horizontal asymptote $y=1$ and intersects it at $x=3$ . (correct answer)
The graph has vertical asymptotes at $x=-2$ and $x=1$ and horizontal asymptote $y=1$ .

Explanation: When you see a rational function, your first moves should be: compare the degrees for the horizontal asymptote, factor numerator and denominator for intercepts and vertical asymptotes, and remember that a graph can cross a horizontal asymptote. Here, numerator and denominator both have degree 2, and both leading coefficients are 1, so the horizontal asymptote is

y=1

. To see whether the graph intersects it, set

f(x)=1

\frac{x^2+x-2}{x^2+1}=1 \Rightarrow x^2+x-2=x^2+1 \Rightarrow x-2=1 \Rightarrow x=3.

So the graph crosses

y=1

(3,1)

. That makes the statement about horizontal asymptote

y=1

and intersection at

x=3

correct. Now the wrong choices: the one saying the graph has horizontal asymptote

y=1

but never intersects it confuses a horizontal asymptote with a vertical barrier; it is only end behavior, so crossing is allowed. The statement with horizontal asymptote

y=0

and x-intercepts at

x=-2

and

x=1

correctly finds the x-intercepts by factoring the numerator:

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

, but the horizontal asymptote is not

0

because the degrees are equal. Finally, the statement claiming vertical asymptotes at

x=-2

and

x=1

mistakes the zeros of the numerator for zeros of the denominator; the denominator

x^2+1

is never zero, so this graph has no vertical asymptotes at all. Study tip: for rational functions, factor everything first, then assign roles — numerator zeros give x-intercepts, denominator zeros give vertical asymptotes, and degree comparison gives the horizontal asymptote. Then always test whether the graph crosses that horizontal asymptote by solving

f(x)=y_{\text{asymptote}}

Question 6

Let $f(x)=\frac{(x-2)(x+3)}{(x-2)^3}$ . Which statement correctly describes the graph near $x=2$ ?

$x=2$ is a vertical asymptote, with $f(x)\to-\infty$ from the left and $f(x)\to+\infty$ from the right.
$x=2$ is a vertical asymptote, with $f(x)\to+\infty$ from the left and $f(x)\to-\infty$ from the right.
$x=2$ is a vertical asymptote, and $f(x)\to+\infty$ as $x\to2$ from both sides. (correct answer)
$x=2$ is a hole, so the graph has no vertical asymptote there.

Explanation: When you see a rational function and a question about behavior near a problematic

x

-value, your first move should always be to simplify the function by canceling common factors. Here,

f(x)=\frac{(x-2)(x+3)}{(x-2)^3}=\frac{x+3}{(x-2)^2} \quad (x\neq 2).

This simplified form is the key. The factor

(x-2)

remains in the denominator, so

x=2

is a vertical asymptote, not a hole. Near

x=2

, the numerator is approximately

2+3=5

, which is positive. The denominator is

(x-2)^2

, which is always positive for

x\neq 2

, whether

x

is slightly less than or slightly greater than

2

. Thus

f(x)\to+\infty

from both sides. That makes the statement "

x=2

is a vertical asymptote, and

f(x)\to+\infty

x\to2

from both sides" correct. Why are the others wrong? The two statements claiming opposite one-sided limits —

-\infty

from one side and

+\infty

from the other — would be true if the denominator had an odd power of

(x-2)

, like

(x-2)^1

, which changes sign across the asymptote. But the squared factor prevents any sign change. The statement calling

x=2

a hole is a common trap: yes, there is a common factor

(x-2)

, but after cancellation a factor still remains in the denominator, so the graph still blows up there. Study tip: always simplify first, then check the remaining denominator. If a zero remains in the denominator, it's an asymptote; if it completely cancels, it's a hole. For the direction, look at the sign of the simplified numerator and denominator on each side.

Question 7

The graph of $f(x)=\frac{(x+1)(x-3)}{(x-2)(x+4)}$ is above the x-axis on which set of intervals?

$(-4,-1)\cup(2,3)$
$(-\infty,-4)\cup(-1,3)\cup(2,\infty)$
$(-4,-1)\cup(-1,2)\cup(3,\infty)$
$(-\infty,-4)\cup(-1,2)\cup(3,\infty)$ (correct answer)

Explanation: When you see a rational function and are asked where its graph is above the

x

-axis, you are really being asked: where is

f(x)>0

? Start by marking the zeros of the numerator and the vertical asymptotes from the denominator. Here, the numerator is zero at

x=-1

and

x=3

; the denominator is zero at

x=-4

and

x=2

. These four numbers split the number line into intervals:

(-\infty,-4)

(-4,-1)

(-1,2)

(2,3)

, and

(3,\infty)

. Test one point in each interval. For

x=-5

, all four factors are negative, so the product is positive. For

x=-2

, the numerator is positive but the denominator is negative, so

f(x)<0

. For

x=0

, both numerator and denominator are negative, so

f(x)>0

. For

x=2.5

, the numerator is negative while the denominator is positive, so

f(x)<0

. For

x=4

, all factors are positive, so

f(x)>0

. Thus the graph is above the axis exactly on

(-\infty,-4)\cup(-1,2)\cup(3,\infty)

. The choice

(-4,-1)\cup(2,3)

is the set where the graph is below the axis. The choice

(-\infty,-4)\cup(-1,3)\cup(2,\infty)

incorrectly includes the negative intervals

(2,3)

and parts of

(-1,3)

. The choice

(-4,-1)\cup(-1,2)\cup(3,\infty)

mistakenly includes the negative interval

(-4,-1)

and drops the positive interval

(-\infty,-4)

. Remember: rational function sign changes can occur at both zeros and vertical asymptotes. Test a point in every interval separated by these critical values, and don't assume a sign pattern without checking.

Question 8

Let $f(x)=3-\frac{4}{x+2}$ . Which statement correctly identifies the asymptotes and y-intercept of the graph?

vertical asymptote at $x=-2$ , horizontal asymptote at $y=3$ , y-intercept at $-1$
vertical asymptote at $x=2$ , horizontal asymptote at $y=3$ , y-intercept at $1$
vertical asymptote at $x=-2$ , horizontal asymptote at $y=-3$ , y-intercept at $1$
vertical asymptote at $x=-2$ , horizontal asymptote at $y=3$ , y-intercept at $1$ (correct answer)

Explanation: Whenever you see a rational function like

f(x)=3-\frac{4}{x+2}

, start by locating where the denominator is zero: that gives the vertical asymptote. Here,

x+2=0

means the vertical asymptote is at

x=-2

. Next, consider end behavior: as

x

becomes very large or very negative,

\frac{4}{x+2}

approaches 0, so the function approaches

3-0=3

. Therefore the horizontal asymptote is

y=3

. Finally, the y-intercept is

f(0)=3-\frac{4}{2}=3-2=1

. So the correct combination is: vertical asymptote at

x=-2

, horizontal asymptote at

y=3

, and y-intercept at

1

. Why are the other options traps? One gives the y-intercept as

-1

; this comes from incorrectly computing

3-4=-1

, forgetting that

4/(x+2)

must be evaluated at

x=0

first, giving

4/2=2

. Another gives the vertical asymptote at

x=2

; that happens when you solve

x+2=0

and mistakenly state

x=2

instead of

x=-2

. Another gives the horizontal asymptote as

y=-3

; this confuses the constant term's sign, but the expression is

3-\frac{4}{x+2}

, so the shift is up 3, not down 3. For any rational function question, find each feature independently: denominator zeros for vertical asymptotes, end behavior for horizontal asymptotes, and

f(0)

for the y-intercept. Checking them separately prevents one small sign error from sinking the whole problem.

Question 9

Let $f(x)=\frac{x^2-4}{x^3-8}$ . Which statement correctly describes the graph?

The graph has a vertical asymptote at $x=2$ , a horizontal asymptote at $y=1$ , and an x-intercept at $x=-2$ .
The graph has a hole at $x=-2$ , a vertical asymptote at $x=2$ , and a horizontal asymptote at $y=0$ .
The graph has a hole at $x=2$ , a horizontal asymptote at $y=0$ , and x-intercepts at $x=2$ and $x=-2$ .
The graph has a hole at $x=2$ , a horizontal asymptote at $y=0$ , and an x-intercept at $x=-2$ . (correct answer)

Explanation: Whenever you see a rational function like this, factor everything first. The graph's features — holes, vertical asymptotes, x-intercepts, and horizontal asymptotes — all come from the factored form. Here,

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

and

x^3-8=(x-2)(x^2+2x+4)

. So

f(x)=\frac{(x-2)(x+2)}{(x-2)(x^2+2x+4)}.

The factor

x-2

cancels, but at

x=2

the original function is undefined. That means the graph has a hole at

x=2

, not a vertical asymptote. After canceling, the simplified function is

\frac{x+2}{x^2+2x+4}.

Now analyze this simplified form: the numerator is zero at

x=-2

, so the x-intercept is at

x=-2

. Since the denominator has degree 2 and the numerator has degree 1, the horizontal asymptote is

y=0

. The choice saying the graph has a hole at

x=2

, a horizontal asymptote at

y=0

, and an x-intercept at

x=-2

is correct. The other choices each contain a trap. The one with a vertical asymptote at

x=2

and horizontal asymptote at

y=1

confuses a cancelled factor with an asymptote and misreads the degrees. The one with a hole at

x=-2

and vertical asymptote at

x=2

swaps the roles:

x=-2

is actually an intercept, and

x=2

is a hole. The one with x-intercepts at both

2

and

-2

forgets that

x=2

is not in the domain, so it cannot be an intercept. Remember: factor, cancel, then use the simplified function to find intercepts and asymptotes — but keep the original domain restrictions for holes.

Question 10

Which function could have a graph with vertical asymptotes at $x=-3$ and $x=2$ , x-intercepts at $x=-1$ and $x=5$ , and a y-intercept at $y=-\frac{5}{6}$ ?

$f(x)=\frac{(x+1)(x-5)}{(x+3)(x-2)}$
$f(x)=-\frac{(x-1)(x+5)}{(x+3)(x-2)}$
$f(x)=-\frac{(x+1)(x-5)}{(x+3)(x-2)}$ (correct answer)
$f(x)=-\frac{(x+1)(x-5)}{(x-3)(x+2)}$

Explanation: When you see a question about rational functions, the key is to connect the graph's features to the factored form. Vertical asymptotes occur where the denominator equals zero, x-intercepts come from setting the numerator to zero, and the y-intercept is found by plugging in

x=0

. The correct function must have denominator factors

(x+3)(x-2)

to give asymptotes at

x=-3

and

x=2

, and numerator factors

(x+1)(x-5)

to give x-intercepts at

x=-1

and

x=5

. That gives the base

\frac{(x+1)(x-5)}{(x+3)(x-2)}

. Now check the y-intercept: at

x=0

, this equals

\frac{(1)(-5)}{(3)(-2)}=\frac{5}{6}

. Since the required y-intercept is

-\frac{5}{6}

, you need a negative sign in front, yielding

-\frac{(x+1)(x-5)}{(x+3)(x-2)}

. Now examine the distractors. The version without the negative sign,

\frac{(x+1)(x-5)}{(x+3)(x-2)}

, has the correct asymptotes and x-intercepts but gives a positive y-intercept, so it fails the last condition. The one with

\frac{-(x-1)(x+5)}{(x+3)(x-2)}

has the correct asymptotes and y-intercept, but its x-intercepts are at

x=1

and

x=-5

, not

-1

and

5

. Finally, the one with

-\frac{(x+1)(x-5)}{(x-3)(x+2)}

has the correct numerator and y-intercept, but its denominator gives asymptotes at

x=3

and

x=-2

, opposite signs. Your strategy: always check the three features in order — asymptotes, intercepts, then y-intercept to fix the sign. This systematic approach eliminates each wrong option efficiently.

Question 11

Let $f(x)=\frac{2x^2+3x-2}{x^2-x-6}$ . Which statement about the graph of $f$ is true?

It has a vertical asymptote at $x=3$ , a hole at $x=-2$ , and a horizontal asymptote at $y=2$ . (correct answer)
It has vertical asymptotes at $x=-2$ and $x=3$ , and a horizontal asymptote at $y=2$ .
It has a vertical asymptote at $x=-3$ , a hole at $x=2$ , and a horizontal asymptote at $y=\frac{1}{2}$ .
It has a vertical asymptote at $x=-2$ , a hole at $x=3$ , and a horizontal asymptote at $y=2$ .

Explanation: Whenever you see a rational function and are asked about asymptotes and holes, your first move should always be to factor the numerator and denominator. That factoring reveals which factors cancel—those become holes—and which remain in the denominator—those become vertical asymptotes. For

f(x)=\frac{2x^2+3x-2}{x^2-x-6}

, factor to get

f(x)=\frac{(2x-1)(x+2)}{(x-3)(x+2)}.

The factor

x+2

appears in both numerator and denominator, so at

x=-2

the function has a hole, not a vertical asymptote. After canceling, the remaining denominator factor

x-3

produces a vertical asymptote at

x=3

. Since the numerator and denominator have the same degree, the horizontal asymptote is the ratio of leading coefficients:

\frac{2}{1}=2

. So the graph has a vertical asymptote at

x=3

, a hole at

x=-2

, and a horizontal asymptote at

y=2

. The other choices show classic traps. Saying there are vertical asymptotes at both

x=-2

and

x=3

treats the canceled factor as an asymptote; it is actually a removable hole. Saying the vertical asymptote is at

x=-3

, the hole at

x=2

, and the horizontal asymptote is

y=\frac12

uses zeros that do not match the factored form at all: the denominator zeros are

3

and

-2

, not

-3

and

2

. Saying there is a vertical asymptote at

x=-2

and a hole at

x=3

swaps the two:

x=3

never cancels, so it cannot be a hole, and

x=-2

cancels, so it cannot be a vertical asymptote. Remember this pattern: factor first; a common factor means a hole, a leftover denominator factor means a vertical asymptote; for equal degrees, the horizontal asymptote comes from leading coefficients. That will keep these straight on exam day.

Question 12

Suppose $f(x)=\frac{kx+6}{x-3}$ has an x-intercept at $x=2$ . Which conclusion about the graph is correct?

$k=3$ and the horizontal asymptote is $y=3$ .
$k=-3$ and the horizontal asymptote is $y=-3$ . (correct answer)
$k=-3$ and the y-intercept is $y=2$ .
$k=2$ and the y-intercept is $y=6$ .

Explanation: When you see a rational function and are told an x-intercept, your first move should be to set the numerator equal to zero at that x-value. An x-intercept at

x=2

means

f(2)=0

, so the numerator

kx+6

must satisfy

k(2)+6=0

. Solving gives

2k=-6

, so

k=-3

. With

k=-3

, the function is

f(x)=\frac{-3x+6}{x-3}

. Since the numerator and denominator have the same degree, the horizontal asymptote is the ratio of their leading coefficients:

-3/1=-3

, so

y=-3

. Also, the y-intercept is

f(0)=\frac{6}{-3}=-2

, not

2

. The choice saying

k=3

and horizontal asymptote

y=3

gets the sign of

k

wrong; substituting

k=3

would give

3(2)+6=12\neq0

. The choice saying

k=-3

and y-intercept

y=2

has the correct

k

but confuses the y-intercept with its positive version — it is actually

-2

. The choice saying

k=2

and y-intercept

y=6

neither satisfies the x-intercept condition nor correctly evaluates

f(0)

, since

k=2

would give

2(2)+6=10\neq0

. A quick study tip: for rational functions, x-intercepts come from the numerator, y-intercepts from plugging in

x=0

, and horizontal asymptotes from comparing degrees. If the degrees match, the asymptote is the leading coefficient ratio — just remember to keep the signs straight.