Algebra 3 Quiz: Asymptotes And Holes

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Asymptotes And HolesQuestion 1 of 12

Let $f(x)=\frac{2x^2+kx+3}{x+1}$ . If the slant asymptote of $f$ is $y=2x+1$ , what is $k$ ?

3

-1

1

5

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Algebra 3 Quiz

Algebra 3 Quiz: Asymptotes And Holes

Practice Asymptotes And Holes 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 Asymptotes And Holes, 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 $f(x)=\frac{2x^2+kx+3}{x+1}$ . If the slant asymptote of $f$ is $y=2x+1$ , what is $k$ ?

$3$ (correct answer)
$-1$
$1$
$5$

Explanation: Whenever you see a rational function whose numerator is one degree higher than the denominator, the slant asymptote is exactly the quotient from polynomial division — the remainder becomes negligible as

x\to\pm\infty

. Sofor

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

, divide:

2x^2+kx+3

divided by

x+1

gives first term

2x

, product

2x^2+2x

, subtracting leaves

(k-2)x+3

. Next term is

k-2

, product

(k-2)x+(k-2)

, and subtracting leaves remainder

5-k

. Thus

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

x

grows,

\frac{5-k}{x+1}\to 0

, so the slant asymptote is

y=2x+(k-2).

You are given

y=2x+1

, so

k-2=1

, meaning

k=3

. If you picked

-1

, you probably made a sign error in the division step, effectively using

k+2

instead of

k-2

. If you picked

1

, you likely assumed the

1

in the asymptote was simply

k

— but

k

is the

x

-coefficient in the numerator, and you must divide to see how it contributes. If you picked

5

, you probably set the remainder

5-k

equal to

0

, or set

k-2

equal to the numerator constant

3

; but the remainder does not need to vanish, and the numerator constant has no direct role in the slant asymptote. So the key habit is: divide first, write the function as quotient plus remainder over denominator, and read the slant asymptote straight from the quotient.

Question 2

Let $f(x)=\frac{x^2-5x+6}{x^2+kx-6}$ . If the graph has a removable discontinuity at $x=3$ , what is $k$ ?

$k=1$
$k=5$
$k=-5$
$k=-1$ (correct answer)

Explanation: A removable discontinuity (a "hole") happens when the numerator and denominator both equal zero at the same

x

-value, so the common factor can be canceled. Here, the hole is at

x=3

, so both the numerator and denominator must be zero when

x=3

. The numerator already is zero at

x=3

3^2-5(3)+6=9-15+6=0

. For the denominator to also be zero at

x=3

, substitute

3

into

x^2+kx-6

3^2+3k-6=0 \quad\Rightarrow\quad 9+3k-6=0 \quad\Rightarrow\quad 3+3k=0 \quad\Rightarrow\quad k=-1.

With

k=-1

, the denominator becomes

x^2-x-6=(x-3)(x+2)

, matching the numerator's factor

x-3

, so the hole exists. The other choices fail because they do not make the denominator zero at

x=3

. For

k=1

, the denominator is

x^2+x-6=(x+3)(x-2)

, which equals

6

x=3

. For

k=5

, the denominator is

x^2+5x-6=(x+6)(x-1)

, which equals

18

x=3

. For

k=-5

, the denominator is

x^2-5x-6=(x-6)(x+1)

, which equals

-12

x=3

. In each of those cases, the function is actually defined at

x=3

, so there is no hole. Remember: a removable discontinuity requires both numerator and denominator to share a zero. Plug the given

x

-value into the denominator, set it equal to zero, and solve for the unknown constant.

Question 3

Which function has a vertical asymptote at $x=-2$ , a horizontal asymptote at $y=1$ , a hole at $x=3$ , and an $x$ -intercept at $x=5$ ?

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

Explanation: Whenever you see a rational-function question, connect each factor to its graph feature: numerator factors create

x

-intercepts, denominator factors create vertical asymptotes, and a factor repeated in numerator and denominator creates a hole. For a horizontal asymptote at

y=1

, the degrees of top and bottom must be equal with matching leading coefficients. In

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

, the factor

x-3

cancels, leaving

\frac{x-5}{x+2}

with

x\neq 3

. That canceled factor is the hole at

x=3

; the remaining denominator

x+2

gives a vertical asymptote at

x=-2

; the numerator zero at

x=5

gives the

x

-intercept;and equal degrees with leading coefficient

1

give

y=1

. This satisfies every condition. The choice

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

cancels

x-5

instead of

x-3

, so it has a hole at

x=5

and intercept at

x=3

— exactly reversed. The choice

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

has no shared factor, so it has no hole; instead it gives two vertical asymptotes,

x=-2

and

x=-3

, and both

x

-intercepts, missing the required removable point at

x=3

. The choice

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

does have the hole at

x=3

, but after canceling

x-3

, its remaining numerator

x+5

makes the

x

-intercept

x=-5

, not

x=5

— a sign-error trap. Strategy: first check for factors common to numerator and denominator — those are holes, not asymptotes. Then inspect the simplified form for intercepts and vertical asymptotes, and confirm the horizontal asymptote with leading coefficients. This order eliminates mismatched options quickly.

Question 4

Let $f(x)=\frac{x^3+2x^2}{x^2-4}$ . Which statement correctly describes the discontinuities and the end-behavior asymptote?

Hole at $x=2$ , vertical asymptote at $x=-2$ , slant asymptote $y=x-2$ .
Hole at $x=-2$ , vertical asymptote at $x=2$ , slant asymptote $y=x+2$ . (correct answer)
Vertical asymptotes at $x=-2$ and at $x=2$ , horizontal asymptote $y=1$ .
Hole at $x=-2$ , vertical asymptote at $x=2$ , horizontal asymptote $y=0$ .

Explanation: When you see a rational function and are asked about discontinuities and end behavior, your first move should always be to factor and simplify. The numerator factors as

x^2(x+2)

and the denominator as

(x-2)(x+2)

, so the function becomes

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

for

x\neq -2

. The canceled factor

x+2

means there is a hole at

x=-2

. The remaining factor in the denominator gives a vertical asymptote at

x=2

. Since the simplified numerator has degree 2 and the denominator has degree 1, there is no horizontal asymptote; instead, there is a slant asymptote. Dividing gives

\frac{x^2}{x-2}=x+2+\frac{4}{x-2}

, so the slant asymptote is

y=x+2

. The choice saying hole at 2 and vertical asymptote at -2 reverses the two factors and also gives the wrong slant asymptote. The choice saying vertical asymptotes at both -2 and 2 with horizontal asymptote 1 ignores the cancellation and incorrectly applies the horizontal asymptote rule to the unsimplified form. The choice with hole at -2 and vertical asymptote at 2 but horizontal asymptote 0 correctly identifies the discontinuities but forgets that a degree difference of exactly one means a slant asymptote, not a horizontal one. Your takeaway: always simplify first. Holes come from canceled factors; vertical asymptotes come from uncanceled denominator zeros; compare degrees only after simplifying to decide horizontal versus slant asymptotes.

Question 5

Let $f(x)=\frac{(x-4)^2}{(x-4)^3(x+7)}$ . Which statement correctly describes the discontinuities of $f$ ?

$f$ has a vertical asymptote at $x=-7$ only; $x=4$ is a hole.
$f$ has vertical asymptotes at $x=4$ and $x=-7$ ; $f$ has no holes. (correct answer)
$f$ has a vertical asymptote at $x=4$ only; $x=-7$ is a hole.
$f$ has holes at $x=4$ and $x=-7$ ; $f$ has no vertical asymptotes.

Explanation: Whenever you see a rational function and are asked about discontinuities, your first move should be to factor and simplify, then compare the simplified denominator to the original. Here,

f(x)=\frac{(x-4)^2}{(x-4)^3(x+7)}

. Cancel two of the three copies of

x-4

, but note that this cancellation is valid only for

x\neq 4

. The simplified form is

f(x)=\frac{1}{(x-4)(x+7)}

away from

x=4

. Because one factor of

x-4

remains in the denominator, the function still becomes unbounded as

x\to 4

, so

x=4

is a vertical asymptote, not a hole. At

x=-7

, the denominator is zero while the numerator is nonzero, so

x=-7

is also a vertical asymptote. There are no holes because no factor cancels completely from the denominator. The choice saying

x=4

is a hole and only

x=-7

is a vertical asymptote confuses "a factor appears in numerator and denominator" with "the factor disappears completely." The choice saying only

x=4

is a vertical asymptote and

x=-7

is a hole misses that

x=-7

makes the denominator zero but not the numerator. The choice saying both are holes and there are no vertical asymptotes treats cancellation as if it removes all discontinuity, but the leftover denominator factor still creates infinite behavior. Study tip: a hole occurs only when a factor cancels entirely and the simplified function is defined at that x-value. If any factor from the denominator remains after simplification, that x is a vertical asymptote.

Question 6

Let $f(x)=\frac{2x^2+kx-5}{3x^2+1}$ . For which value(s) of $k$ does the graph have the horizontal asymptote $y=\frac{2}{3}$ ?

Every real number $k$ (correct answer)
$k=0$ only
$k=\frac{5}{3}$ only
No such value of $k$ exists.

Explanation: Whenever you see a rational function and are asked about a horizontal asymptote, your first move should be to compare the degrees of the numerator and denominator, focusing only on the leading terms as

x \to \pm\infty

. Here both numerator and denominator are degree 2: the leading terms are

2x^2

and

3x^2

. Since degrees match, the horizontal asymptote is

y=\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}} =\frac{2}{3}.

This result holds for every real value of

k

, because the

kx

term is lower-order; as

x^2

dominates, it disappears in the limit. The constant

-5

also doesn't matter. So the correct choice is "Every real number

k

." Why are the others wrong? "

k=0

only" might seem plausible if you think the middle term must be eliminated, but a linear term is negligible compared with

x^2

at infinity, so it does not affect the asymptote. "

k=\frac{5}{3}

only" appears to confusethe horizontal asymptote with solving for

k

from constants; but there isno equation involving

k

in the end behavior. "No such value of

k

exists" is also false: theunction has exactly that asymptote regardless of what

k

is. . For your study strategy: when determining horizontal asymptotes, divide through by the highest power of

x

in the denominator, or simply compare leading coefficients. Ignore all lower-degree terms—they are irrelevant to end behavior. This insta saves time and prevents traps on problems like this.

Question 7

Let $f(x)=\frac{x^2-3x-10}{x^2-2x-8}$ . Which statement is true?

$f$ has a vertical asymptote at $x=-2$ and a hole at $\left(4,\frac{7}{6}\right)$ .
$f$ has a vertical asymptote at $x=-4$ and a hole at $\left(2,-\frac{7}{6}\right)$ .
$f$ has a vertical asymptote at $x=4$ and a hole at $\left(-2,-\frac{7}{6}\right)$ .
$f$ has a vertical asymptote at $x=4$ and a hole at $\left(-2,\frac{7}{6}\right)$ . (correct answer)

Explanation: Whenever you see a rational function, factor the numerator and denominator before deciding where vertical asymptotes and holes are. Here,

x^2-3x-10=(x-5)(x+2),\quad x^2-2x-8=(x-4)(x+2).

f(x)=\frac{(x-5)(x+2)}{(x-4)(x+2)}=\frac{x-5}{x-4},\quad x\ne -2,4.

The common factor

(x+2)

cancels, so

x=-2

is a hole, not a vertical asymptote. Substitute

x=-2

into the simplified form:

\frac{-2-5}{-2-4}=\frac{-7}{-6}=\frac76.

Thus the hole is

\left(-2,\frac76\right)

. The remaining denominator factor

x-4

is zero at

x=4

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

x=4

. The statement that gives a vertical asymptote at

x=4

and a hole at

\left(-2,\frac76\right)

is therefore correct. The choice saying there is an asymptote at

x=-2

and a hole at

\left(4,\frac76\right)

swaps the roles of the two

x

-values. The choice with an asymptote at

x=-4

and a hole at

\left(2,-\frac76\right)

uses zeros that are not factors of the denominator. Finally, the choice with the asymptote at

x=4

and hole at

\left(-2,-\frac76\right)

has the right

x

-locations but misses that the two negatives cancel, giving

\frac76

, not

-\frac76

. Remember: factor first, cancel common factors to find holes, and evaluate the simplified function at the hole's

x

-value. The remaining denominator zero gives the vertical asymptote.

Question 8

The graph of $f(x)=\frac{2x^2+3x+1}{x^2+1}$ has horizontal asymptote $y=2$ . At how many points does the graph intersect the line $y=2$ ?

$0$
$2$
$1$ (correct answer)
infinitely many

Explanation: Whenever you see a rational function with a horizontal asymptote, remember that the asymptote describes end behavior, not a barrier. The graph can cross the horizontal asymptote at finite points, so the number of intersections must be found by solving the equation. Set

f(x)=2

\frac{2x^2+3x+1}{x^2+1}=2

Multiply both sides by

x^2+1

2x^2+3x+1=2x^2+2

Subtract

2x^2

from both sides:

3x+1=2

3x=1

, giving

x=\frac13

. There is exactly one intersection point:

\left(\frac13,2\right)

. The choice 0 is the common trap: many students assume a horizontal asymptote is a line the graph can never touch. In fact, a graph may cross its horizontal asymptote finitely many times before leveling off. The choice 2 might come from counting the leading terms or thinking the asymptote is approached on both sides, but the equation shows only one solution. The choice infinitely many would require the function to equal

2

for every

x

, which would mean the numerator is exactly twice the denominator for all

x

; here

3x+1

is not identically

0

, so that is impossible. For exam day: whenever asked about intersections with an asymptote, don't rely on the graph's shape—solve the equation explicitly. The horizontal asymptote is about limits as

x\to\infty

, not about prohibiting crossings.

Question 9

Which function has no vertical asymptote and has the x-axis as its horizontal asymptote?

$f(x)=\frac{x^2+1}{x^2-4}$
$f(x)=\frac{x+1}{x^2+4}$ (correct answer)
$f(x)=\frac{x^2+1}{x^2+4}$
$f(x)=\frac{x+1}{x^2-4}$

Explanation: Whenever you see a rational function question about asymptotes, start by checking two things: does the denominator have real zeros? And how do the degrees of numerator and denominator compare? The horizontal asymptote is the x-axis,

y=0

, exactly when the numerator has a smaller degree than the denominator. because

x^2+4

has no real zeros — it is always positive — so it never makes the denominator zero. Its numerator has degree

1

, denominator degree

2

, so the x-axis is its horizontal asymptote. That is why it satisfies both conditions. loses the x-axis because the degrees are equal: the numerator and denominator both have degree

2

, so their leading coefficients give

y=\frac{1}{1}=1

, not

y=0

. Finally,

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

does have the x-axis as its horizontal asymptote, since degree

1

is less than degree

2

, but its denominator factors as

(x-2)(x+2)

, producing vertical asymptotes at

x=2

and

x=-2

. So it fails the "no vertical asymptote" condition. The useful pattern: these two conditions are independent — denominator zeros cause vertical asymptotes, while degree comparison controls the horizontal asymptote. Check each condition separately before you choose.

Question 10

Let $f(x)=\frac{x^2-1}{x^3-x^2+x-1}$ . Which statement is true?

$f$ has a vertical asymptote at $x=1$ and no holes.
$f$ has vertical asymptotes at $x=1$ and $x=-1$ , and no holes.
$f$ has a hole at $x=1$ and no vertical asymptotes. (correct answer)
$f$ has holes at $x=1$ and $x=-1$ , and no vertical asymptotes.

Explanation: When you see a rational function and are asked about asymptotes or holes, your first step is to factor completely. The difference between a hole and a vertical asymptote comes down to whether a factor cancels or stays in the denominator. Factor the numerator:

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

. Factor the denominator by grouping:

x^3-x^2+x-1 = x^2(x-1)+1(x-1) = (x-1)(x^2+1)

. So

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

. The

(x-1)

cancels, so

x=1

is a hole (removable discontinuity). After cancellation, the denominator is

x^2+1

, which is never zero for real numbers, so there are no vertical asymptotes. Thus, "a hole at

x=1

and no vertical asymptotes" is correct. Now, the wrong choices. The choice claiming "a vertical asymptote at

x=1

and no holes" ignores the cancellation — you must simplify before deciding. The choice with "vertical asymptotes at

x=1

and

x=-1

" confuses numerator zeros with denominator zeros; at

x=-1

, the denominator is

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

, so it's just an x-intercept. The choice with "holes at

x=1

and

x=-1

" also fails because a hole requires both numerator and denominator to be zero; at

x=-1

, the denominator isn't zero. Your takeaway: Factor and cancel first. Cancelled factors create holes; remaining denominator zeros create vertical asymptotes. Also, a denominator like

x^2+1

never equals zero, so it contributes no asymptotes.

Question 11

Which statement describes the graph of $f(x)=\frac{x^3+1}{x^2+1}$ ?

It has a horizontal asymptote $y=1$ and a vertical asymptote at $x=-1$ .
It has a slant asymptote $y=x+1$ and no vertical asymptotes.
It has a horizontal asymptote $y=0$ and no vertical asymptotes.
It has a slant asymptote $y=x$ and no vertical asymptotes. (correct answer)

Explanation: Whenever you see a question about asymptotes of a rational function, start by comparing the degrees of the numerator and denominator, then check whether the denominator can equal zero. For

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

, the denominator

x^2+1

is never zero because

x^2\ge 0

, so there are no vertical asymptotes. The numerator has degree 3, exactly one more than the denominator's degree 2, so the graph has a slant asymptote. Performing polynomial division:

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

x\to\pm\infty

, the fraction approaches

0

, so the slant asymptote is

y=x

. The choice claiming a horizontal asymptote

y=1

and a vertical asymptote at

x=-1

is wrong because

x=-1

makes the numerator zero, not the denominator, and the denominator has no real zeros. Also, a horizontal asymptote cannot exist when the numerator's degree exceeds the denominator's. The choice with slant asymptote

y=x+1

comes from a division slip—the quotient is exactly

x

, not

x+1

. The choice with horizontal asymptote

y=0

confuses this with the case where the numerator has lower degree than the denominator. For your study, remember the degree comparison: deg numerator < deg denominator gives

y=0

; equal gives

y=

leading coefficient ratio; exactly one greater gives a slant asymptote found by division. Always check whether the denominator actually has real zeros—if not, there are no vertical asymptotes.

Question 12

Consider $f(x)=\frac{(x+1)(x-2)^2}{(x+1)(x-2)(x+3)}$ . Which statement is true?

$f$ has a vertical asymptote at $x=-3$ and holes at $x=-1$ and $x=2$ . (correct answer)
$f$ has vertical asymptotes at $x=-1$ and $x=2$ and a hole at $x=-3$ .
$f$ has a vertical asymptote at $x=-3$ and a hole at $x=-1$ only.
$f$ has vertical asymptotes at $x=-3$ , $x=-1$ , and $x=2$ , and no holes.

Explanation: When you see a rational function like this, your first move should be to factor and look for cancellation. Zeros that cancel become holes; zeros of the denominator that remain become vertical asymptotes. Here the numerator is

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

and the denominator is

(x+1)(x-2)(x+3)

. Canceling the common factors

(x+1)

and one

(x-2)

leaves

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

, but only after noting that

x=-1

and

x=2

are excluded from the domain. Because those factors cancel completely, they produce holes, not asymptotes. The factor

(x+3)

does not cancel, so

x=-3

is a genuine vertical asymptote. Thus the correct statement is the one saying

f

has a vertical asymptote at

x=-3

and holes at

x=-1

and

x=2

. The claim that

f

has vertical asymptotes at

x=-1

and

x=2

and a hole at

x=-3

reverses the roles: canceling factors cannot create asymptotes, and an uncanceled denominator zero cannot be a hole. The statement with a vertical asymptote at

x=-3

and a hole only at

x=-1

forgets that

x=2

is also a hole after canceling

(x-2)

. And the statement listing vertical asymptotes at

x=-3

x=-1

, and

x=2

with no holes ignores cancellation entirely — a common trap when you forget to simplify before analyzing. On exam day, always simplify the function first and note the excluded

x

-values before you classify discontinuities.