Math 3 Quiz: Rational Function Features

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Rational Function FeaturesQuestion 1 of 14

A rational function has the form $h(x) = \frac{ax^2 + bx + c}{(x-3)^2(x+1)}$ where $a$ , $b$ , and $c$ are constants. If the function has a horizontal asymptote at $y = 2$ , which additional condition would create a hole in the graph?

The numerator must have

(x-3)

as a factor, requiring

9a + 3b + c = 0

to be satisfied.

The numerator must have

(x+1)

as a factor, requiring

a - b + c = 0

to be satisfied.

The numerator must have

(x-3)^2

as a factor, requiring both

9a + 3b + c = 0

and

6a + b = 0

The numerator must equal zero at

x = 3

, but this creates a vertical asymptote rather than a hole.

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Math 3 Quiz

Math 3 Quiz: Rational Function Features

Practice Rational Function Features in Math 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 Rational Function Features, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 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 numerator must have $(x-3)$ as a factor, requiring $9a + 3b + c = 0$ to be satisfied. (correct answer)
The numerator must have $(x+1)$ as a factor, requiring $a - b + c = 0$ to be satisfied.
The numerator must have $(x-3)^2$ as a factor, requiring both $9a + 3b + c = 0$ and $6a + b = 0$ .
The numerator must equal zero at $x = 3$ , but this creates a vertical asymptote rather than a hole.

Explanation: Since the horizontal asymptote is

y = 2

, we need

a = 2

. For a hole to exist, a factor in the denominator must cancel with the numerator. The denominator has factors

(x-3)^2

and

(x+1)

. If the numerator has factor

(x-3)

, then one factor of

(x-3)

cancels, leaving

(x-3)

in the denominator (still a vertical asymptote) and creating a hole. For

(x-3)

to be a factor of the numerator, we need the numerator to equal zero when

x = 3

a(3)^2 + b(3) + c = 9a + 3b + c = 0

. Choice B would create a hole but at

x = -1

. Choice C would completely cancel

(x-3)^2

, removing the vertical asymptote entirely. Choice D is incorrect about the result.

Question 2

Consider the piecewise-defined function where $f(x) = \frac{x^2 - 9}{x + 3}$ for $x \neq -3$ and $f(-3) = 5$ . How does this compare to the rational function $g(x) = \frac{x^2 - 9}{x + 3}$ that is undefined at $x = -3$ ?

Both functions have identical behavior including the same type of discontinuity at $x = -3$ with matching limits.
Function $f$ has a jump discontinuity at $x = -3$ while $g$ has a removable discontinuity at the same point. (correct answer)
Function $f$ has a removable discontinuity at $x = -3$ while $g$ has a hole, but both approach the same limit.
Function $f$ has a removable discontinuity at $x = -3$ while $g$ has a hole, with different limiting behaviors.

Explanation: For both functions,

\frac{x^2-9}{x+3} = \frac{(x-3)(x+3)}{x+3} = x-3

when

x \neq -3

. So

\lim_{x \to -3} f(x) = \lim_{x \to -3} g(x) = -3-3 = -6

. Function

g

has a hole (removable discontinuity) at

x = -3

since the limit exists but the function is undefined there. Function

f

has

f(-3) = 5 \neq -6

, so while the limit exists, the function value differs from the limit, creating a jump discontinuity (the function 'jumps' from the limit value to the defined value). Choice A ignores that

f(-3) = 5

. Choice C incorrectly categorizes both as the same type of discontinuity. Choice D correctly identifies the discontinuity types but wrongly claims different limits.

Question 3

The rational function $h(x) = \frac{x^3 - 8}{x^2 + 2x + 4}$ appears to have a denominator that doesn't factor over the reals. However, there's a hidden relationship. What is the correct analysis of this function?

The function has no real vertical asymptotes since the denominator has no real zeros, and approaches $y = 0$ as $x \to \infty$ .
The function simplifies to $h(x) = x - 2$ because the numerator and denominator share a common factor. (correct answer)
The function has no real vertical asymptotes and no horizontal asymptote since the numerator degree exceeds the denominator degree.
The function has complex vertical asymptotes but behaves like $y = x$ for large values due to polynomial division.

Explanation: The key insight is recognizing that

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

using the difference of cubes formula:

a^3 - b^3 = (a-b)(a^2+ab+b^2)

. So

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

(for

x \neq

the complex roots of

x^2+2x+4=0

). Since the denominator has no real zeros, there are no holes or asymptotes in the real plane - the function is simply the line

y = x-2

. Choice A incorrectly assumes the original form determines the behavior. Choice C correctly identifies no real asymptotes but wrong about the simplified form. Choice D mentions irrelevant complex asymptotes and incorrect end behavior.

Question 4

Consider the rational function $f(x) = \frac{x^2 - 9}{x^2 - 2x - 3}$ . After simplifying this function, what is the behavior at $x = 3$ ?

There is a vertical asymptote at $x = 3$
There is a removable discontinuity (hole) at $x = 3$ (correct answer)
The function is continuous at $x = 3$
There is a horizontal asymptote at $x = 3$

Explanation: First, factor the numerator and denominator:

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

. Since

(x-3)

appears in both numerator and denominator, it cancels out, leaving

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

for

x \neq 3

. This creates a removable discontinuity (hole) at

x = 3

. Choice A is wrong because the factor cancels rather than creating an asymptote. Choice C is wrong because the original function is undefined at

x = 3

. Choice D is wrong because horizontal asymptotes are not located at specific x-values.

Question 5

A rational function has the form $r(x) = \frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials. If $r(x)$ has vertical asymptotes at $x = -2$ and $x = 5$ , and a hole at $x = 3$ , which expression could represent $Q(x)$ ?

$(x+2)(x-5)(x-3)$ (correct answer)
$(x+2)(x-5)$
$(x+2)^2(x-5)(x-3)$
$(x-2)(x+5)(x-3)$

Explanation: For vertical asymptotes at

x = -2

and

x = 5

, the denominator

Q(x)

must have factors

(x+2)

and

(x-5)

that do not completely cancel with the numerator. For a hole at

x = 3

, both the numerator and denominator must have the factor

(x-3)

, and this factor must cancel completely. Therefore,

Q(x)

must contain

(x+2)

(x-5)

, and

(x-3)

. Choice A satisfies this. Choice B is missing the

(x-3)

factor needed for the hole. Choice C would work but has an extra factor. Choice D has wrong signs for the asymptote factors.

Question 6

For the rational function $k(x) = \frac{x^2 - 1}{x^3 - x^2 - x + 1}$ , determine the total number of vertical asymptotes and holes combined.

One vertical asymptote and one hole (correct answer)
Two vertical asymptotes and no holes
One vertical asymptote and no holes
No vertical asymptotes and one hole

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

. The function becomes

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

. The factor

(x+1)

cancels completely, creating a hole at

x = -1

. One factor of

(x-1)

cancels, leaving

(x-1)

in the denominator, creating a vertical asymptote at

x = 1

. So there is one vertical asymptote and one hole.

Question 7

The rational function $g(x) = \frac{2x^3 + x^2 - 8x - 4}{x^2 + 3x + 2}$ can be simplified by factoring. What is the horizontal asymptote of the simplified function?

The function has no horizontal asymptote because the degree of the numerator exceeds the degree of the denominator. (correct answer)
The horizontal asymptote is $y = 2$ based on the ratio of the leading coefficients after simplification.
The horizontal asymptote is $y = 0$ since the denominator degree exceeds the numerator degree after cancellation.
The horizontal asymptote is $y = 1$ determined by the remaining linear factors in the simplified expression.

Explanation: First, factor the denominator:

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

. For the numerator, factor by grouping:

2x^3 + x^2 - 8x - 4 = x^2(2x+1) - 4(2x+1) = (x^2-4)(2x+1) = (x-2)(x+2)(2x+1)

. So

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

. After canceling

(x+2)

, we get

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

. The simplified function has degree 2 in the numerator and degree 1 in the denominator, so there is no horizontal asymptote. Choices B, C, and D all incorrectly assume a horizontal asymptote exists.

Question 8

For the rational function $k(x) = \frac{3x^3 - 12x^2 + 9x}{x^3 - 6x^2 + 11x - 6}$ , what happens as $x$ approaches positive infinity?

The function approaches $y = 3$ because the ratio of leading coefficients determines the horizontal asymptote behavior. (correct answer)
The function approaches $y = 0$ since after factoring and simplification, the denominator degree exceeds the numerator degree.
The function increases without bound because the simplified form has a higher degree numerator than denominator.
The function approaches $y = 1$ based on the behavior of the simplified rational expression at infinity.

Explanation: For large values of

x

, the behavior is determined by the leading terms. Both numerator and denominator have degree 3, so the horizontal asymptote is the ratio of leading coefficients:

\frac{3x^3}{x^3} = 3

. Even though this function can be factored (numerator:

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

; denominator factors as

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

), and factors cancel to give

\frac{3x}{x-2}

after removing holes, the end behavior is still determined by

y = 3

. Choice B incorrectly assumes the degrees change after factoring. Choice C is wrong about the direction. Choice D gives an incorrect asymptote value.

Question 9

The function $m(x) = \frac{x^3 + 2x^2 - 9x - 18}{x^2 + x - 6}$ has a specific type of discontinuity. After performing polynomial long division, what can be concluded about the function's behavior?

The function has a slant asymptote $y = x + 1$ and a vertical asymptote at $x = 2$ only.
The function has a slant asymptote $y = x + 1$ and a hole at $x = -3$ with no vertical asymptotes.
The function has a slant asymptote $y = x + 1$ , a hole at $x = -3$ , and a vertical asymptote at $x = 2$ . (correct answer)
The function has vertical asymptotes at both $x = -3$ and $x = 2$ with a slant asymptote $y = x + 1$ .

Explanation: First, factor the denominator:

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

. Factor the numerator by grouping:

x^3 + 2x^2 - 9x - 18 = x^2(x+2) - 9(x+2) = (x^2-9)(x+2) = (x-3)(x+3)(x+2)

. So

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

. The factor

(x+3)

cancels, creating a hole at

x = -3

. This gives

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

. Performing polynomial long division:

x^2-x-6 = (x-2)(x+1) - 4

, so

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

. This shows a slant asymptote

y = x+1

and vertical asymptote at

x = 2

. Choice A misses the hole. Choice B misses the vertical asymptote. Choice D incorrectly includes

x = -3

as a vertical asymptote.

Question 10

A rational function has the form $g(x) = \frac{P(x)}{(x+2)^3(x-1)^2}$ where $P(x)$ is a polynomial. If $g(x)$ has exactly one vertical asymptote, what must be true about $P(x)$ ?

$P(x)$ must have $(x+2)^2$ as a factor but not $(x+2)^3$ , leaving one factor to create the asymptote.
$P(x)$ must have $(x-1)^2$ as a factor but cannot have $(x+2)$ as a factor to maintain the asymptote.
$P(x)$ must have either $(x+2)^3$ or $(x-1)^2$ as a factor, but not both simultaneously.
$P(x)$ must have $(x+2)^2(x-1)^2$ as a factor, leaving only $(x+2)$ in the denominator for the asymptote. (correct answer)

Explanation: The denominator has potential vertical asymptotes at

x = -2

and

x = 1

. For exactly one vertical asymptote, we must eliminate one of these. If

P(x)

has

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

as a factor, then

g(x) = \frac{(x+2)^2(x-1)^2 \cdot Q(x)}{(x+2)^3(x-1)^2} = \frac{Q(x)}{x+2}

for some polynomial

Q(x)

. This leaves only one vertical asymptote at

x = -2

. Choice A leaves

(x+2)

in the denominator but doesn't address

x = 1

, giving two asymptotes. Choice B leaves

(x+2)^3

in the denominator, still giving two asymptotes. Choice C is too vague and doesn't specify the correct multiplicities needed.

Question 11

A rational function $p(x) = \frac{(x-a)^2(x-b)}{(x-a)(x-c)(x-d)}$ where $a$ , $b$ , $c$ , and $d$ are distinct real numbers. Which statement best describes the key features of this function?

The function has a hole at $x = a$ , vertical asymptotes at $x = c$ and $x = d$ , and touches the x-axis at $x = b$ .
The function has vertical asymptotes at $x = a$ , $x = c$ , and $x = d$ , with a simple zero at $x = b$ .
The function has a hole at $x = a$ , vertical asymptotes at $x = c$ and $x = d$ , and crosses the x-axis at $x = b$ . (correct answer)
The function has a double zero at $x = a$ , vertical asymptotes at $x = c$ and $x = d$ , and crosses the x-axis at $x = b$ .

Explanation: After canceling the common factor

(x-a)

, we get

p(x) = \frac{(x-a)(x-b)}{(x-c)(x-d)}

. The canceled factor creates a hole at

x = a

. The remaining denominator factors

(x-c)

and

(x-d)

create vertical asymptotes at

x = c

and

x = d

. The remaining numerator factor

(x-b)

has multiplicity 1, so the function crosses (not touches) the x-axis at

x = b

. Choice A incorrectly states the function touches the x-axis at

x = b

(which would require even multiplicity). Choice B ignores the cancellation and incorrectly claims a vertical asymptote at

x = a

. Choice D incorrectly claims

x = a

is a zero rather than a hole.

Question 12

Consider the rational function $r(x) = \frac{x^4 - 5x^3 + 6x^2}{x^3 - 6x^2 + 9x}$ . What is the most complete description of this function's key features?

After factoring, the function simplifies to $r(x) = \frac{x(x-2)}{x-3}$ with a hole at $x = 0$ and vertical asymptote at $x = 3$ .
After factoring, the function simplifies to $r(x) = \frac{x(x-2)}{x-3}$ with holes at $x = 0$ and $x = 3$ , and a vertical asymptote at $x = 3$ .
After factoring, the function simplifies to $r(x) = \frac{x-2}{x-3}$ with holes at $x = 0$ and $x = 3$ , and a vertical asymptote at $x = 3$ .
After factoring, the function simplifies to $r(x) = \frac{x-2}{x-3}$ with a hole at $x = 0$ and vertical asymptote at $x = 3$ . (correct answer)

Explanation: Factor the numerator:

x^4 - 5x^3 + 6x^2 = x^2(x^2 - 5x + 6) = x^2(x-2)(x-3)

. Factor the denominator:

x^3 - 6x^2 + 9x = x(x^2 - 6x + 9) = x(x-3)^2

. So

r(x) = \frac{x^2(x-2)(x-3)}{x(x-3)^2} = \frac{x(x-2)}{x-3}

after canceling

x

and one factor of

(x-3)

. The canceled factor

x

creates a hole at

x = 0

. The remaining

(x-3)

in the denominator creates a vertical asymptote at

x = 3

. Choice A keeps the extra

x

factor incorrectly. Choice B incorrectly claims

x = 3

is both a hole and asymptote. Choice C keeps the wrong factors after cancellation.

Question 13

Consider the rational function $f(x) = \frac{x^2 - 4x - 5}{x^2 - 6x + 5}$ . After factoring both numerator and denominator, which statement correctly describes the behavior of this function?

The function has vertical asymptotes at $x = 1$ and $x = 5$ , with no holes or removable discontinuities.
The function has a vertical asymptote at $x = 1$ and a hole at $x = 5$ where the function is undefined. (correct answer)
The function has a vertical asymptote at $x = 5$ and a hole at $x = 1$ where the function is undefined.
The function has holes at both $x = 1$ and $x = 5$ with no vertical asymptotes present.

Explanation: Factoring the numerator:

x^2 - 4x - 5 = (x-5)(x+1)

. Factoring the denominator:

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

. So

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

. The factor

(x-5)

cancels, creating a hole at

x = 5

. The remaining denominator factor

(x-1)

creates a vertical asymptote at

x = 1

. Choice A incorrectly identifies both zeros of the original denominator as vertical asymptotes. Choice C reverses which discontinuity is which. Choice D incorrectly claims both are holes.

Question 14

A rational function $f(x) = \frac{ax^2 + bx + c}{dx^2 + ex + f}$ has a horizontal asymptote at $y = 3$ . If $d = 2$ , what must be true about the value of $a$ ?

$a = 6$ (correct answer)
$a = 3$
$a = \frac{3}{2}$
$a$ can be any nonzero value

Explanation: For a rational function where the numerator and denominator have the same degree, the horizontal asymptote is

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

. Here, the horizontal asymptote is

y = \frac{a}{d} = \frac{a}{2} = 3

. Solving for

a

a = 6

. Choice B gives

y = \frac{3}{2}

. Choice C gives

y = \frac{3/2}{2} = \frac{3}{4}

. Choice D is incorrect because the horizontal asymptote depends specifically on the ratio of leading coefficients.