Asymptotes & Holes - Algebra 3

1. 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 $x=5$?
   Answer: $f(x)=\frac{(x-3)(x-5)}{(x+2)(x-3)}$ 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.
2. Let $f(x)=\dfrac{x^2-(k+1)x+k}{x^2-5x+6}$. If the graph of $f$ has exactly one vertical asymptote and exactly one hole, what is the sum of all possible values of $k$?
   Answer: 5 The denominator factors as $(x-2)(x-3)$. For exactly one vertical asymptote and one hole, the numerator must share exactly one of these factors. Substituting $x=2$ gives $2-k=0$, so $k=2$. Substituting $x=3$ gives $6-2k=0$, so $k=3$. Both values are valid, and their sum is $5$. The distractors list a single valid value or an incorrect sum.
3. Let $f(x)=\dfrac{x^2+kx-8}{x^2-2x-8}$. If the graph has no vertical asymptote, which $x$-values are holes?
   Answer: $x=-2$ and $x=4$ The denominator factors as $(x-4)(x+2)$. For there to be no vertical asymptote, the numerator must contain both factors, so $x^2+kx-8=(x-4)(x+2)=x^2-2x-8$, which gives $k=-2$. Then the function is equal to $1$ wherever it is defined, with holes at the two excluded values $x=4$ and $x=-2$. The other choices either reverse the signs of the factors or list only one of the two holes.
4. Let $f(x)=\frac{x^2-3x-10}{x^2-2x-8}$. Which statement is true?
   Answer: $f$ has a vertical asymptote at $x=4$ and a hole at $\left(-2,\frac{7}{6}\right)$. Whenever you see a rational function, factor the numerator and denominator before deciding where vertical asymptotes and holes are. Here, $$x^2-3x-10=(x-5)(x+2),\quad x^2-2x-8=(x-4)(x+2).$$ So $$f(x)=\frac{(x-5)(x+2)}{(x-4)(x+2)}=\frac{x-5}{x-4},\quad x\ne -2,4.$$ 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.
5. Consider $f(x)=\dfrac{(x-1)(x+2)^2}{(x-1)^2(x+2)(x-3)}$. Which statement about the graph of $f$ is correct?
   Answer: Vertical asymptotes at $x=1$ and $x=3$; hole at $x=-2$; horizontal asymptote $y=0$. One factor of $(x-1)$ and one factor of $(x+2)$ cancel, leaving $\frac{x+2}{(x-1)(x-3)}$. The original domain excludes $x=1,-2,3$. Since $x=-2$ cancels completely, it is a hole; $x=1$ and $x=3$ remain in the denominator, so they are vertical asymptotes. The degree of the numerator is less than the degree of the denominator, so the horizontal asymptote is $y=0$.
6. Let $f(x)=\dfrac{x^3-1}{x^2-1}$. Which statement about the graph of $f$ is correct?
   Answer: Vertical asymptote at $x=-1$, hole at $x=1$, slant asymptote $y=x$. The numerator factors as $(x-1)(x^2+x+1)$ and the denominator as $(x-1)(x+1)$. Canceling $x-1$ leaves $\frac{x^2+x+1}{x+1}$, so $x=1$ is a hole and $x=-1$ is a vertical asymptote. Long division gives quotient $x$ with remainder $1$, so the slant asymptote is $y=x$. There is no horizontal asymptote because the degree of the numerator is one greater than the degree of the denominator.
7. Which statement describes the graph of $f(x)=\frac{x^3+1}{x^2+1}$?
   Answer: It has a slant asymptote $y=x$ and no vertical asymptotes. 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: $\frac{x^3+1}{x^2+1}=x+\frac{-x+1}{x^2+1}$ As $x\to\pm\infty$, 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.
8. Consider $f(x)=\frac{(x+1)(x-2)^2}{(x+1)(x-2)(x+3)}$. Which statement is true?
   Answer: $f$ has a vertical asymptote at $x=-3$ and holes at $x=-1$ and $x=2$. 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.
9. Which of the following functions has $y=2x-5$ as a slant asymptote and no vertical asymptote?
   Answer: $f(x)=\dfrac{2x^3-5x^2+x-2}{x^2+1}$ Dividing $2x^3-5x^2+x-2$ by $x^2+1$ gives quotient $2x-5$ and remainder $x+3$, so the slant asymptote is $y=2x-5$. The denominator $x^2+1$ has no real zeros, so there are no vertical asymptotes. The first function has equal degrees and a horizontal asymptote, the third has quotient $2x-7$, and the fourth has the correct quotient but vertical asymptotes at $x=\pm1$.
10. Let $f(x)=\frac{(x-4)^2}{(x-4)^3(x+7)}$. Which statement correctly describes the discontinuities of $f$?
    Answer: $f$ has vertical asymptotes at $x=4$ and $x=-7$; $f$ has no holes. 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.
11. Let $f(x)=\frac{2x^2+kx+3}{x+1}$. If the slant asymptote of $f$ is $y=2x+1$, what is $k$?
    Answer: $3$ 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 $$f(x)=2x+(k-2)+\frac{5-k}{x+1}.$$ As $x$ 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.
12. Let $f(x)=\dfrac{x^2-5x+6}{x^3-7x^2+14x-8}$. Which statement correctly describes the graph of $f$?
    Answer: Vertical asymptotes at $x=1$ and $x=4$, hole at $x=2$, horizontal asymptote $y=0$. The numerator factors as $(x-2)(x-3)$ and the denominator factors as $(x-1)(x-2)(x-4)$. After canceling $x-2$, the denominator still contains $x-1$ and $x-4$, so those are vertical asymptotes, while $x=2$ is a hole. Since the denominator has greater degree than the numerator, the horizontal asymptote is $y=0$. The other choices fail to cancel the common factor, mistake the zero $x=3$ for a hole, or ignore the horizontal asymptote.
13. Let $f(x)=\frac{2x^2+kx-5}{3x^2+1}$. For which value(s) of $k$ does the graph have the horizontal asymptote $y=\frac{2}{3}$?
    Answer: Every real number $k$ 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.
14. Let $f(x)=\dfrac{(x+2)(x^2-1)}{(x+2)(x-3)}$. Which statement correctly describes the graph of $f$?
    Answer: Slant asymptote $y=x+3$; hole at $x=-2$. The factor $x+2$ cancels, so $x=-2$ is a hole. The reduced function is $\frac{x^2-1}{x-3}$. Dividing $x^2-1$ by $x-3$ gives quotient $x+3$ and remainder $8$, so the slant asymptote is $y=x+3$. The degree of the numerator is one greater than the degree of the denominator, so there is no horizontal asymptote.
15. Let $f(x)=\dfrac{(k-1)x^2+2x-5}{x^2+x-2}$. If the graph has horizontal asymptote $y=3$, which statement about the graph is true?
    Answer: Hole at $x=1$, vertical asymptote at $x=-2$, horizontal asymptote $y=3$. For the horizontal asymptote to be $y=3$, the leading coefficient of the numerator must be $3$, so $k-1=3$ and $k=4$. Then the numerator is $3x^2+2x-5=(3x+5)(x-1)$. The denominator is $(x+2)(x-1)$, so $x-1$ cancels, giving a hole at $x=1$, while $x=-2$ remains a vertical asymptote. The horizontal asymptote remains $y=3$.
16. Let $f(x)=\dfrac{ax^2+bx-3}{x^2-1}$. If the graph has horizontal asymptote $y=2$, vertical asymptote at $x=-1$, and a hole at $x=1$, what is the $y$-coordinate of the hole?
    Answer: $\frac{5}{2}$ The horizontal asymptote $y=2$ requires the leading coefficients to match, so $a=2$. A hole at $x=1$ requires the numerator to be zero at $x=1$: $2+b-3=0$, so $b=1$. Then the numerator is $2x^2+x-3=(2x+3)(x-1)$, and the denominator is $(x-1)(x+1)$. The reduced function is $\frac{2x+3}{x+1}$, and its value at $x=1$ is $\frac{5}{2}$. The distractors come from incorrect values of $b$ or evaluation errors.
17. Let $f(x)=\frac{x^3+2x^2}{x^2-4}$. Which statement correctly describes the discontinuities and the end-behavior asymptote?
    Answer: Hole at $x=-2$, vertical asymptote at $x=2$, slant asymptote $y=x+2$. 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.
18. Let $f(x)=\dfrac{x^3-2x^2-5x+6}{x^2-4}$. Which statement correctly describes the graph of $f$?
    Answer: Vertical asymptote at $x=2$, hole at $x=-2$, slant asymptote $y=x-2$. The numerator factors as $(x-1)(x-3)(x+2)$ and the denominator as $(x-2)(x+2)$. Canceling $x+2$ gives a hole at $x=-2$ and leaves $x-2$ in the denominator, so $x=2$ is a vertical asymptote. The reduced function is $\frac{(x-1)(x-3)}{x-2}=\frac{x^2-4x+3}{x-2}$, and long division gives $x-2-\frac{1}{x-2}$, so the slant asymptote is $y=x-2$.
19. Let $f(x)=\dfrac{x^2+kx-8}{x^2-2x-3}$. If the graph has a removable discontinuity at $x=-1$, what is the $y$-coordinate of the hole?
    Answer: $\frac{9}{4}$ The denominator factors as $(x-3)(x+1)$. For a hole at $x=-1$, the numerator must also have $x+1$ as a factor, so $1-k-8=0$, giving $k=-7$. The numerator becomes $x^2-7x-8=(x-8)(x+1)$, so the reduced function is $\frac{x-8}{x-3}$. Evaluating at $x=-1$ gives $\frac{-9}{-4}=\frac{9}{4}$. The distractors come from sign errors or incorrect reduced forms.
20. A student simplifies $f(x)=\dfrac{x^2-4}{x^2-x-2}$ as $\dfrac{x+2}{x+1}$ and then states: I. The graph has a vertical asymptote at $x=-1$. II. The graph has a hole at $x=2$. III. The graph has no horizontal asymptote because the numerator and denominator of the simplified form have equal degrees. Which of the student's statements are correct?
    Answer: I and II only The original function factors as $\frac{(x-2)(x+2)}{(x-2)(x+1)}$. The factor $x-2$ cancels, so $x=2$ is a hole and $x=-1$ remains a vertical asymptote. Because the simplified numerator and denominator have equal degree, the horizontal asymptote is $y=1$, not $y=\text{none}$. Therefore statements I and II are correct, while III is false.

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