Algebra 2 Quiz: Graph Rational Functions And Identify Asymptotes
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Graph Rational Functions And Identify AsymptotesQuestion 1 of 20

For f(x)=x2−9x2−4x−5,f(x)=\frac{x^2-9}{x^2-4x-5}, identify the zeros and vertical asymptotes. (Factor to find features.)​

Zeros: x=3,−3x=3,-3; VA: x=5,−1x=5,-1
Zeros: x=5,−1x=5,-1; VA: x=3,−3x=3,-3
Zeros: x=3x=3 only; VA: x=5,−1x=5,-1
Zeros: x=3,−3x=3,-3; VA: x=4,−5x=4,-5
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Algebra 2 Quiz

Algebra 2 Quiz: Graph Rational Functions And Identify Asymptotes

Practice Graph Rational Functions And Identify Asymptotes in Algebra 2 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 Graph Rational Functions And Identify Asymptotes, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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

For f(x)=x2−9x2−4x−5,f(x)=\frac{x^2-9}{x^2-4x-5}, identify the zeros and vertical asymptotes. (Factor to find features.)​

  1. Zeros: x=3,−3x=3,-3; VA: x=5,−1x=5,-1 (correct answer)
  2. Zeros: x=5,−1x=5,-1; VA: x=3,−3x=3,-3
  3. Zeros: x=3x=3 only; VA: x=5,−1x=5,-1
  4. Zeros: x=3,−3x=3,-3; VA: x=4,−5x=4,-5
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x) = (x²-9)/(x²-4x-5), first factor: numerator x²-9 = (x-3)(x+3), and denominator x²-4x-5 = (x-5)(x+1). Zeros occur when (x-3)(x+3) = 0, giving x = 3 and x = -3. Vertical asymptotes occur when (x-5)(x+1) = 0, giving x = 5 and x = -1. No common factors exist, so no holes. Choice A correctly identifies zeros at x = 3, -3 and vertical asymptotes at x = 5, -1. Choice B incorrectly swaps zeros and VAs—a common mistake when not carefully tracking which features come from numerator vs denominator. The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y = 0; degrees equal → y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically!

Question 2

For f(x)=x2−1x+1,f(x)=\frac{x^2-1}{x+1}, does the graph have a vertical asymptote at x=−1x=-1? Identify any hole and give its coordinate.

  1. Yes. VA at x=−1x=-1; no hole.
  2. No. There is a hole at (−1,0)(-1,0).
  3. No. There is a hole at (−1,−2)(-1,-2). (correct answer)
  4. Yes. VA at x=−1x=-1 and also a hole at (−1,0)(-1,0).
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x) = (x2x^2 - 1)/(x + 1) = (x-1)(x+1)/(x+1), cancel (x+1), leaving x-1 with hole at x=-1 (y=-1-1=-2, so (-1,-2)); no VA since canceled. Choice C correctly states no VA at x=-1, but a hole at (-1,-2). A distractor like choice A treats it as VA without canceling, but simplifying reveals the removable discontinuity—excellent observation! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y = 0; degrees equal → y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically!

Question 3

Graph the rational function f(x)=(x−2)(x+3)(x−5)(x+1)f(x)=\frac{(x-2)(x+3)}{(x-5)(x+1)} identifying its zeros, vertical asymptotes, and horizontal asymptote (end behavior). Choose the option that lists the correct features and matches a reasonable sketch description.

  1. Zeros: x=2,−3x=2,-3; VAs: x=5,−1x=5,-1; HA: y=1y=1. Sketch crosses the xx-axis at x=2x=2 and x=−3x=-3, has dashed vertical lines at x=5x=5 and x=−1x=-1, and approaches y=1y=1 as x→±∞x\to\pm\infty. (correct answer)
  2. Zeros: x=−2,3x=-2,3; VAs: x=5,−1x=5,-1; HA: y=−1y=-1. Sketch crosses at x=−2x=-2 and x=3x=3 and approaches y=−1y=-1.
  3. Zeros: x=2,−3x=2,-3; VAs: x=5x=5 only; HA: y=0y=0. Sketch approaches the xx-axis as x→±∞x\to\pm\infty.
  4. Zeros: x=5,−1x=5,-1; VAs: x=2,−3x=2,-3; HA: y=1y=1. Sketch crosses at x=5,−1x=5,-1 and has vertical asymptotes at x=2,−3x=2,-3.
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x) = (x-2)(x+3)/((x-5)(x+1)), the zeros are at x=2 and x=-3 since the numerator factors set to zero, vertical asymptotes at x=5 and x=-1 from the denominator, and since degrees are equal (both 2), the horizontal asymptote is y=1 (ratio of leading coefficients 1/1). Choice B correctly identifies zeros at x=2 and x=-3, vertical asymptotes at x=5 and x=-1, and horizontal asymptote y=1, with a sketch that crosses the x-axis at those zeros and approaches y=1 at infinity. A common distractor like choice A swaps zeros and asymptotes, mistakenly assigning numerator roots to asymptotes, but remember, zeros come from numerator only after checking for holes. The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y = 0; degrees equal → y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically!

Question 4

For the rational function f(x)=(x−2)(x+1)(x−4)(x+3),f(x)=\frac{(x-2)(x+1)}{(x-4)(x+3)}, identify the zeros, vertical asymptotes, and horizontal asymptote (end behavior).​

  1. Zeros: x=4,−3x=4,-3; VA: x=2,−1x=2,-1; HA: y=1y=1
  2. Zeros: x=2,−1x=2,-1; VA: x=4,−3x=4,-3; HA: y=1y=1 (correct answer)
  3. Zeros: x=2,−1x=2,-1; VA: x=4,−3x=4,-3; HA: y=0y=0
  4. Zeros: x=2,−1x=2,-1; VA: x=4x=4 only; HA: y=1y=1
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x) = (x-2)(x+1)/(x-4)(x+3), zeros occur when (x-2)(x+1) = 0, giving x = 2 and x = -1. Vertical asymptotes occur when (x-4)(x+3) = 0, giving x = 4 and x = -3. Since both numerator and denominator have degree 2, the horizontal asymptote is y = 1/1 = 1 (ratio of leading coefficients). Choice B correctly identifies zeros at x = 2, -1, vertical asymptotes at x = 4, -3, and horizontal asymptote y = 1. Choice A incorrectly swaps the zeros and vertical asymptotes—remember, zeros come from numerator, VAs from denominator! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y = 0; degrees equal → y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically!

Question 5

Graph f(x)=(x−2)(x+4)x(x−3)f(x)=\frac{(x-2)(x+4)}{x(x-3)} identifying zeros, vertical asymptotes, and the horizontal asymptote.

  1. Zeros: x=2,−4x=2,-4; VA: x=0,3x=0,3; HA: y=1y=1 (correct answer)
  2. Zeros: x=0,3x=0,3; VA: x=2,−4x=2,-4; HA: y=1y=1
  3. Zeros: x=2,4x=2,4; VA: x=0,3x=0,3; HA: y=1y=1
  4. Zeros: x=2,−4x=2,-4; VA: x=0,3x=0,3; HA: y=0y=0
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x)=p(x)q(x)f(x) = \frac{p(x)}{q(x)} have distinctive features: zeros where p(x)=0p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x)=0q(x) = 0 (denominator equals zero, graph shoots to ±∞±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x−2)(x - 2)), that creates a hole (removable discontinuity) at x=2x = 2, not a zero or asymptote—the common factor cancels! For f(x)=(x−2)(x+4)x(x−3)f(x) = \frac{(x-2)(x+4)}{x(x-3)}, no common factors, so zeros at x=2x=2 and x=−4x=-4, vertical asymptotes at x=0x=0 and x=3x=3, and horizontal asymptote y=1y=1 since degrees equal. Choice C correctly identifies zeros at x=2,−4x=2,-4, vertical asymptotes at x=0,3x=0,3, and horizontal asymptote y=1y=1. A distractor like Choice B swaps zeros and asymptotes, but remember to assign numerator roots to zeros and denominator to asymptotes. The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg⁡(num)<deg⁡(den)→y=0\deg(\text{num}) < \deg(\text{den}) \to y = 0; degrees equal →y=\to y = ratio of leading coefficients; deg⁡(num)>deg⁡(den)→\deg(\text{num}) > \deg(\text{den}) \to no HA. (4) OBLIQUE ASYMPTOTE: if deg⁡(num)=deg⁡(den)+1\deg(\text{num}) = \deg(\text{den}) + 1, divide to find it. Follow these steps systematically!

Question 6

For f(x)=x2−1x−1f(x)=\frac{x^2-1}{x-1} does the graph have a vertical asymptote at x=1x=1? Identify any hole and any zero(s).

  1. Yes; VA at x=1x=1 and zero at x=−1x=-1
  2. No; hole at (1,2)(1,2) and zero at x=−1x=-1 (correct answer)
  3. No; hole at (1,0)(1,0) and zeros at x=1,−1x=1,-1
  4. Yes; VA at x=−1x=-1 and hole at (1,0)(1,0)
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! Factoring gives (x-1)(x+1)/(x-1), so cancel (x-1) for simplified x+1 with hole at x=1 where y=2, zero at x=-1, and no vertical asymptote at x=1. Choice B correctly states no vertical asymptote at x=1, with hole at (1,2) and zero at x=-1. A distractor like Choice A treats x=1 as a vertical asymptote, forgetting to check for common factors and cancellation. Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x - 3) appears in both, it cancels, creating a hole at x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x = 3 into the simplified function. Holes are easy to miss—always check for common factors! Example: (x² - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2) has a hole at x = 2 (cancels), leaving simplified f(x) = x + 2 with a gap at x = 2.

Question 7

For f(x)=(x−3)(x+1)(x−3)(x−2),f(x)=\frac{(x-3)(x+1)}{(x-3)(x-2)}, identify the zero(s), vertical asymptote(s), and any hole (removable discontinuity).

  1. Zero: x=−1x=-1; VA: x=2x=2; hole at x=3x=3 (correct answer)
  2. Zero: x=−1,3x=-1,3; VA: x=2x=2; no hole
  3. Zero: x=−1x=-1; VA: x=2,3x=2,3; no hole
  4. Zero: x=2x=2; VA: x=−1x=-1; hole at x=3x=3
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x) = [(x-3)(x+1)] / [(x-3)(x-2)], common (x-3) cancels, leaving (x+1)/(x-2) with hole at x=3; zero at x=-1, VA at x=2. Choice A correctly lists zero at x=-1, VA at x=2, hole at x=3. Distractors like choice B ignore the hole, listing zero at x=3 incorrectly—remember, holes mean undefined, not zero! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y = 0; degrees equal → y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically! Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x - 3) appears in both, it cancels, creating a hole at x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x = 3 into the simplified function. Holes are easy to miss—always check for common factors! Example: (x² - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2) has a hole at x = 2 (cancels), leaving simplified f(x) = x + 2 with a gap at x = 2. You're a pro at holes now!

Question 8

Graph f(x)=(x−3)(x+2)(x−1)(x+4)f(x)=\dfrac{(x-3)(x+2)}{(x-1)(x+4)} by identifying its zeros, vertical asymptotes, and horizontal asymptote.​

  1. Zeros: x=3,−2x=3,-2; VA: x=1,−4x=1,-4; HA: y=1y=1 (correct answer)
  2. Zeros: x=1,−4x=1,-4; VA: x=3,−2x=3,-2; HA: y=1y=1
  3. Zeros: x=3,−2x=3,-2; VA: x=1,−4x=1,-4; HA: y=0y=0
  4. Zeros: x=3,−2x=3,-2; VA: x=1x=1; HA: y=1y=1
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Horizontal asymptotes depend on degree comparison: (1) if numerator degree less than denominator degree, HA is y = 0 (graph flattens toward x-axis as x → ±∞), (2) if degrees equal, HA is y = (numerator leading coefficient)/(denominator leading coefficient) (graph approaches this horizontal line), (3) if numerator degree exceeds denominator by exactly 1, there's an oblique (slant) asymptote found by polynomial division, (4) if numerator degree exceeds by 2+, no horizontal or oblique asymptote—end behavior is more like a polynomial. These degree rules determine long-term graph behavior! For f(x) = (x-3)(x+2)/(x-1)(x+4), zeros come from (x-3)(x+2) = 0, giving x = 3 and x = -2; vertical asymptotes from (x-1)(x+4) = 0, giving x = 1 and x = -4; both polynomials have degree 2 with leading coefficient 1, so HA is y = 1/1 = 1. Choice A correctly identifies zeros at x = 3, -2, vertical asymptotes at x = 1, -4, and horizontal asymptote y = 1. Choice B reverses zeros and VAs—a common error when students forget which feature comes from which part of the fraction! Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x - 3) appears in both, it cancels, creating a hole at x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x = 3 into the simplified function. Holes are easy to miss—always check for common factors!

Question 9

For the rational function f(x)=2x2−8x2+1,f(x)=\frac{2x^2-8}{x^2+1}, what are the zeros and the horizontal asymptote? (Then you could sketch using these features.)

  1. Zeros: x=±2x=\pm 2; HA: y=2y=2. (correct answer)
  2. Zeros: x=±2x=\pm 2; HA: y=0y=0.
  3. Zeros: x=±1x=\pm 1; HA: y=2y=2.
  4. Zeros: none; HA: y=12y=\tfrac{1}{2}.
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Horizontal asymptotes depend on degree comparison: (1) if numerator degree less than denominator degree, HA is y=0y = 0 (graph flattens toward xx-axis as x→±∞x \to \pm \infty), (2) if degrees equal, HA is y=numerator leading coefficientdenominator leading coefficienty = \frac{\text{numerator leading coefficient}}{\text{denominator leading coefficient}} (graph approaches this horizontal line), (3) if numerator degree exceeds denominator by exactly 1, there's an oblique (slant) asymptote found by polynomial division, (4) if numerator degree exceeds by 2+, no horizontal or oblique asymptote—end behavior is more like a polynomial. These degree rules determine long-term graph behavior! For f(x)=2x2−8x2+1f(x) = \frac{2x^2 - 8}{x^2 + 1}, factor numerator as 2(x2−4)=2(x−2)(x+2)2(x^2 - 4) = 2(x-2)(x+2), so zeros at x=±2x=\pm 2; denominator x2+1=0x^2 + 1 = 0 has no real roots (no VAs); degrees equal (both 2), so HA y=2/1=2y=2/1=2, and the graph crosses x-axis at ±2\pm 2 while approaching y=2y=2 horizontally. Choice A correctly identifies zeros at x=±2x=\pm 2 and HA y=2y=2, allowing an accurate sketch without vertical asymptotes. A distractor like choice B uses y=0y=0, which would apply if deg(num) < deg(den), but here degrees match, so use the leading coefficient ratio instead—keep practicing those rules! Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x−3)(x - 3) appears in both, it cancels, creating a hole at x=3x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x=3x = 3 into the simplified function. Holes are easy to miss—always check for common factors! Example: x2−4x−2=(x+2)(x−2)x−2\frac{x^2 - 4}{x - 2} = \frac{(x + 2)(x - 2)}{x - 2} has a hole at x=2x = 2 (cancels), leaving simplified f(x)=x+2f(x) = x + 2 with a gap at x=2x = 2.

Question 10

Graph f(x)=x+1x−2f(x)=\frac{x+1}{x-2} identifying the zero, the vertical asymptote, and the horizontal asymptote.

  1. Zero: x=−1x=-1; VA: x=2x=2; HA: y=1y=1 (correct answer)
  2. Zero: x=2x=2; VA: x=−1x=-1; HA: y=1y=1
  3. Zero: x=−1x=-1; VA: x=2x=2; HA: y=0y=0
  4. Zero: x=−1x=-1; VA: x=2x=2; HA: y=−1y=-1
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x) = (x + 1)/(x - 2), we find the zero by setting the numerator equal to zero: x + 1 = 0, so x = -1. The vertical asymptote occurs where the denominator equals zero: x - 2 = 0, so x = 2. For the horizontal asymptote, both numerator and denominator have degree 1 (equal degrees), so HA = (leading coefficient of numerator)/(leading coefficient of denominator) = 1/1 = 1. Choice A correctly identifies zero at x = -1, vertical asymptote at x = 2, and horizontal asymptote at y = 1. Choice B incorrectly swaps the zero and vertical asymptote—remember, zeros come from numerator, vertical asymptotes from denominator! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y = 0; degrees equal → y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically! When graphing, plot the zero as an x-intercept, draw the vertical asymptote as a dashed vertical line, and sketch the horizontal asymptote as a dashed horizontal line that the graph approaches as x → ±∞.

Question 11

For the rational function f(x)=x2−9x2−4f(x)=\dfrac{x^2-9}{x^2-4}, identify the zeros, vertical asymptotes, and horizontal asymptote.​

  1. Zeros: x=±2x=\pm 2; VA: x=±3x=\pm 3; HA: y=1y=1
  2. Zeros: x=±3x=\pm 3; VA: x=±2x=\pm 2; HA: y=1y=1 (correct answer)
  3. Zeros: x=±3x=\pm 3; VA: x=±2x=\pm 2; HA: y=0y=0
  4. Zeros: x=±3x=\pm 3; VA: x=±2x=\pm 2; HA: y=94y=\tfrac{9}{4}
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x) = (x²-9)/(x²-4), factor to get f(x) = (x+3)(x-3)/[(x+2)(x-2)]; zeros occur when x²-9 = 0, so x = ±3; vertical asymptotes when x²-4 = 0, so x = ±2; since both numerator and denominator have degree 2 with leading coefficient 1, the horizontal asymptote is y = 1/1 = 1. Choice B correctly identifies zeros at x = ±3, vertical asymptotes at x = ±2, and horizontal asymptote y = 1. Choice A swaps zeros and VAs—remember to factor difference of squares: x²-9 = (x+3)(x-3) for zeros, x²-4 = (x+2)(x-2) for VAs! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y = 0; degrees equal → y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically!

Question 12

For f(x)=(x−5)(x+1)x2−25f(x)=\dfrac{(x-5)(x+1)}{x^2-25}, find the vertical asymptote(s) and any hole(s).​

  1. VA: x=5x=5; hole at x=−5x=-5
  2. VA: x=−5x=-5; hole at x=5x=5 (correct answer)
  3. VA: x=±5x=\pm 5; hole: none
  4. VA: none; hole at x=5x=5 and x=−5x=-5
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x) = (x-5)(x+1)/(x²-25), factor the denominator: x²-25 = (x-5)(x+5), so f(x) = (x-5)(x+1)/[(x-5)(x+5)]; the factor (x-5) appears in both numerator and denominator and cancels, creating a hole at x = 5; after canceling, f(x) = (x+1)/(x+5) for x ≠ 5, which has a vertical asymptote only at x = -5. Choice B correctly identifies VA at x = -5 and a hole at x = 5. Choice C incorrectly claims VAs at both x = ±5—remember to check for common factors that create holes! Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x - 3) appears in both, it cancels, creating a hole at x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x = 3 into the simplified function. Holes are easy to miss—always check for common factors! Example: (x² - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2) has a hole at x = 2 (cancels), leaving simplified f(x) = x + 2 with a gap at x = 2.

Question 13

Identify the vertical asymptote(s) and any hole(s) of f(x)=(x−1)(x+2)(x−1)(x−3)f(x)=\dfrac{(x-1)(x+2)}{(x-1)(x-3)}.​

  1. VA: x=1,3x=1,3; hole: none
  2. VA: x=3x=3; hole at x=1x=1 (correct answer)
  3. VA: x=1x=1; hole at x=3x=3
  4. VA: none; hole at x=1x=1 and x=3x=3
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x) = (x-1)(x+2)/[(x-1)(x-3)], notice that (x-1) appears in both numerator and denominator—this common factor cancels, creating a hole at x = 1, not a vertical asymptote; after canceling, we get f(x) = (x+2)/(x-3) for x ≠ 1, which has a vertical asymptote only at x = 3 (where the simplified denominator equals zero). Choice B correctly identifies VA at x = 3 and a hole at x = 1. Choice A incorrectly treats x = 1 as a vertical asymptote—remember, common factors create holes, not asymptotes! Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x - 3) appears in both, it cancels, creating a hole at x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x = 3 into the simplified function. Holes are easy to miss—always check for common factors! Example: (x² - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2) has a hole at x = 2 (cancels), leaving simplified f(x) = x + 2 with a gap at x = 2.

Question 14

Identify the vertical asymptote(s) of f(x)=x+1(x−3)(x+2).f(x)=\frac{x+1}{(x-3)(x+2)}.​​​

  1. Vertical asymptotes: x=−1x=-1
  2. Vertical asymptotes: x=3x=3 only
  3. Vertical asymptotes: x=3x=3 and x=−2x=-2 (correct answer)
  4. Vertical asymptotes: x=−2x=-2 and x=1x=1
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x) = (x + 1)/[(x - 3)(x + 2)], vertical asymptotes occur where the denominator equals zero. Setting (x - 3)(x + 2) = 0, we get x - 3 = 0 or x + 2 = 0, which gives x = 3 or x = -2. Since the numerator (x + 1) doesn't share any factors with the denominator, there are no holes—both values create vertical asymptotes. Choice C correctly identifies vertical asymptotes at x = 3 and x = -2. Choice B incorrectly lists only x = 3, missing the second asymptote at x = -2—remember to find all values that make the denominator zero! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y = 0; degrees equal → y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically! When the denominator is already factored, finding vertical asymptotes is straightforward—just set each factor equal to zero and solve.

Question 15

Describe the end behavior by identifying the horizontal asymptote of f(x)=2x−5x2+1.f(x)=\frac{2x-5}{x^2+1}.

  1. Horizontal asymptote: y=2y=2
  2. Horizontal asymptote: y=−5y=-5
  3. Horizontal asymptote: y=0y=0 (correct answer)
  4. No horizontal asymptote; slant asymptote y=2x−5y=2x-5
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Horizontal asymptotes depend on degree comparison: (1) if numerator degree less than denominator degree, HA is y=0y = 0 (graph flattens toward x-axis as x→±∞x \to \pm \infty), (2) if degrees equal, HA is y=(y = (numerator leading coefficient)/()/(denominator leading coefficient)) (graph approaches this horizontal line), (3) if numerator degree exceeds denominator by exactly 1, there's an oblique (slant) asymptote found by polynomial division, (4) if numerator degree exceeds by 2+, no horizontal or oblique asymptote—end behavior is more like a polynomial. These degree rules determine long-term graph behavior! For f(x)=2x−5x2+1f(x) = \frac{2x-5}{x^2+1}, numerator degree 1 is less than denominator degree 2, so horizontal asymptote is y=0y=0. Choice C correctly applies the degree comparison rule for numerator degree < denominator degree to find y=0y=0. A distractor like Choice D confuses it with oblique asymptote criteria, but oblique requires numerator degree exactly one higher, not lower. The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y=0y = 0; degrees equal → y=y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically!

Question 16

Which rational function has a horizontal asymptote at y=−2y = -2 and vertical asymptotes at x=1x = 1 and x=−4x = -4?

  1. f(x)=−2x2+7x−6(x−1)(x+4)f(x) = \frac{-2x^2 + 7x - 6}{(x-1)(x+4)} because this form clearly shows the desired asymptotic behavior
  2. f(x)=−2x2+x+8x2+3x−4f(x) = \frac{-2x^2 + x + 8}{x^2 + 3x - 4} because the denominator factors to (x−1)(x+4)(x-1)(x+4) giving the vertical asymptotes
  3. f(x)=−2x2+5x−3x2+3x−4f(x) = \frac{-2x^2 + 5x - 3}{x^2 + 3x - 4} because the leading coefficient ratio gives the horizontal asymptote (correct answer)
  4. f(x)=−2x3+5x2−3xx2+3x−4f(x) = \frac{-2x^3 + 5x^2 - 3x}{x^2 + 3x - 4} because higher degree numerator creates the horizontal asymptote
Explanation: When analyzing rational functions for asymptotes, you need to examine both the numerator and denominator carefully. Vertical asymptotes occur where the denominator equals zero (and the numerator doesn't), while horizontal asymptotes depend on the degrees and leading coefficients of the numerator and denominator. For vertical asymptotes at x=1x = 1 and x=−4x = -4, the denominator must factor as (x−1)(x+4)=x2+3x−4(x-1)(x+4) = x^2 + 3x - 4. You can verify this by expanding or check that the given expression x2+3x−4x^2 + 3x - 4 factors correctly. For the horizontal asymptote at y=−2y = -2, you need the numerator and denominator to have the same degree (both quadratic), with the ratio of leading coefficients equal to −2-2. Since the denominator's leading coefficient is 1, the numerator needs a leading coefficient of −2-2. Choice C gives f(x)=−2x2+5x−3x2+3x−4f(x) = \frac{-2x^2 + 5x - 3}{x^2 + 3x - 4}. The denominator factors to (x−1)(x+4)(x-1)(x+4) providing the correct vertical asymptotes, and the ratio of leading coefficients is −21=−2\frac{-2}{1} = -2, giving the horizontal asymptote y=−2y = -2. Choice A has the wrong numerator for the required horizontal asymptote. Choice B's numerator has leading coefficient −2-2, but you'd need to verify the specific form doesn't create unwanted cancellations. Choice D has a cubic numerator over a quadratic denominator, which creates no horizontal asymptote (the function grows without bound). Study tip: Always check both conditions separately—factor the denominator for vertical asymptotes, then compare degrees and leading coefficients for horizontal asymptotes.

Question 17

For f(x)=(x+2)(x−1)x2−4,f(x)=\frac{(x+2)(x-1)}{x^2-4}, find the zeros, vertical asymptote(s), and any hole. (Use factorization to decide whether a common factor creates a hole.)

  1. Zero: x=1x=1; VA: x=−2x=-2; hole at x=2x=2.
  2. Zeros: x=−2,1x=-2,1; VAs: x=−2,2x=-2,2; no holes.
  3. Zero: x=−2x=-2; VA: x=2x=2; hole at x=−2x=-2.
  4. Zero: x=1x=1; VA: x=2x=2; hole at x=−2x=-2. (correct answer)
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For f(x)=(x+2)(x−1)x2−4=(x+2)(x−1)(x+2)(x−2),f(x) = \frac{(x+2)(x-1)}{x^2 - 4} = \frac{(x+2)(x-1)}{(x+2)(x-2)}, cancel (x+2)(x+2), leaving x−1x−2\frac{x-1}{x-2} with hole at x=−2x=-2; zero at x=1x=1 from simplified numerator, VA at x=2x=2. Choice D correctly identifies zero at x=1x=1, VA at x=2x=2, and hole at x=−2x=-2. A distractor like choice A confuses the hole with a zero or VA, but after canceling, x=−2x=-2 is neither—simplify to see clearly! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y = 0; degrees equal → y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically!

Question 18

Describe the end behavior of f(x)=2x−5x2+1.f(x)=\frac{2x-5}{x^2+1}. Which statement is correct?

  1. As x→±∞x\to\pm\infty, f(x)→0f(x)\to 0 (horizontal asymptote y=0y=0). (correct answer)
  2. There is no horizontal asymptote because the degree of the numerator is greater than the degree of the denominator.
  3. As x→±∞x\to\pm\infty, f(x)→2f(x)\to 2 (horizontal asymptote y=2y=2).
  4. As x→±∞x\to\pm\infty, f(x)→12f(x)\to \frac{1}{2} (horizontal asymptote y=12y=\frac{1}{2}).
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Horizontal asymptotes depend on degree comparison: (1) if numerator degree less than denominator degree, HA is y=0y = 0 (graph flattens toward x-axis as x→±∞x \to \pm\infty), (2) if degrees equal, HA is y=y = (numerator leading coefficient)/(denominator leading coefficient) (graph approaches this horizontal line), (3) if numerator degree exceeds denominator by exactly 1, there's an oblique (slant) asymptote found by polynomial division, (4) if numerator degree exceeds by 2+, no horizontal or oblique asymptote—end behavior is more like a polynomial. These degree rules determine long-term graph behavior! For f(x)=2x−5x2+1f(x) = \frac{2x -5}{x^2 +1}, deg(num)=1 < deg(den)=2, so end behavior approaches y=0y=0 as x→±∞x\to\pm\infty. Choice B correctly states this with HA y=0y=0. Distractors like choice A might miscompare degrees or confuse coefficients, but remember: lower num degree means y=0y=0—simple! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y=0y = 0; degrees equal → y=y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically! Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x−3)(x - 3) appears in both, it cancels, creating a hole at x=3x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x=3x = 3 into the simplified function. Holes are easy to miss—always check for common factors! Example: x2−4x−2=(x+2)(x−2)x−2\frac{x^2 - 4}{x - 2} = \frac{(x + 2)(x - 2)}{x - 2} has a hole at x=2x = 2 (cancels), leaving simplified f(x)=x+2f(x) = x + 2 with a gap at x=2x = 2. Fantastic progress!

Question 19

Sketch f(x)=x+2x−1f(x)=\frac{x+2}{x-1} showing all asymptotes and intercepts. Which set of features is correct?

  1. Zero at x=−2x=-2; VA: x=1x=1; HA: y=0y=0
  2. Zero at x=1x=1; VA: x=−2x=-2; HA: y=1y=1
  3. Zero at x=2x=2; VA: x=−1x=-1; HA: y=1y=1
  4. Zero at x=−2x=-2; VA: x=1x=1; HA: y=1y=1 (correct answer)
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x)=p(x)q(x)f(x) = \frac{p(x)}{q(x)} have distinctive features: zeros where p(x)=0p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x)=0q(x) = 0 (denominator equals zero, graph shoots to ±∞\pm\infty), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x−2)(x - 2)), that creates a hole (removable discontinuity) at x=2x = 2, not a zero or asymptote—the common factor cancels! For f(x)=x+2x−1f(x) = \frac{x+2}{x-1}, zero at x=−2x=-2 (num=0), VA at x=1x=1 (den=0, no common factors), HA at y=1y=1 since degrees equal (1=11=1, coeffs 1/11/1). Choice A correctly identifies zero at x=−2x=-2, VA at x=1x=1, HA at y=1y=1. Distractors like choice C change HA to y=0y=0, perhaps thinking deg⁡(num)<deg⁡(den)\deg(\text{num})<\deg(\text{den}), but degrees match—compare carefully! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg⁡(num)<deg⁡(den)→y=0\deg(\text{num}) < \deg(\text{den}) \rightarrow y = 0; degrees equal →y=\rightarrow y = ratio of leading coefficients; deg⁡(num)>deg⁡(den)→\deg(\text{num}) > \deg(\text{den}) \rightarrow no HA. (4) OBLIQUE ASYMPTOTE: if deg⁡(num)=deg⁡(den)+1\deg(\text{num}) = \deg(\text{den}) + 1, divide to find it. Follow these steps systematically! Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x−3)(x - 3) appears in both, it cancels, creating a hole at x=3x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x=3x = 3 into the simplified function. Holes are easy to miss—always check for common factors! Example: (x2−4)/(x−2)=(x+2)(x−2)x−2(x^2 - 4)/(x - 2) = \frac{(x + 2)(x - 2)}{x - 2} has a hole at x=2x = 2 (cancels), leaving simplified f(x)=x+2f(x) = x + 2 with a gap at x=2x = 2. Keep up the awesome work!

Question 20

For f(x)=x2+5x+6x+2,f(x)=\frac{x^2+5x+6}{x+2}, determine whether there is a vertical asymptote or a hole at x=−2x=-2, and identify any remaining asymptotes.

  1. Vertical asymptote at x=−2x=-2; horizontal asymptote y=1y=1
  2. Vertical asymptote at x=−2x=-2; oblique asymptote y=x+3y=x+3
  3. Hole at x=−2x=-2; horizontal asymptote y=0y=0
  4. Hole at x=−2x=-2; oblique asymptote y=x+3y=x+3 (correct answer)
Explanation: This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Horizontal asymptotes depend on degree comparison: (1) if numerator degree less than denominator degree, HA is y=0y = 0 (graph flattens toward x-axis as x→±∞x \to \pm \infty), (2) if degrees equal, HA is y=y = (numerator leading coefficient)/(denominator leading coefficient) (graph approaches this horizontal line), (3) if numerator degree exceeds denominator by exactly 1, there's an oblique (slant) asymptote found by polynomial division, (4) if numerator degree exceeds by 2+, no horizontal or oblique asymptote—end behavior is more like a polynomial. These degree rules determine long-term graph behavior! For f(x)=x2+5x+6x+2f(x) = \frac{x^2 +5x +6}{x+2}, factoring num to (x+2)(x+3)(x+2)(x+3) reveals a common (x+2)(x+2), so hole at x=−2x=-2; simplified to x+3x+3, but original degrees 2>12>1 by 1 suggest oblique y=x+3y=x+3 via division (exact match with hole). Choice B correctly identifies the hole at x=−2x=-2 and oblique asymptote y=x+3y=x+3. Distractors like choice A or D mistake the hole for a VA, but canceling factors create holes, not asymptotes—always factor first! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y=0y = 0; degrees equal → y=y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically! Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x−3)(x - 3) appears in both, it cancels, creating a hole at x=3x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x=3x = 3 into the simplified function. Holes are easy to miss—always check for common factors! Example: x2−4x−2=(x+2)(x−2)x−2\frac{x^2 - 4}{x - 2} = \frac{(x + 2)(x - 2)}{x - 2} has a hole at x=2x = 2 (cancels), leaving simplified f(x)=x+2f(x) = x + 2 with a gap at x=2x = 2. You're mastering this—keep going!