Algebra 2 Quiz: Graph Rational Functions And Identify Asymptotes
20 questions · exam conditions
0:00
Graph Rational Functions And Identify AsymptotesQuestion 1 of 20

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.

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.
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.
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.
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.
← Back to quizzes

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

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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!

Question 9

Two rational functions are given: f(x)=x2−1x+1f(x) = \frac{x^2 - 1}{x + 1} and g(x)=x3−xx2+xg(x) = \frac{x^3 - x}{x^2 + x}. Which statement correctly compares their simplified forms and asymptotic behavior?

  1. Both functions simplify to x−1x - 1 after factoring, so they have identical graphs with no asymptotes
  2. Both functions have the same oblique asymptote y=x−1y = x - 1, but different hole locations and domains
  3. f(x)f(x) has a vertical asymptote at x=−1x = -1 and g(x)g(x) has vertical asymptotes at x=0x = 0 and x=−1x = -1
  4. f(x)f(x) simplifies to x−1x - 1 with a hole at x=−1x = -1, while g(x)g(x) simplifies to x−1x - 1 with holes at x=0x = 0 and x=−1x = -1 (correct answer)
Explanation: When analyzing rational functions, your first step should always be to factor both the numerator and denominator completely, then look for common factors that can be cancelled. This reveals the function's simplified form and identifies any holes (removable discontinuities). For f(x)=x2−1x+1f(x) = \frac{x^2 - 1}{x + 1}, factor the numerator as a difference of squares: x2−1=(x−1)(x+1)x^2 - 1 = (x-1)(x+1). This gives us f(x)=(x−1)(x+1)x+1f(x) = \frac{(x-1)(x+1)}{x+1}. The common factor (x+1)(x+1) cancels, leaving f(x)=x−1f(x) = x - 1 with a hole at x=−1x = -1 (where the cancelled factor equals zero). For g(x)=x3−xx2+xg(x) = \frac{x^3 - x}{x^2 + x}, factor out xx from both parts: g(x)=x(x2−1)x(x+1)=x(x−1)(x+1)x(x+1)g(x) = \frac{x(x^2 - 1)}{x(x + 1)} = \frac{x(x-1)(x+1)}{x(x+1)}. Two common factors cancel: xx and (x+1)(x+1), leaving g(x)=x−1g(x) = x - 1 with holes at x=0x = 0 and x=−1x = -1. Choice A is wrong because holes aren't the same as "no asymptotes" — the functions have holes, not asymptotes at these points. Choice B incorrectly calls y=x−1y = x - 1 an oblique asymptote when it's actually the simplified function itself. Choice C is wrong because neither function has vertical asymptotes; the apparent asymptote locations are actually holes due to the cancelled factors. Study tip: Always factor completely before concluding a rational function has a vertical asymptote. If a factor cancels from numerator and denominator, it creates a hole, not an asymptote.

Question 10

For f(x)=x2−4x−3,f(x)=\frac{x^2-4}{x-3}, identify the zero(s), vertical asymptote(s), and oblique asymptote.​

  1. Zeros: x=±2x=\pm 2; VA: x=3x=3; oblique asymptote y=x+3y=x+3 (correct answer)
  2. Zeros: x=3x=3; VA: x=±2x=\pm 2; oblique asymptote y=x+3y=x+3
  3. Zeros: x=±2x=\pm 2; VA: x=3x=3; horizontal asymptote y=1y=1
  4. Zeros: x=±2x=\pm 2; VA: none; oblique asymptote y=x−3y=x-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). 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) = (x²-4)/(x-3), factor the numerator: x²-4 = (x-2)(x+2). No common factors exist with denominator (x-3). Zeros occur when (x-2)(x+2) = 0, giving x = 2 and x = -2. Vertical asymptote occurs when x-3 = 0, giving x = 3. Since numerator has degree 2 and denominator has degree 1 (degree difference = 1), there's an oblique asymptote. Perform polynomial division: (x²-4)÷(x-3) = x+3 with remainder 5, so f(x) = x+3 + 5/(x-3). The oblique asymptote is y = x+3. Choice A correctly identifies zeros at x = ±2, VA at x = 3, and oblique asymptote y = x+3. Choice D incorrectly claims no VA and gives the wrong oblique asymptote equation. 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 11

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 12

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

  1. Zeros: x=2,−1x=2,-1; VA: x=4,−3x=4,-3; HA: y=1y=1 (correct answer)
  2. Zeros: x=2,−1x=2,-1; VA: x=4,−3x=4,-3; HA: y=0y=0
  3. Zeros: x=4,−3x=4,-3; VA: x=2,−1x=2,-1; HA: y=1y=1
  4. Zeros: x=2,−1x=2,-1; VA: x=4x=4; 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)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+1)(x−4)(x+3)f(x) = \frac{(x-2)(x+1)}{(x-4)(x+3)}, the zeros are at x=2x=2 and x=−1x=-1 where the numerator is zero, vertical asymptotes at x=4x=4 and x=−3x=-3 where the denominator is zero (no common factors), and since degrees are equal, the horizontal asymptote is y=1y=1 from the ratio of leading coefficients. Choice B correctly identifies zeros at x=2,−1x=2,-1, vertical asymptotes at x=4,−3x=4,-3, and horizontal asymptote y=1y=1. A common distractor like Choice A swaps zeros and asymptotes, mistakenly assigning denominator roots to zeros, but remember, zeros come from the numerator only. 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: if deg⁡(num)<deg⁡(den)\deg(\text{num}) < \deg(\text{den}) then y=0y = 0; if degrees equal then y=y = ratio of leading coefficients; if deg⁡(num)>deg⁡(den)\deg(\text{num}) > \deg(\text{den}) then 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 13

Describe the end behavior of f(x)=3x−1x2+4.f(x)=\frac{3x-1}{x^2+4}. Which statement is correct?​

  1. As x→±∞x\to\pm\infty, f(x)→3f(x)\to 3 (horizontal asymptote y=3y=3).
  2. As x→±∞x\to\pm\infty, f(x)→0f(x)\to 0 (horizontal asymptote y=0y=0). (correct answer)
  3. As x→±∞x\to\pm\infty, f(x)→13f(x)\to \frac{1}{3} (horizontal asymptote y=13y=\frac{1}{3}).
  4. As x→±∞x\to\pm\infty, f(x)f(x) approaches the slant asymptote y=3x−1y=3x-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) = (3x-1)/(x²+4), the numerator has degree 1 and denominator has degree 2. Since degree(numerator) < degree(denominator), the horizontal asymptote is y = 0. As x → ±∞, the denominator grows much faster than the numerator, causing f(x) → 0. Think of it as 3x/x² ≈ 3/x → 0 for large |x|. Choice B correctly states that as x → ±∞, f(x) → 0 with horizontal asymptote y = 0. Choice A incorrectly claims y = 3, likely confusing this with the case of equal degrees, while Choice D incorrectly suggests a slant asymptote when the numerator degree is actually less than the denominator degree. 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 14

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 15

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 16

For f(x)=x2−9x+2,f(x)=\frac{x^2-9}{x+2}, identify the zero(s), vertical asymptote(s), and oblique asymptote.​​​

  1. Zeros: x=±3x=\pm 3; VA: x=−2x=-2; oblique asymptote: y=x−2y=x-2 (correct answer)
  2. Zeros: x=−2x=-2; VAs: x=±3x=\pm 3; oblique asymptote: y=x−2y=x-2
  3. Zeros: x=±3x=\pm 3; VA: x=−2x=-2; horizontal asymptote: y=1y=1
  4. Zeros: x=±3x=\pm 3; VA: x=2x=2; oblique asymptote: y=x−2y=x-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). 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)/(x + 2), first factor the numerator: x² - 9 = (x + 3)(x - 3). So f(x) = (x + 3)(x - 3)/(x + 2). Zeros occur when numerator = 0: (x + 3)(x - 3) = 0 gives x = -3 and x = 3. The vertical asymptote occurs when denominator = 0: x + 2 = 0 gives x = -2. Since the numerator has degree 2 and denominator has degree 1, and deg(num) = deg(den) + 1, there's an oblique asymptote. Dividing x² - 9 by x + 2: First term: x² ÷ x = x. Multiply: x(x + 2) = x² + 2x. Subtract: (x² - 9) - (x² + 2x) = -2x - 9. Second term: -2x ÷ x = -2. Multiply: -2(x + 2) = -2x - 4. Subtract: (-2x - 9) - (-2x - 4) = -5. So f(x) = x - 2 + (-5)/(x + 2), giving oblique asymptote y = x - 2. Choice A correctly identifies zeros at x = ±3, VA at x = -2, and oblique asymptote y = x - 2. Choice D incorrectly has VA at x = 2 instead of x = -2—remember to solve x + 2 = 0, not confuse the sign! 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! Always factor completely before identifying features—this helps spot patterns and makes calculations easier.

Question 17

Find the oblique (slant) asymptote of f(x)=x2+3x+2x+1.f(x)=\frac{x^2+3x+2}{x+1}.​

  1. y=x+2y=x+2 (correct answer)
  2. y=x+1y=x+1
  3. y=1y=1
  4. No oblique asymptote; horizontal asymptote 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) = (x²+3x+2)/(x+1), the numerator has degree 2 and denominator has degree 1, so degree(num) = degree(den) + 1, indicating an oblique asymptote. To find it, perform polynomial division: (x²+3x+2)÷(x+1). Using long division or factoring first: x²+3x+2 = (x+1)(x+2), so f(x) = (x+1)(x+2)/(x+1) = x+2 (except at x = -1 where there's a hole). The oblique asymptote is y = x+2. Choice A correctly identifies the oblique asymptote as y = x+2. Choice B incorrectly gives y = x+1, likely from a division error, while Choice D incorrectly claims a horizontal asymptote when the degree difference clearly indicates an oblique asymptote. 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

Sketch f(x)=x+1(x−2)(x+3)f(x)=\frac{x+1}{(x-2)(x+3)} showing all vertical asymptotes, the horizontal asymptote, and the zero. Which option correctly identifies these features?

  1. Zero: x=−1x=-1; VAs: x=1x=1 and x=−3x=-3; HA: y=0y=0.
  2. Zero: x=−1x=-1; VAs: x=2,−3x=2,-3; HA: y=0y=0. (correct answer)
  3. Zero: x=−1x=-1; VA: x=2x=2 only; HA: y=0y=0.
  4. Zero: x=1x=1; VAs: x=2,−3x=2,-3; 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+1)/((x-2)(x+3)), zero at x=-1 from numerator; VAs at x=2 and x=-3 from denominator; deg num 1 < deg den 2, so HA y=0. Choice A correctly identifies zero at x=-1, VAs at x=2 and -3, and HA y=0 for sketching. A distractor like choice B shifts values, perhaps misfactoring, but double-check roots—keep up the great work! 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 19

For the rational function f(x)=x2+x−6x−2,f(x)=\frac{x^2+x-6}{x-2}, what is the correct description of the discontinuity at x=2x=2 and the end behavior asymptote?​

  1. Vertical asymptote at x=2x=2; horizontal asymptote y=1y=1
  2. Hole at x=2x=2; horizontal asymptote y=1y=1
  3. Hole at x=2x=2; oblique asymptote y=x+3y=x+3 (correct answer)
  4. Vertical asymptote at x=2x=2; oblique asymptote y=x+3y=x+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). 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) = (x²+x-6)/(x-2), first factor the numerator: x²+x-6 = (x+3)(x-2). Notice (x-2) appears in both numerator and denominator! This common factor cancels, creating a hole at x = 2, not a vertical asymptote. After canceling, f(x) = x+3 (except at x = 2). Since the original function has degree 2 numerator and degree 1 denominator (degree difference = 1), we perform polynomial division: (x²+x-6)÷(x-2) = x+3 with remainder 0, confirming the oblique asymptote y = x+3. Choice C correctly identifies a hole at x = 2 and oblique asymptote y = x+3. Choice A incorrectly claims a vertical asymptote at x = 2—missing that the common factor creates a hole instead. 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 20

For f(x)=x2−4x2−x−6,f(x)=\frac{x^2-4}{x^2-x-6}, identify the zeros and vertical asymptotes. (Factor to find them.)

  1. Zeros: x=2,−2x=2,-2; VA: x=3,−2x=3,-2
  2. Zeros: x=2,−2x=2,-2; VA: x=3,−2x=3,-2 with a hole at x=−2x=-2
  3. Zeros: x=2,−2x=2,-2; VA: x=3,−2x=3,-2 and HA: y=0y=0
  4. Zeros: x=2x=2 only; VA: x=3x=3 only (since x+2x+2 cancels) (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 = 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) = (x² - 4)/(x² - x - 6), we first factor completely: numerator x² - 4 = (x - 2)(x + 2), and denominator x² - x - 6 = (x - 3)(x + 2). This gives f(x) = [(x - 2)(x + 2)]/[(x - 3)(x + 2)]. Notice the common factor (x + 2) in both numerator and denominator—this cancels, creating a hole at x = -2, not a zero or vertical asymptote! After canceling, we get the simplified function f(x) = (x - 2)/(x - 3), which has zero at x = 2 and vertical asymptote at x = 3. Choice D correctly identifies that x = 2 is the only zero and x = 3 is the only vertical asymptote, recognizing that the (x + 2) factor cancels. Choice A incorrectly treats x = -2 as both a zero and a vertical asymptote, missing that the common factor creates a hole instead. 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.