Unit 1 · Topic 1.9

Rational Functions and Vertical Asymptotes

A rational function has a vertical asymptote x = a where (x − a) divides the denominator more times than the numerator.

Practice 2: Multiple RepresentationsPractice 3: Communication and Reasoning

Part 1 · Hook

Why this matters

Squeeze a gas into a smaller and smaller space and its pressure climbs without limit. On a graph that is a curve racing up beside a vertical line it never touches: a vertical asymptote. The exam asks you to describe it from each side with one-sided limits, and it takes points for loose notation.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Which inputs make x+2x2−4x\frac{x + 2}{x^2 - 4x} undefined?

  1. x=0x = 0 and x=4x = 4
  2. x=−2x = -2
  3. x=4x = 4
  4. x=±2x = \pm 2
Show the answer

x(x−4)=0x(x - 4) = 0.

  • Correct: x=0x = 0 and x=4x = 4:
  • x=−2x = -2:
  • x=4x = 4:
  • x=±2x = \pm 2:

2. At a zero of multiplicity 22, the graph of a polynomial

  1. touches the axis without crossing
  2. crosses the axis
  3. has a break
  4. is vertical
Show the answer

An even power does not change sign.

  • Correct: touches the axis without crossing:
  • crosses the axis:
  • has a break:
  • is vertical:

3. What is limx→∞3x2−3x2+2\lim_{x \to \infty} \frac{3x^2 - 3}{x^2 + 2}?

  1. 33
  2. 00
  3. ∞\infty
  4. −32-\frac{3}{2}
Show the answer

Equal degrees: 31\frac{3}{1}.

  • Correct: 33:
  • 00:
  • ∞\infty:
  • −32-\frac{3}{2}:

Part 4 · See it

See it first

The graph of r(x) = x/((x − 1)(x + 2)) with dashed vertical asymptotes x = −2 and x = 1. As x approaches −2 from the left the outputs decrease without bound, and from the right they increase without bound. As x approaches 1 from the left the outputs decrease without bound, and from the right they increase without bound.
r(x)r(x)=x(x−1)(x+2)= \frac{x}{(x - 1)(x + 2)} has vertical asymptotes x=−2x = -2 and x=1x = 1; at x=1x = 1, limx→1−r(x)=−∞\lim_{x \to 1^-} r(x) = -\infty and limx→1+r(x)=∞\lim_{x \to 1^+} r(x) = \infty. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. (x−a)(x - a) is a factor of the denominator, with higher multiplicity than in the numeratorx=ax = a is a vertical asymptote: outputs grow without bound in size near aa
  2. The factor appears equally often (or more) in the numeratorit cancels and there is no vertical asymptote at aa
  3. You test the signs of the factors just left and just right of aayou get each one-sided limit, ∞\infty or −∞-\infty
  4. The leftover power of (x−a)(x - a) is evenboth one-sided limits have the same sign

Part 6 · Key ideas

Key ideas

  • Vertical asymptote x=ax = a: the outputs increase or decrease without bound as xx approaches aa from one side or both.
  • For a rational function: a zero of the denominator with greater multiplicity than in the numerator.
  • One-sided limits: limx→a−\lim_{x \to a^-} from the left, limx→a+\lim_{x \to a^+} from the right, each =∞= \infty or =−∞= -\infty.
  • Find the sign with a test input just beside aa, such as a−0.1a - 0.1 and a+0.1a + 0.1.

Part 7 · Misconception

A common mistake

The wrong idea: Every zero of the denominator is a vertical asymptote.

What actually happens: Cancel common factors first. x−3(x−3)(x+2)\frac{x - 3}{(x - 3)(x + 2)} has a zero denominator at x=3x = 3, but the factor cancels and the outputs approach 15\frac{1}{5} there. Its only vertical asymptote is x=−2x = -2.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of f(x)f(x)=x+1(x−2)2= \frac{x + 1}{(x - 2)^2}

−6−5−4−3−2−112345678−2−112345678xyf

1. What is the vertical asymptote of the graph of ff?

  1. x=−1x = -1
  2. x=2x = 2
  3. y=0y = 0
  4. x=4x = 4
Show the answer

x=2x = 2 makes the denominator 00 while the numerator is 33. The outputs grow without bound near 22, so x=2x = 2 is a vertical asymptote.

  • x=−1x = -1: At x=−1x = -1 the numerator is 00: that is a zero of ff, not an asymptote.
  • Correct: x=2x = 2: Right: the denominator is 00 at x=2x = 2 and the numerator is not.
  • y=0y = 0: y=0y = 0 is the horizontal asymptote.
  • x=4x = 4: (x−2)2(x - 2)^2 is 00 at x=2x = 2; squaring does not change where.

2. Which pair of statements describes ff near x=2x = 2?

  1. limx→2−f(x)=∞\lim_{x \to 2^-} f(x) = \infty and limx→2+f(x)=∞\lim_{x \to 2^+} f(x) = \infty
  2. limx→2−f(x)=−∞\lim_{x \to 2^-} f(x) = -\infty and limx→2+f(x)=∞\lim_{x \to 2^+} f(x) = \infty
  3. limx→2−f(x)=∞\lim_{x \to 2^-} f(x) = \infty and limx→2+f(x)=−∞\lim_{x \to 2^+} f(x) = -\infty
  4. f(2)=∞f(2) = \infty
Show the answer

Near x=2x = 2 the numerator is near 33 (positive) and (x−2)2(x - 2)^2 is a small positive number on both sides, so f(x)f(x) is large and positive on both sides.

  • Correct: limx→2−f(x)=∞\lim_{x \to 2^-} f(x) = \infty and limx→2+f(x)=∞\lim_{x \to 2^+} f(x) = \infty: Right: both sides rise without bound.
  • limx→2−f(x)=−∞\lim_{x \to 2^-} f(x) = -\infty and limx→2+f(x)=∞\lim_{x \to 2^+} f(x) = \infty: That pattern goes with an odd power in the denominator; here (x−2)2(x - 2)^2 is never negative.
  • limx→2−f(x)=∞\lim_{x \to 2^-} f(x) = \infty and limx→2+f(x)=−∞\lim_{x \to 2^+} f(x) = -\infty: The right side also rises: f(2.1)=3.10.01=310f(2.1) = \frac{3.1}{0.01} = 310.
  • f(2)=∞f(2) = \infty: f(2)f(2) is undefined; describe what happens near 22 with one-sided limits.

3. Why does ff have the same one-sided behavior on both sides of x=2x = 2?

  1. Because near x=2x = 2 the numerator is positive and (x−2)2(x - 2)^2 is positive on both sides
  2. Because x=2x = 2 is a zero of the numerator, so both sides behave the same way
  3. Because ff has a horizontal asymptote, so both ends of the graph behave alike
  4. Because the graph is symmetric about the vertical line x=2x = 2
Show the answer

The sign of ff near the asymptote comes from the signs of its factors. An even power of (x−2)(x - 2) stays positive, so both one-sided limits have the sign of the numerator, ++.

  • Correct: Because near x=2x = 2 the numerator is positive and (x−2)2(x - 2)^2 is positive on both sides: Right: the denominator's factor has an even power, so it does not change sign at 22.
  • Because x=2x = 2 is a zero of the numerator, so both sides behave the same way: The numerator is 33 at x=2x = 2, not 00.
  • Because ff has a horizontal asymptote, so both ends of the graph behave alike: End behavior says nothing about the outputs near x=2x = 2.
  • Because the graph is symmetric about the vertical line x=2x = 2: It is not: f(1)=2f(1) = 2 but f(3)=4f(3) = 4.

Table

Outputs near x=1x = 1

Values of g(x) = (x^2 + 1)/(x - 1) for inputs close to 1, rounded to two decimal places
xx0.90.990.9991.0011.011.1
g(x)g(x)-18.10-198.01-1998.002002.00202.0122.10

4. The table gives values of g(x)=x2+1x−1g(x) = \frac{x^2 + 1}{x - 1} near x=1x = 1. Which describes gg as xx approaches 11 from the left?

  1. limx→1−g(x)=∞\lim_{x \to 1^-} g(x) = \infty
  2. limx→1−g(x)=−∞\lim_{x \to 1^-} g(x) = -\infty
  3. limx→1−g(x)=2\lim_{x \to 1^-} g(x) = 2
  4. limx→1+g(x)=−∞\lim_{x \to 1^+} g(x) = -\infty
Show the answer

For xx just below 11, the top is near 22 and the bottom is a small negative number, so g(x)g(x) is a large negative number: limx→1−g(x)=−∞\lim_{x \to 1^-} g(x) = -\infty.

  • limx→1−g(x)=∞\lim_{x \to 1^-} g(x) = \infty: The values to the left of 11 are negative: −18.10-18.10, −198.01-198.01, −1998.00-1998.00.
  • Correct: limx→1−g(x)=−∞\lim_{x \to 1^-} g(x) = -\infty: Right: the outputs are negative and grow without bound in size.
  • limx→1−g(x)=2\lim_{x \to 1^-} g(x) = 2: 22 is the numerator's value at x=1x = 1, not the limit.
  • limx→1+g(x)=−∞\lim_{x \to 1^+} g(x) = -\infty: The notation 1+1^+ means from the right; the question asks about the left.

5. The table gives values of g(x)=x2+1x−1g(x) = \frac{x^2 + 1}{x - 1} near x=1x = 1. Which describes gg as xx approaches 11 from the right?

  1. limx→1+g(x)=∞\lim_{x \to 1^+} g(x) = \infty
  2. limx→1+g(x)=−∞\lim_{x \to 1^+} g(x) = -\infty
  3. limx→1+g(x)=0\lim_{x \to 1^+} g(x) = 0
  4. g(1)=∞g(1) = \infty
Show the answer

Just right of 11, x−1x - 1 is a small positive number and the top is near 22: limx→1+g(x)=∞\lim_{x \to 1^+} g(x) = \infty.

  • Correct: limx→1+g(x)=∞\lim_{x \to 1^+} g(x) = \infty: Right: the outputs 202.01202.01, 2002.002002.00 grow without bound.
  • limx→1+g(x)=−∞\lim_{x \to 1^+} g(x) = -\infty: To the right of 11 the bottom is positive, so the outputs are positive.
  • limx→1+g(x)=0\lim_{x \to 1^+} g(x) = 0: The outputs grow as xx gets closer to 11, not shrink.
  • g(1)=∞g(1) = \infty: g(1)g(1) is undefined. Use a one-sided limit.

6. Which vertical asymptotes does h(x)h(x)=x−3(x−3)(x+2)= \frac{x - 3}{(x - 3)(x + 2)} have?

  1. x=3x = 3 and x=−2x = -2
  2. x=−2x = -2
  3. x=3x = 3
  4. y=0y = 0
Show the answer

A vertical asymptote sits at a zero of the denominator whose multiplicity is greater than in the numerator. x−3x - 3 appears once in each, so x=3x = 3 is not an asymptote; x+2x + 2 appears only below, so x=−2x = -2 is.

  • x=3x = 3 and x=−2x = -2: The factor x−3x - 3 cancels: near x=3x = 3 the outputs approach 15\frac{1}{5} instead of growing.
  • Correct: x=−2x = -2: Right: after cancelling, 1x+2\frac{1}{x + 2} still has a zero denominator at −2-2.
  • x=3x = 3: At x=3x = 3 the numerator is 00 too, with the same multiplicity, so there is no asymptote there.
  • y=0y = 0: That is the horizontal asymptote.

Part 9 · Summary

Summary

A rational function has a vertical asymptote x=ax = a where (x−a)(x - a) divides the denominator more times than the numerator. Describe it from each side with one-sided limits, limx→a−f(x)\lim_{x \to a^-} f(x) and limx→a+f(x)\lim_{x \to a^+} f(x), each ∞\infty or −∞-\infty, found from the signs of the factors just beside aa.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic