AP Precalculus · Lesson 2

End behavior and rational functions

Know Know it

For large inputs, the leading term of a polynomial outweighs the rest, so it decides the end behavior. Even degree: both ends go the same way. Odd degree: the ends go opposite ways. A positive leading coefficient sends the right end up.

A rational function is a ratio r(x)=p(x)q(x)r(x)=\frac{p(x)}{q(x)} of polynomials. Its end behavior comes from the ratio of the leading terms. If the numerator has the lower degree, the horizontal asymptote is y=0y=0. If the degrees match, it is y=y= the ratio of the leading coefficients. If the numerator degree is one more, there is a slant asymptote instead.

Factor first. A zero of the denominator gives a vertical asymptote if it does not cancel. If the same factor appears at least as many times in the numerator, it cancels and leaves a hole. The zeros of rr are the zeros of the numerator that are not zeros of the denominator.

Trap Cancel before you call it an asymptote

In (x−2)(x+1)(x−2)(x−3)\frac{(x-2)(x+1)}{(x-2)(x-3)}, the denominator is 0 at x=2x=2, but the factor x−2x-2 cancels. The graph has a hole at x=2x=2, not a vertical asymptote. Only x=3x=3 is a vertical asymptote.

Worked example Easy

Let f(x)=−2x5+4x2−7f(x)=-2x^5+4x^2-7. What happens to f(x)f(x) as xx decreases without bound?

  1. f(x)f(x) decreases without bound
  2. f(x)f(x) approaches -7
  3. f(x)f(x) approaches 0
  4. f(x)f(x) increases without boundAnswer

How to solve it

  1. Find the leading term: −2x5-2x^5.
  2. For xx far to the left, x5x^5 is hugely negative.
  3. Multiplying by −2-2 makes it hugely positive, so f(x)f(x) increases without bound.

Why each choice is right or wrong

  • A. Incorrect. Decreasing without bound is the right-hand behavior; on the left, x5x^5 is negative and times -2 becomes positive.
  • B. Incorrect. The value -7 is the yy-intercept, f(0)f(0), and says nothing about very large inputs.
  • C. Incorrect. Approaching 0 describes some rational functions, but a nonconstant polynomial never levels off.
  • D. Correct. The leading term −2x5-2x^5 rules: for a large negative xx, x5x^5 is a large negative number, and −2-2 times it is a large positive number.

Explain Explain it

Task verbs: determine, justify

To justify an asymptote or a hole, show the factored form and say which factor cancels and which one stays in the denominator.

Worked example Medium

Let r(x)=3x2−12x2−x−2r(x)=\frac{3x^2-12}{x^2-x-2}. Which statement lists the vertical asymptotes and holes of the graph of rr?

  1. Vertical asymptote x=−1x=-1; hole at x=2x=2Answer
  2. Vertical asymptotes x=−1x=-1 and x=2x=2; no hole
  3. Vertical asymptote x=2x=2; hole at x=−1x=-1
  4. Vertical asymptote x=1x=1; hole at x=−2x=-2

How to solve it

  1. Factor: 3x2−12=3(x−2)(x+2)3x^2-12=3(x-2)(x+2) and x2−x−2=(x−2)(x+1)x^2-x-2=(x-2)(x+1).
  2. The factor x−2x-2 appears once on top and once below, so it cancels: a hole at x=2x=2, at the point (2,4)(2,4).
  3. The factor x+1x+1 stays in the denominator: a vertical asymptote at x=−1x=-1.

Why each choice is right or wrong

  • A. Correct. r(x)=3(x−2)(x+2)(x−2)(x+1)r(x)=\frac{3(x-2)(x+2)}{(x-2)(x+1)}. The factor x−2x-2 cancels, leaving a hole at x=2x=2, and x+1x+1 stays below, giving the asymptote x=−1x=-1.
  • B. Incorrect. Calling x=2x=2 an asymptote misses that the factor x−2x-2 is in the numerator too and cancels.
  • C. Incorrect. This swaps the roles: x−2x-2 is the factor that cancels, and x+1x+1 is the one left in the denominator.
  • D. Incorrect. The signs are flipped: the denominator is 0 at x=2x=2 and x=−1x=-1, not at x=1x=1 or x=−2x=-2.

Justify Your turn

Let r(x)=x+4x2−16r(x)=\frac{x+4}{x^2-16}. Justify why the graph of rr has a vertical asymptote at x=4x=4 but only a hole at x=−4x=-4.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Factors the denominator as (x−4)(x+4)(x-4)(x+4) and says the factor x+4x+4 cancels, so x=−4x=-4 is a hole
  • Says the factor x−4x-4 stays in the denominator, so the outputs grow without bound near x=4x=4

Model answer

Factoring gives r(x)=x+4(x−4)(x+4)r(x)=\frac{x+4}{(x-4)(x+4)}. The factor x+4x+4 cancels, so for x≠−4x \ne -4 the function equals 1x−4\frac{1}{x-4}, and x=−4x=-4 is only a hole. The factor x−4x-4 does not cancel: near x=4x=4 the numerator is close to 8 while the denominator is close to 0, so the outputs grow without bound and x=4x=4 is a vertical asymptote.

Apply Apply it

Practice these topics. AP Practice questions on exactly what this lesson covers, with an explanation for every answer.