Unit 1 · Topic 1.7

Rational Functions and End Behavior

A rational function's end behavior matches the quotient of its leading terms.

Practice 2: Multiple RepresentationsPractice 3: Communication and Reasoning

Part 1 · Hook

Why this matters

Order more posters and the average cost of each one drops, but never below the cost of the paper and ink. That floor is a horizontal asymptote. Functions that divide one polynomial by another level off, grow without bound, or hug a slanted line, and the leading terms tell you which.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is limx→−∞(2x3−50x2)\lim_{x \to -\infty} (2x^3 - 50x^2)?

  1. −∞-\infty
  2. ∞\infty
  3. 22
  4. 00
Show the answer

The leading term 2x32x^3 is negative and huge for large negative xx.

  • Correct: −∞-\infty:
  • ∞\infty:
  • 22:
  • 00:

2. Which is equal to 2x2+3x\frac{2x^2 + 3}{x} for x≠0x \ne 0?

  1. 2x+3x2x + \frac{3}{x}
  2. 2x+32x + 3
  3. 2x2+3x2x^2 + \frac{3}{x}
  4. 2x+31\frac{2x + 3}{1}
Show the answer

Divide each term by xx.

  • Correct: 2x+3x2x + \frac{3}{x}:
  • 2x+32x + 3:
  • 2x2+3x2x^2 + \frac{3}{x}:
  • 2x+31\frac{2x + 3}{1}:

Part 4 · See it

See it first

The graph of r(x) = (2x² − 8)/(x² + 1). Its zeros are x = −2 and x = 2 and its y-intercept is (0, −8). For large positive and negative x the graph gets close to the dashed horizontal asymptote y = 2, the quotient of the leading terms 2x²/x².
r(x)r(x)=2x2−8x2+1= \frac{2x^2 - 8}{x^2 + 1} levels off at the horizontal asymptote y=2y = 2, the quotient of its leading terms 2x2x2\frac{2x^2}{x^2}. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. |x||x| grows large in p(x)q(x)\frac{p(x)}{q(x)}each polynomial behaves like its leading term, so the function behaves like their quotient
  2. The quotient of leading terms is a constant cc (equal degrees)limx→±∞r(x)=c\lim_{x \to \pm\infty} r(x) = c and y=cy = c is a horizontal asymptote
  3. The denominator has the higher degreethe quotient goes to 00: the horizontal asymptote is y=0y = 0
  4. The quotient of leading terms is linear (top degree one more)the graph approaches a slant asymptote y=mx+by = mx + b

Part 6 · Key ideas

Key ideas

  • A rational function is r(x)=p(x)q(x)r(x) = \frac{p(x)}{q(x)} with polynomials pp and qq.
  • End behavior comes from the quotient of the leading terms, written in limit notation.
  • Horizontal asymptote y=cy = c when limx→∞r(x)=c\lim_{x \to \infty} r(x) = c or limx→−∞r(x)=c\lim_{x \to -\infty} r(x) = c: equal degrees give the ratio of leading coefficients; a bigger bottom degree gives y=0y = 0.
  • Slant asymptote when the top's degree is one more: rewrite as a line plus a fraction that goes to 00.

Part 7 · Misconception

A common mistake

The wrong idea: A graph can never cross its horizontal asymptote.

What actually happens: A horizontal asymptote describes the ends only. 5x+2x2+1\frac{5x + 2}{x^2 + 1} has asymptote y=0y = 0 yet crosses it at x=−0.4x = -0.4.

Part 8 · Check yourself

Check yourself

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

Graph

The graph of s(x)s(x)=3x2−3x2+2= \frac{3x^2 - 3}{x^2 + 2}

−10−8−6−4−2246810−2−11234xys

1. Which describes the right end of the graph of ss?

  1. limx→∞s(x)=∞\lim_{x \to \infty} s(x) = \infty
  2. limx→∞s(x)=0\lim_{x \to \infty} s(x) = 0
  3. limx→∞s(x)=3\lim_{x \to \infty} s(x) = 3
  4. limx→∞s(x)\lim_{x \to \infty} s(x)=−1.5= -1.5
Show the answer

For large xx, s(x)s(x) behaves like 3x2x2=3\frac{3x^2}{x^2} = 3: limx→∞s(x)=3\lim_{x \to \infty} s(x) = 3.

  • limx→∞s(x)=∞\lim_{x \to \infty} s(x) = \infty: Top and bottom both have degree 22, so their quotient levels off.
  • limx→∞s(x)=0\lim_{x \to \infty} s(x) = 0: 00 is the limit when the denominator has the higher degree; here the degrees are equal.
  • Correct: limx→∞s(x)=3\lim_{x \to \infty} s(x) = 3: Right: the quotient of the leading terms is 3x2x2=3\frac{3x^2}{x^2} = 3.
  • limx→∞s(x)\lim_{x \to \infty} s(x)=−1.5= -1.5: −1.5-1.5 is s(0)s(0), the yy-intercept.

2. What is the horizontal asymptote of the graph of ss?

  1. y=3y = 3
  2. y=0y = 0
  3. x=3x = 3
  4. y=−32y = -\frac{3}{2}
Show the answer

limx→±∞s(x)=3\lim_{x \to \pm\infty} s(x) = 3, so the line y=3y = 3 is a horizontal asymptote.

  • Correct: y=3y = 3: Right: the outputs approach 33 as x→±∞x \to \pm\infty.
  • y=0y = 0: y=0y = 0 would need the denominator to have the higher degree.
  • x=3x = 3: A horizontal asymptote is a horizontal line, y=cy = c.
  • y=−32y = -\frac{3}{2}: That is the ratio of the constant terms, s(0)s(0); end behavior uses the leading terms.

3. For large values of |x||x|, are the outputs of ss above or below 33?

  1. Above 33, because the graph rises toward the asymptote
  2. Below 33, because s(x)=3−9x2+2s(x) = 3 - \frac{9}{x^2 + 2}
  3. Exactly 33 once xx is large enough
  4. Above on the right and below on the left
Show the answer

3x2−3=3(x2+2)−93x^2 - 3 = 3(x^2 + 2) - 9, so s(x)=3−9x2+2s(x) = 3 - \frac{9}{x^2 + 2}. The fraction is positive and shrinks toward 00: the outputs climb toward 33 from below.

  • Above 33, because the graph rises toward the asymptote: Rising toward y=3y = 3 from below means the outputs are under 33.
  • Correct: Below 33, because s(x)=3−9x2+2s(x) = 3 - \frac{9}{x^2 + 2}: Right: the subtracted fraction is positive, so s(x)<3s(x) < 3 for every xx.
  • Exactly 33 once xx is large enough: 9x2+2\frac{9}{x^2 + 2} gets small but never reaches 00.
  • Above on the right and below on the left: ss is even: s(−x)=s(x)s(-x) = s(x), so both sides behave alike.

Table

Outputs far from the origin

Values of g(x) = (5x + 2)/(x^2 + 1), rounded to four decimal places
xx-100-10101001000
g(x)g(x)-0.0498-0.47520.51490.05020.0050

4. The table gives values of g(x)=5x+2x2+1g(x) = \frac{5x + 2}{x^2 + 1}. What is limx→∞g(x)\lim_{x \to \infty} g(x)?

  1. 55
  2. 00
  3. ∞\infty
  4. 22
Show the answer

5xx2=5x\frac{5x}{x^2} = \frac{5}{x}, which goes to 00. The table agrees: 0.51490.5149, 0.05020.0502, 0.00500.0050.

  • 55: 55 is the leading coefficient of the top; the bottom's degree is higher.
  • Correct: 00: Right: the outputs shrink toward 00 as xx grows.
  • ∞\infty: The outputs get smaller, not larger: 0.00500.0050 at x=1000x = 1000.
  • 22: 22 is the constant term of the numerator.

5. The table gives values of g(x)=5x+2x2+1g(x) = \frac{5x + 2}{x^2 + 1}. Why does gg have the horizontal asymptote y=0y = 0?

  1. Because g(0)=2g(0) = 2
  2. Because the numerator has a zero at x=−0.4x = -0.4
  3. Because the denominator has the higher degree
  4. Because the outputs are negative when x<0x < 0
Show the answer

When the denominator's degree is greater, the quotient of leading terms (5x\frac{5}{x} here) goes to 00 at both ends.

  • Because g(0)=2g(0) = 2: The value at 00 does not describe the ends of the graph.
  • Because the numerator has a zero at x=−0.4x = -0.4: One zero says nothing about large |x||x|.
  • Correct: Because the denominator has the higher degree: Right: x2x^2 outgrows 5x5x, so the quotient shrinks toward 00.
  • Because the outputs are negative when x<0x < 0: The sign tells which side of 00 the outputs are on, not why they approach 00.

6. Find the slant asymptote of h(x)=x2+3x+1xh(x) = \frac{x^2 + 3x + 1}{x}.

Type an expression for y=y =.

Show the answer

h(x)=x+3+1xh(x) = x + 3 + \frac{1}{x}. As x→±∞x \to \pm\infty, 1x→0\frac{1}{x} \to 0, so the graph approaches the line y=x+3y = x + 3.

  • Answer: y=y = x+3x + 3

Part 9 · Summary

Summary

A rational function's end behavior matches the quotient of its leading terms. Equal degrees give a horizontal asymptote at the ratio of the leading coefficients; a larger denominator degree gives y=0y = 0; a numerator one degree larger gives a slant asymptote. State the end behavior in limit notation, such as limx→∞r(x)=3\lim_{x \to \infty} r(x) = 3.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic