AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/1/1-7)
Unit 1 · Topic 1.7
1.7 Rational Functions and End Behavior
A rational function is a fraction with polynomials on top and bottom. For inputs far from zero, the ratio of the two leading terms decides what the graph does: level off at a horizontal asymptote, follow a slanted line, or grow without bound.
Key terms
- rational function
- horizontal asymptote
- slant (oblique) asymptote
- leading terms
- end behavior
What a rational function measures
A rational function has the form r(x) = p(x)/q(x), where p and q are polynomials and q is not the zero polynomial. For each input, it compares the size of the numerator with the size of the denominator.
Far to the left or right, each polynomial acts like its leading term. So the end behavior of r matches the end behavior of (leading term of p) / (leading term of q).
Three cases
Compare degrees and simplify the quotient of leading terms.
| Case | Quotient of leading terms | End behavior |
|---|---|---|
| Degree of top < degree of bottom | Constant over a power of x, like 4/x | Horizontal asymptote y = 0 |
| Degrees equal | A constant, like 6/3 = 2 | Horizontal asymptote y = (ratio of leading coefficients) |
| Degree of top > degree of bottom | A nonconstant polynomial | Grows or falls without bound like that polynomial |
Horizontal asymptotes and limit notation
A horizontal asymptote y = b means the outputs get as close as you like to b, and stay close, as x → ∞ or x → −∞. You write lim (x→∞) r(x) = b or lim (x→−∞) r(x) = b.
A horizontal asymptote describes the ends only. The graph is allowed to cross it in the middle. For example, r(x) = x/(x² + 1) crosses its asymptote y = 0 at x = 0.
Slant asymptotes
When the degree of the numerator is exactly one more than the denominator's, the quotient of leading terms is linear, and the graph approaches a slanted line. That line is a slant (oblique) asymptote.
The quotient of leading terms tells you the slope, but to get the exact line you need polynomial long division (topic 1.11). Divide, then drop the remainder part: what's left is the asymptote.
If the numerator's degree is two or more higher, there's no line to approach; the graph behaves like a higher-degree polynomial at the ends. For instance, (x³ + 1)/x behaves like x² at both ends.
End behavior in context
Horizontal asymptotes often have a real meaning. Suppose making x phone cases costs 500 dollars for equipment plus 3 dollars per case. The average cost per case is C(x) = (500 + 3x)/x. The leading terms give 3x/x = 3, so lim (x→∞) C(x) = 3.
In words: as you make more and more cases, the average cost per case gets closer and closer to 3 dollars, because the 500-dollar setup cost is spread over more items. It never actually reaches 3, since 500/x is always positive.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Equal degrees
Find the horizontal asymptote of r(x) = (6x² − x)/(3x² + 5), and write the end behavior with limits.
Show the solutionHide the solution
- Step 1: Leading terms: 6x² on top, 3x² on the bottom. The degrees are equal.
- Step 2: The quotient is 6x²/3x² = 2, a constant.
- Step 3: So the outputs approach 2 at both ends.
Answer: Horizontal asymptote y = 2; lim (x→∞) r(x) = 2 and lim (x→−∞) r(x) = 2.
- Example 2
Bigger denominator
Describe the end behavior of r(x) = (4x + 1)/(x² − 9).
Show the solutionHide the solution
- Step 1: Leading terms: 4x on top, x² on the bottom. The bottom has the higher degree.
- Step 2: The quotient is 4x/x² = 4/x. As x → ±∞, 4/x gets closer and closer to 0.
Answer: Horizontal asymptote y = 0; lim (x→∞) r(x) = 0 and lim (x→−∞) r(x) = 0.
- Example 3
Finding a slant asymptote
Find the slant asymptote of r(x) = (2x² + 3x − 1)/(x − 1).
Show the solutionHide the solution
- Step 1: The top has degree 2 and the bottom degree 1, one more on top, so expect a slant asymptote. The quotient of leading terms, 2x²/x = 2x, says the slope will be 2.
- Step 2: Long division: 2x² ÷ x = 2x. Multiply 2x(x − 1) = 2x² − 2x and subtract: (2x² + 3x) − (2x² − 2x) = 5x. Bring down −1 to get 5x − 1.
- Step 3: 5x ÷ x = 5. Multiply 5(x − 1) = 5x − 5 and subtract: (5x − 1) − (5x − 5) = 4.
- Step 4: So r(x) = 2x + 5 + 4/(x − 1). As x → ±∞, 4/(x − 1) → 0, so r(x) gets close to 2x + 5.
Answer: y = 2x + 5.
Common mistakes
- Reading the leading coefficient from the first term written. In (5 − 3x²)/(2x² + x), the leading terms are −3x² and 2x², so the asymptote is y = −3/2.
- Saying a graph can never cross its horizontal asymptote. It can cross in the middle; the asymptote only describes the ends.
- Using the quotient of leading terms (2x) as the slant asymptote. It gives the slope, but you need long division to get the full line, like y = 2x + 5.
On the exam
- Asymptote questions are very common in multiple choice. Know the three degree cases cold, and be ready to write the limit statements.
- On free response, a full answer about end behavior uses limit notation, such as lim (x→∞) r(x) = 2.
Connected topics
Videos
Check yourself
4 questions on 1.7 Rational Functions and End Behavior. Pick an answer to see if you got it, and why.
Let r(x) = (6x² − x + 1)/(3x² + 4x). Which of the following is an equation of the horizontal asymptote of the graph of r?
Let r(x) = ((2x − 1)(x + 3))/((1 − x)(4x + 5)). Which of the following is an equation of the horizontal asymptote of the graph of r?
Let r(x) = (−2x⁴ + x)/(x² + 3). Which of the following describes the end behavior of r?
Let r(x) = (x² + 3x − 1)/(x − 2). Which of the following is an equation of the slant asymptote of the graph of r?
0 of 4 answered