End Behaviour of Rational Functions

Comparing the degrees, top and bottom.

11 problems · about 22 minutes · AP Precalculus 1.12

What this lesson teaches

The student determines the horizontal or slant asymptote by comparing numerator and denominator degrees.

End behaviour of a rational function is decided by comparing the two degrees.
Bottom heavier gives y = 0; equal degrees give the ratio of leading coefficients.
Top heavier gives no horizontal asymptote, and a slant one when it exceeds by exactly 1.

Straightforward practice. Get the method working first.

4 problems

(9x² + 2)/(3x² − 7). Horizontal asymptote y = ?
Answer 3
Why y = 3.
5/(x + 2). Horizontal asymptote y = ?
Answer 0
Why Bottom heavier.
(4x² − 1)/(x² + 3). Horizontal asymptote y = ?
Answer 4
Why 4 over 1.
(x³ + 1)/(x + 2). How many horizontal asymptotes?
Answer 0
Why Top heavier by two.

The same ideas with more to keep track of.

3 problems

Sort each function by its end behaviour.
Answer Approaches y = 0: 1/x, x/(x³ + 2) · Approaches a non-zero y: 2x/(x + 1) · No horizontal asymptote: x²/(x + 1)
Why Compare the two degrees each time.
(2x + 7)/(5x − 1). Horizontal asymptote y = ? Give it as a decimal.
Answer 0.4
Why 2 over 5.
Top degree exceeds bottom by exactly 1. Is there a slant asymptote? 1 for yes, 0 for no.
Answer 1
Why Exactly one more.

Mixed problems. Work out what kind of question it is before you start.

4 problems

(x² + 5)/(x² + 5). Horizontal asymptote y = ?
Answer 1
Why The coefficients match.
Can a graph cross its horizontal asymptote? 1 for yes, 0 for no.
Answer 1
Why It describes the far ends, not the middle.
Bottom Heavy: For x / (x² + 1), what is the horizontal asymptote? Give the y value.
Answer 0
Why y = 0.
Equal Degrees: For (6x + 1) / (3x − 2), what is the horizontal asymptote? Give the y value.
Answer 2
Why y = 2.