When the numerator degree is exactly one more than the denominator, there is a slant asymptote rather than a horizontal one.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Finding it
Divide the numerator by the denominator. The polynomial part of the answer is the asymptote.
The remainder fades
The remaining proper fraction approaches zero far out, which is exactly why the line describes the end behaviour.
Never both
A rational function has a horizontal asymptote or a slant one, never both. The degree comparison decides which.
They may be crossed
Any end-behaviour asymptote can be crossed near the origin. Only the far-out behaviour is being described.
A full sketch
Find the domain, the vertical asymptotes and holes, the end behaviour, and the intercepts. Those four together fix the graph.
When the end behaviour is a line
If the numerator's degree is exactly one more than the denominator's, the graph approaches a slanted line at both ends. That line is the slant, or oblique, asymptote.
Find it by division
Divide the numerator by the denominator. The quotient is the asymptote and the remainder term vanishes as x grows. Polynomial division is doing genuine graphical work here, not just algebra.
You cannot have both kinds
A rational function has either a horizontal asymptote or a slant one, never both, because the degree comparison falls into exactly one case. Checking the degrees first tells you which to look for.
Sketch the asymptotes first
Draw the vertical asymptotes, then the slant or horizontal one, then plot the intercepts. The asymptotes form a frame the curve must fit inside, which makes an accurate sketch quick.
Step 2: Try It Yourself
Tap and try it out.
- Point(3, 0.33)
Step 3: Watch an Example
One step at a time.
Watch Kofi Find a Slant Asymptote
Kofi analyses f(x) = (x² + 1) ÷ (x − 1).
- Step 1
The numerator has degree 2 and the denominator degree 1, a difference of exactly one.
Step 4: Your Turn
Practice makes it stick.
The Degrees
Problem 1 of 2
Numerator degree 3, denominator degree 2. Slant asymptote? 1 yes, 0 no.
The Vertical
Problem 2 of 2
f(x) = (x² + 1) ÷ (x − 1). At which x is the vertical asymptote?
Complete the Picture
1 of 8
Numerator degree 2, denominator degree 1. Slant asymptote? 1 yes, 0 no.
2 of 8
Numerator degree 1, denominator degree 1. Slant asymptote? 1 yes, 0 no.
3 of 8
Numerator degree 4, denominator degree 2. Slant asymptote? 1 yes, 0 no.
4 of 8
Can a rational function have both a horizontal and a slant asymptote? 1 yes, 0 no.
5 of 8
f(x) = (6x + 1) ÷ (2x − 5). Horizontal asymptote value?
6 of 8
f(x) = 5 ÷ (x + 3). Vertical asymptote at which x?
7 of 8
Sort each degree comparison by the end behaviour it gives.
Tap something to move it.
- Empty
- Empty
8 of 8
f(x) = (x + 2) ÷ (x² + 1). Horizontal asymptote value?
Step 5: Quick Check
Show what you know.
Question 1 of 2
Numerator degree 3, denominator degree 2. Slant asymptote? 1 yes, 0 no.
Question 2 of 2
How do you find a slant asymptote?
What You Learned
- A numerator degree exactly one higher gives a slant asymptote.
- Divide, and the polynomial part is the asymptote.
- A function has a horizontal or a slant asymptote, never both.