A rational function is one polynomial divided by another. It is undefined wherever the denominator is zero.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Vertical asymptotes
These occur where the denominator is zero and the numerator is not. The values run away there.
Holes
If a factor cancels from top and bottom, the graph has a hole rather than an asymptote. Factor before deciding.
Horizontal asymptotes
Compare degrees. Equal gives the ratio of leading coefficients, a smaller top gives y = 0, and a larger top gives none.
Horizontal asymptotes may be crossed
They describe long-run behaviour only. A curve can cross one near the origin and still approach it far out.
The domain
The domain is every real number except the zeros of the denominator, whether they give holes or asymptotes.
The domain excludes zeros of the denominator
Finding those values is the first step in any rational problem. They remain excluded even when a factor later cancels, because cancelling changes the formula and not the original function.
Asymptote or hole
A denominator zero that cancels gives a hole; one that does not gives a vertical asymptote. Factoring both parts is what distinguishes them, and inspection alone cannot.
End behaviour from the degrees
Lower numerator degree gives a horizontal asymptote at zero; equal degrees give the ratio of leading coefficients; higher gives no horizontal asymptote at all.
A horizontal asymptote can be crossed
It describes long-run behaviour only, so the graph may cross it in the middle of the domain. A vertical asymptote can never be crossed, and the difference is worth stating.
Step 2: Try It Yourself
Tap and try it out.
- Point(3, 0.67)
Step 3: Watch an Example
One step at a time.
Watch Diego Find Both Asymptotes
Diego analyses f(x) = (2x + 1) ÷ (x − 3).
- Step 1
Setting the denominator to zero gives x = 3.
Step 4: Your Turn
Practice makes it stick.
The Break
Problem 1 of 2
f(x) = 1 ÷ (x − 7). At which x is the vertical asymptote?
The Far End
Problem 2 of 2
f(x) = (6x + 1) ÷ (2x − 5). What is the horizontal asymptote value?
Find the Breaks
1 of 8
f(x) = 5 ÷ (x + 2). Vertical asymptote at which x?
2 of 8
f(x) = (4x) ÷ (x + 1). Horizontal asymptote value?
3 of 8
f(x) = (x + 2) ÷ (x² + 1). Horizontal asymptote value?
4 of 8
f(x) = (x² − 4) ÷ (x − 2). Asymptote or hole at x = 2? 1 asymptote, 2 hole.
5 of 8
f(x) = (9x²) ÷ (3x² + 1). Horizontal asymptote value?
6 of 8
f(x) = 1 ÷ x. Horizontal asymptote value?
7 of 8
Sort each situation by what appears at that x value.
Tap something to move it.
- Empty
- Empty
8 of 8
f(x) = 3 ÷ (x − 8). Vertical asymptote at which x?
Step 5: Quick Check
Show what you know.
Question 1 of 2
f(x) = (8x + 3) ÷ (4x − 1). What is the horizontal asymptote value?
Question 2 of 2
What produces a hole rather than an asymptote?
What You Learned
- A rational function is undefined where its denominator is zero.
- A cancelling factor gives a hole; otherwise there is a vertical asymptote.
- Compare degrees for the horizontal asymptote.