Math · College algebra · Worked example
See a graph cross its horizontal asymptote
Find the horizontal asymptote of f(x) = x/(x² + 1). Does the graph ever cross it?
Compare the degrees
The numerator has degree 1 and the denominator degree 2, so the values shrink toward 0 far out: the horizontal asymptote is y = 0.
Look for vertical asymptotes
x² + 1 is never zero, so the domain is all real numbers and there is no vertical asymptote.
Find where the graph meets y = 0
f(x) = 0 when the numerator is zero, at x = 0. The graph crosses its horizontal asymptote at the origin: negative for x < 0, positive for x > 0.
Describe the shape
The graph starts near 0 on the far left, dips to a low point at (−1, −1/2), rises through the origin to a peak at (1, 1/2), and falls back toward 0 on the right.
Result
The horizontal asymptote is y = 0, and the graph crosses it at the origin.
Your turn
Does f(x) = (2x² − 8)/(x² + 1) cross its horizontal asymptote?
Show the answer and explanation
No. The asymptote is y = 2, and f(x) = 2 has no solution.
Equal degrees give y = 2/1 = 2. Setting (2x² − 8)/(x² + 1) = 2 gives 2x² − 8 = 2x² + 2, so −8 = 2, which is false: the graph stays below y = 2.
Keep exploring
In Graph, trace far to the right. At x = 10, f(10) = 10/101 ≈ 0.099: positive and shrinking toward the asymptote from above.
Return to the concept →Sources and scope
Authored study material. Tool results depend on the stated inputs and model assumptions.
Try in the workspace
Open the example inputs, change a value and keep a useful result on your board.
See the crossing in Graph Check the values in Math Open worked example on a board Asymptotes and holes in Math ReferenceYour existing work stays on this device. Examples open as editable copies.