2 · IB Maths
Rational functions and asymptotes
The graph of y = (ax + b) ÷ (cx + d) is a hyperbola that hugs two asymptotes. Change the four coefficients and see the asymptotes, the intercepts and the equivalent shifted form move with them.
Predict first
Start from the default settings, y = (x − 2) ÷ (x − 1), with the vertical asymptote at x = 1. If you change d from −1.0 to −3.0, where does the vertical asymptote go?
The denominator is cx + d = x − 3, which is zero at x = −d ÷ c = 3. Only d has changed, so the horizontal asymptote y = a ÷ c stays at 1.
Check it in the model: Move the d slider from −1.0 to −3.0 and read the Vertical asymptote readout.
y = (ax + b) ÷ (cx + d) vertical asymptote horizontal asymptote intercepts (dots)
vertical asymptote: x = −d ÷ c
horizontal asymptote: y = a ÷ c
x-intercept: x = −b ÷ a
y-intercept: y = b ÷ d
y = k ÷ (x − h) + m, h = −d ÷ c, m = a ÷ c, k = (bc − ad) ÷ c²
Horizontal asymptote:
Intercepts:
Shifted form:
Trace:
Try this
- Choose the preset y = 1 ÷ x, then change d to 1.0. Which asymptote moves, and what is its new equation?
- Set a = 1.0, b = 0, c = 1.0, d = −2.0 (so y = x ÷ (x − 2)). Predict the horizontal asymptote and both intercepts before you read them off, then drag the trace x towards 2 from each side.
- Make ad = bc, for example a = 1, b = 2, c = 1, d = 2. What happens to the curve, and why is there no asymptote at x = −d ÷ c?
Exam tip: state asymptotes as equations (x = −d ÷ c and y = a ÷ c), and find the vertical one by setting the denominator to zero. If the numerator also vanishes there, you have a hole, not an asymptote.