A reciprocal in disguise: A function y = (ax + b)/(cx + d) is a transformed reciprocal — same two-branch hyperbola shape, but with its own asymptotes. The vertical asymptote is where the denominator = 0.
IB-style question — the vertical asymptote
Find the vertical asymptote of y = (2x + 1)/(x − 4).
Step by step
- Set the denominator to zero.
Final answer
Vertical asymptote x = 4 (the graph is undefined there).
Domain excludes the vertical asymptote: The domain is all x except the value that makes the denominator zero (here x ≠ 4).
Denominator zero = forbidden x: Solve cx + d = 0 for the vertical asymptote; that same x is excluded from the domain.
IB-style question — vertical asymptote & domain
State the vertical asymptote and domain of y = (3x − 2)/(2x + 6).
Step by step
- Denominator zero.
Final answer
Vertical asymptote x = −3; domain x ≠ −3.
One vertical asymptote here: An (ax + b)/(cx + d) function has exactly one vertical asymptote (one value makes the linear denominator zero).
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Ratio of the leading coefficients: As x → ±∞, the +b and +d barely matter, so y → a/c. The horizontal asymptote is y = a/c — the ratio of the x-coefficients on top and bottom.
IB-style question — horizontal asymptote
Find the horizontal asymptote of y = (2x + 1)/(x − 4).
Step by step
- Compare the leading coefficients (a = 2, c = 1).
Final answer
Horizontal asymptote y = 2.
Why a/c?: Divide top and bottom by x: (2 + 1/x)/(1 − 4/x) → 2/1 as the 1/x terms vanish.
Numerator zero → x-intercept: The x-intercept is where the numerator = 0 (a fraction is zero only when its top is zero). The y-intercept is the value at x = 0 (b/d). Then sketch the two branches around the asymptotes.
IB-style question — all the features
For y = (2x + 1)/(x − 4), find the x- and y-intercepts.
Step by step
- x-intercept: numerator zero.
- y-intercept: x = 0.
Final answer
x-intercept (−1/2, 0); y-intercept (0, −1/4). With asymptotes x = 4 and y = 2, the sketch is complete.