A hyperbola in two branches: The reciprocal function y = 1/x is a hyperbola with two separate branches — one in the top-right, one in the bottom-left. As x grows, y shrinks toward 0; as x nears 0, y shoots off to ±∞.
It never touches the axes: 1/x is never 0 (no x-intercept) and is undefined at x = 0 (no y-intercept). The axes are its asymptotes.
Big in, small out: Large x → tiny y; tiny x → huge y. The two are reciprocals, so the curve hugs both axes.
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x ≠ 0, y ≠ 0: For y = 1/x: the domain is x ≠ 0 (can't divide by zero) and the range is y ≠ 0 (the output is never 0). The asymptotes are the lines x = 0 (vertical) and y = 0 (horizontal).
IB-style question — state the features
State the domain, range and asymptotes of y = 1/x.
Step by step
- Denominator zero is banned.
- Output never reaches 0.
Final answer
Domain x ≠ 0, range y ≠ 0, asymptotes x = 0 and y = 0.
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Asymptotes are dashed guides: Draw the asymptotes first (here the two axes); the branches then hug them.
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The asymptotes move with the shift: y = 1/(x − h) + k is y = 1/x slid h right and k up. So the vertical asymptote becomes x = h and the horizontal asymptote becomes y = k.
IB-style question — shifted reciprocal
State the asymptotes of y = 1/(x − 3) + 2.
Step by step
- Vertical: denominator zero.
- Horizontal: the +2 raises the level.
Final answer
Vertical asymptote x = 3, horizontal asymptote y = 2.
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Sign of h: 1/(x − 3) shifts right 3 (vertical asymptote x = +3); 1/(x + 3) shifts left (x = −3).
Asymptotes, then the two branches: To sketch: draw the asymptotes (dashed), find any intercepts (set x = 0 for the y-intercept, set y = 0 for the x-intercept), then draw the two branches hugging the asymptotes.
IB-style question — intercepts of a shifted reciprocal
Find the y-intercept of y = 1/(x − 3) + 2.
Step by step
- Set x = 0.
- Simplify.
Final answer
y-intercept at (0, 5/3).
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Reciprocals have at most one of each intercept: A shifted reciprocal crosses each axis at most once (sometimes not at all if an asymptote is in the way).