Hyperbola & Exponential Graphs (Grade 11)

y = a/(x + p) + q and y = a·b^x+p + q — locating the asymptotes, finding intercepts, sketching, and stating domain and range.

1The hyperbola

y = ax + p + q

Vertical asymptote: x = −p
Horizontal asymptote: y = q

An asymptote is a line the graph approaches but never touches. Draw it dashed — marks are given for it.

y = 4x − 1 + 2, with asymptotes x = 1 and y = 2 shown dashed.

Worked ExampleSketch y = 4x − 1 + 2

1
Here p = −1 and q = 2.
Compare with ax + p + q.
2
Asymptotes: x = 1 and y = 2.
x = −p = 1.
3
y-intercept: x = 0 → y = 4−1 + 2 = −2.
4
x-intercept: y = 0 → 4x − 1 = −2 → 4 = −2(x − 1).
5
−2 = x − 1, so x = −1.
Check on the sketch: the curve does cross at (−1 ; 0). ✓

Domain: x ≠ −p · Range: y ≠ q

⚠️

The domain of a hyperbola is x ∈ ℝ, x ≠ 1 — you must exclude the vertical asymptote, because dividing by zero is undefined.

y = a·b^x+p + q

Horizontal asymptote: y = q

y = 2^x − 1, with the horizontal asymptote y = −1.

Worked ExampleSketch y = 2^x − 1

1
q = −1, so the asymptote is y = −1.
The whole graph is shifted down 1.
2
y-intercept: x = 0 → y = 2⁰ − 1 = 0.
It passes through the origin.
3
x-intercept: 2^x = 1 → x = 0.
Same point here.
4
Range: y > −1.
2^x is always positive, so y is always above the asymptote.

💡

b > 1 gives a rising curve; 0 < b < 1 gives a falling one. Either way the graph never crosses its horizontal asymptote.

Work each one out, then click to reveal the answer.

1
Write down the asymptotes of y = 4x − 1 + 2.
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x = 1 and y = 2
2
Determine the y-intercept of y = 4x − 1 + 2.
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4−1 + 2 = −2
3
Determine the x-intercept of y = 4x − 1 + 2.
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4 = −2(x − 1) → x = −1
4
Write down the domain and range of y = 4x − 1 + 2.
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Domain x ∈ ℝ, x ≠ 1; range y ∈ ℝ, y ≠ 2
5
Write down the equation of the asymptote of y = 2^x − 1.
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y = −1
6
Determine the y-intercept of y = 2^x − 1.
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2⁰ − 1 = 0
7
Write down the range of y = 2^x − 1.
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y > −1
8
Describe the transformation from y = 2^x to y = 2^x − 1.
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A translation of 1 unit downwards