Hyperbola & Exponential Graphs — Grade 11 Maths (CAPS)

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 = a/(x + p) + q

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

An asymptote is a line the graph gets closer and closer to as x (or y) runs far out. The hyperbolas and exponential graphs in this lesson never cross theirs — some other graphs do, so the test is that the curve APPROACHES the line, not that it can never meet it. Draw it dashed — marks are given for it.

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

Worked ExampleSketch y = 4/(x − 1) + 2

1
Here p = −1 and q = 2.
Compare with a/(x + 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 → 4/(x − 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 such as y = 2/(x − 1) + 3 is x ∈ ℝ, x ≠ 1 — you must exclude the vertical asymptote, because dividing by zero is undefined.

2The axes of symmetry of a hyperbola

A hyperbola y = a/(x + p) + q has TWO axes of symmetry. Both pass through the point where the asymptotes cross, (−p ; q), one with gradient 1 and one with gradient −1:

y = (x + p) + q and y = −(x + p) + q

Worked ExampleDetermine the axes of symmetry of y = 2/(x − 1) + 3

1
The asymptotes are x = 1 and y = 3, so they cross at (1 ; 3).
p = −1 and q = 3.
2
Gradient 1 through (1 ; 3): y = x + c, and 3 = 1 + c gives c = 2, so y = x + 2.
Substitute the crossing point.
3
Gradient −1 through (1 ; 3): y = −x + c, and 3 = −1 + c gives c = 4, so y = −x + 4.
The two axes are perpendicular: 1 × (−1) = −1.

3The exponential graph

y = a·b^{x + p} + q (b > 0, b ≠ 1, a ≠ 0)

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.

💡

Which way does it go? Read b AND the sign of a.
For a > 0: b > 1 rises · 0 < b < 1 falls.
For a < 0 the curve is reflected in its asymptote, so it does the opposite: b > 1 falls · 0 < b < 1 rises.
Either way the graph never crosses its horizontal asymptote, and it lies ABOVE it when a > 0, BELOW it when a < 0.

4The third parameter: p shifts the graph sideways

Grade 10 gave you a and q. Grade 11 adds p, and p is the one that trips people up because it works backwards.

y = a · b^{x + p} + q shifts the basic graph p units LEFT when p is positive

y = 2^{x + 3} is y = 2^x moved 3 left — not 3 right.
y = 2^{x − 3} is y = 2^x moved 3 right.
The horizontal asymptote does not move: it is still y = q, because p only slides the graph sideways.

Worked ExampleFor y = 3 · 2^{x + 1} − 4, write down the asymptote, the y-intercept and the effect of p.

1
Asymptote: y = −4
q gives the horizontal asymptote; p has no effect on it.
2
p = +1, so the graph of y = 3 · 2^x − 4 is shifted 1 unit LEFT.
A positive p moves the graph in the negative x direction.
3
y-intercept: let x = 0 → y = 3 · 2¹ − 4 = 6 − 4 = 2
Substitute x = 0 into the full equation.
4
So the curve passes through (0 ; 2) and flattens towards y = −4 on the left.

⚠️

Reading y = 2^{x + 3} as a shift 3 to the right because the sign is positive. Think about which x gives the same output as before: to get 2⁰ you now need x = −3, so every point has moved left.

Practice exercises

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

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Write down the asymptotes of y = 4/(x − 1) + 2.
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x = 1 and y = 2
2
Determine the y-intercept of y = 4/(x − 1) + 2.
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4/−1 + 2 = −2
3
Determine the x-intercept of y = 4/(x − 1) + 2.
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4 = −2(x − 1) → x = −1
4
Write down the domain and range of y = 4/(x − 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
9
For y = 2^{x − 3} + 1, write down the equation of the asymptote and describe the transformation from y = 2^x.
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Asymptote y = 1 ✓; the graph is shifted 3 units RIGHT ✓ and 1 unit UP ✓
10
Determine the y-intercept of y = 4 · 3^{x + 1} − 5.
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x = 0: y = 4 × 3¹ − 5 ✓ = 12 − 5 ✓ = 7 ✓
11
Explain why the value of p does NOT change the horizontal asymptote of an exponential graph.
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p is a horizontal shift ✓; the asymptote is a horizontal line whose height is set by q only, so sliding the graph sideways leaves it at y = q ✓