Practice Maths

Graphs of Exponential Functions

Key Terms

The basic exponential function is y = ax where a > 0 and a ≠ 1.
If a > 1: the function is increasing (exponential growth) — e.g. y = 2x.
If 0 < a < 1: the function is decreasing (exponential decay) — e.g. y = (1/2)x.
All graphs of y = ax pass through (0, 1) — the y-intercept is always 1.
Horizontal asymptote
: y = 0 (the x-axis). The curve approaches but never touches it.
The general transformed form is y = b · ax−h + c:
    — b: vertical dilation (and reflection if b < 0)
    — h: horizontal shift (right if h > 0)
    — c: vertical shift, and the asymptote becomes y = c
Key Features of y = ax
DomainAll real x   (−∞, ∞)
Rangey > 0   (0, ∞)
y-intercept(0, 1)
x-interceptNone
Asymptotey = 0 (horizontal)
Behaviour (a>1)Increasing; as x→−∞, y→0; as x→+∞, y→+∞
Behaviour (0<a<1)Decreasing; as x→+∞, y→0; as x→−∞, y→+∞

Graph of y = 2x (blue, growth) and y = (1/2)x (orange, decay) — both pass through (0, 1); asymptote y = 0

x y 1 2 −1 −2 1 2 4 y=2x y=(1/2)x asymptote y=0 (0,1)
Hot Tip When identifying the asymptote of a transformed function y = b · ax + c, look at the vertical shift c. The asymptote is always y = c. The y-intercept of y = b · ax + c is found by substituting x = 0: y = b · a0 + c = b + c.

Worked Example 1 — Sketching a transformed exponential

Question: For y = 3 × 2x − 6, find: (a) the asymptote, (b) the y-intercept, (c) the x-intercept, (d) describe the shape.

(a) Asymptote: c = −6, so the asymptote is y = −6.

(b) y-intercept: x = 0: y = 3 × 1 − 6 = −3. Point: (0, −3).

(c) x-intercept: Set y = 0: 0 = 3 × 2x − 6 → 2x = 2 → x = 1. Point: (1, 0).

(d) Since b = 3 > 0 and base 2 > 1, the function is increasing, approaching y = −6 from above as x → −∞.

Worked Example 2 — Identifying growth vs decay and reading transformations

Question: State whether each function is growth or decay, and identify any transformations from y = 2x.

(a) y = 2−x: Writing as y = (2−1)x = (1/2)x. Base = 1/2 < 1: decay. This is a reflection of y = 2x in the y-axis.

(b) y = 2x+3: Base = 2 > 1: growth. Horizontal shift left 3. y-intercept: y = 23 = 8.

Why y = ax Has No x-Intercept and Why the Range Is y > 0

For the equation ax = 0 to have a solution, you would need some power of a positive number to equal zero. But multiplying any positive number by itself any number of times always produces a positive result — you can never reach zero. This is why y = ax has no x-intercept and why y = 0 (the x-axis) is an asymptote. As x → −∞, ax approaches 0 from above but never equals 0. The range is therefore strictly y > 0, not y ≥ 0. Many students write y ≥ 0 and lose marks — the boundary value 0 is never actually achieved.

Growth vs Decay — Understanding the Base

Whether y = ax represents growth or decay depends entirely on whether a > 1 or 0 < a < 1. Consider what happens when x increases by 1: the function multiplies by a. If a = 2, it doubles; if a = 1/2, it halves. Both types pass through (0, 1) because a0 = 1 for any valid base. An exponential decay function like y = (1/2)x is identical to y = 2−x — this shows that a decaying exponential is just a growing exponential reflected in the y-axis. The base a = 1 is excluded because 1x = 1 for all x, which is a constant, not an exponential function.

Exam Tip: For a transformed function y = b · ax + c, the horizontal asymptote shifts to y = c (not y = 0). The y-intercept is found by substituting x = 0: y = b · a0 + c = b + c. Students frequently forget to add c when computing the y-intercept. Always substitute x = 0 into the full transformed function, not just the base function.

Transformations and Their Effect on Key Features

Every transformation of y = ax changes specific features in predictable ways. A vertical shift by c (i.e. y = ax + c) moves the asymptote to y = c and the y-intercept to (0, 1+c). A horizontal shift by h (i.e. y = ax−h) shifts the graph right by h units but does NOT move the asymptote — it remains y = 0 unless there is also a vertical shift. A vertical dilation by b (i.e. y = b·ax) stretches the graph vertically: the asymptote stays at y = 0 but the y-intercept moves to (0, b). A reflection across the x-axis (b < 0) flips the curve below the asymptote, so the range becomes y < c (not y > c).

Finding x-Intercepts by Solving Exponential Equations

To find where y = 3 × 2x − 6 crosses the x-axis: set y = 0, giving 3 × 2x = 6, so 2x = 2, thus x = 1. Not every exponential function has an x-intercept: y = 2x + 3 has asymptote y = 3, so the curve sits entirely above y = 3 and never reaches y = 0. To check for an x-intercept, set y = 0 and decide whether the resulting exponential equation has a solution. If the asymptote is positive and the curve is above it, there is no x-intercept.

Exam Tip: A common trap is confusing y = 4 × 2x with y = (4×2)x = 8x. They are not the same: 4 × 2x means multiply the output by 4, whereas 8x raises a different base. Related: y = 4 × 2x and y = 2x+2 are equivalent because 2x+2 = 4 × 2x (using index law am+n = am × an). Recognising such equivalences is a useful skill for matching equations to graphs.

A Brief Introduction to the Number e

The natural base e ≈ 2.718 appears throughout higher mathematics because it has a unique property: y = ex is its own derivative (d/dx(ex) = ex). This means the rate of change of ex equals its own value at every point, making it the natural base for modelling continuous growth. Continuous compound interest uses the formula A = Pert. At Year 11 level, it is sufficient to know e exists, that e ≈ 2.718, and that y = ex behaves like any exponential growth function with a > 1, passing through (0, 1) with asymptote y = 0.

Mastery Practice

See Answers ➔
  1. Fluency

    For each exponential function, state whether it shows growth or decay, and give the y-intercept.

    1. (a) y = 3x
    2. (b) y = (0.5)x
    3. (c) y = (3/4)x
    4. (d) y = 5x
    5. (e) y = (1.2)x
  2. Fluency

    Complete the table of values for each function, then sketch the graph.

    1. (a) y = 2x for x = −2, −1, 0, 1, 2, 3
    2. (b) y = 3x for x = −2, −1, 0, 1, 2
    3. (c) y = (1/3)x for x = −2, −1, 0, 1, 2
  3. Fluency

    For each function, state the asymptote, y-intercept, and domain and range.

    1. (a) y = 2x + 3
    2. (b) y = 3x − 5
    3. (c) y = −2x
    4. (d) y = 4 × 2x
  4. Fluency

    Find the x-intercept (if it exists) of each function by solving y = 0.

    1. (a) y = 2x − 4
    2. (b) y = 3 × 2x − 12
    3. (c) y = 2x − 1/4
    4. (d) y = −2x + 8
  5. Understanding

    Describe the transformations applied to y = 2x to obtain each of the following. Then state the asymptote and y-intercept.

    Method: Compare each function to y = b · 2x−h + c and identify b, h, c.
    1. (a) y = 2x−3
    2. (b) y = 2 × 2x + 1
    3. (c) y = −2x+1 + 4
    4. (d) y = 2−x
  6. Understanding

    Match the graph description to the correct function.

    Functions: A) y = 2x   B) y = (1/2)x   C) y = 2x + 2   D) y = −2x
    1. (a) Increasing, y-intercept (0, 1), asymptote y = 0.
    2. (b) Decreasing, y-intercept (0, 1), asymptote y = 0.
    3. (c) Increasing, y-intercept (0, 3), asymptote y = 2.
    4. (d) Decreasing (reflected), y-intercept (0, −1), asymptote y = 0.
  7. Understanding

    Finding the equation from a graph.

    1. (a) An exponential function passes through (0, 5) and (1, 10). Write the equation in the form y = b × 2x.
    2. (b) An exponential function has asymptote y = 2, y-intercept (0, 5), and passes through (1, 11). Write the equation in the form y = b × 3x + 2.
    3. (c) An exponential decay graph has equation y = A × (1/2)x and passes through (2, 3). Find A.
  8. Understanding

    Comparing rates of growth.

    1. (a) For y = 2x and y = 3x, which grows faster for large positive x? Justify using a table of values for x = 0, 1, 2, 3, 5.
    2. (b) At what x value does 3x first exceed 2x by more than 100?
    3. (c) Sketch y = 2x and y = 3x on the same axes and describe the key differences.
  9. Problem Solving

    Exponential growth and decay context.

    Challenge. A population of bacteria is modelled by P = 500 × 2t, where t is time in hours.
    1. (a) What is the initial population (at t = 0)?
    2. (b) What is the population after 4 hours?
    3. (c) Describe the asymptote of this model and explain what it means in context.
    4. (d) Sketch the graph of P vs t for 0 ≤ t ≤ 5, labelling two key points.
  10. Problem Solving

    Investigating transformations.

    Challenge.
    1. (a) The function y = 2x+k passes through (0, 8). Find k and write the equation in two equivalent forms.
    2. (b) Show that y = 4 × 2x and y = 2x+2 are the same function.
    3. (c) The graph of y = a × bx passes through (0, 3) and (2, 75). Find a and b and state whether this shows growth or decay.