Open Math›Year 10›Non-linear relationships›Exponential graphs

Exponential graphs

Evaluate and sketch exponential functions of the form

y = a^x

, identify growth and decay, and describe the effect of changing the base.

Worked examples

Evaluating an exponential function and finding the y-intercept

Straightforward

Problem

For the function

y = 2^x

, evaluate

y

when

x = -1, 0, 1, 2, 3

and state the y-intercept.

Evaluate

y

for each value of

x

using

y = 2^x

\begin{array}{c|c} x & y = 2^x \\ \hline -1 & 2^{-1} = \dfrac{1}{2} \\ 0 & 2^0 = 1 \\ 1 & 2^1 = 2 \\ 2 & 2^2 = 4 \\ 3 & 2^3 = 8 \end{array}

Identify the y-intercept by finding the value of

y

when

x = 0

When

x = 0

y = 2^0 = 1

. The y-intercept is

(0, 1)

Observe the behaviour of the function.

x

increases,

y

doubles each time — this is exponential growth. As

x

decreases,

y

approaches 0 but never reaches it. The x-axis (

y = 0

) is the horizontal asymptote.

Answer

The y-intercept is

1

. The table of values is:

x = -1 \Rightarrow y = \tfrac{1}{2}

;

x = 0 \Rightarrow y = 1

;

x = 1 \Rightarrow y = 2

;

x = 2 \Rightarrow y = 4

;

x = 3 \Rightarrow y = 8

Identifying growth or decay and finding the horizontal asymptote

Moderate

Problem

For the function

y = (0.3)^x

: (a) state whether the graph shows growth or decay and give a reason, (b) state the y-intercept, and (c) write the equation of the horizontal asymptote.

Identify the base and compare it to 1.

The base is

a = 0.3

. Since

0 < 0.3 < 1

, the graph shows exponential decay — as

x

increases,

y

decreases toward 0.

Find the y-intercept by substituting

x = 0

y = (0.3)^0 = 1

. The y-intercept is

(0, 1)

. This is true for all exponential functions

y = a^x

regardless of the base.

Determine the horizontal asymptote.

x \to \infty

(0.3)^x \to 0

but never equals 0. As

x \to -\infty

(0.3)^x \to \infty

. The horizontal asymptote is

y = 0

Answer

(a) Exponential decay, because

0 < 0.3 < 1

. (b) The y-intercept is

1

. (c) The horizontal asymptote is

y = 0

Finding the base from a point on the graph and describing the effect of changing the base

Challenging

Problem

The graph of

y = a^x

passes through the point

(4, 16)

. Find the value of

a

, state whether the graph shows growth or decay, and describe what happens to the graph when

a

is increased to 3.

Substitute the point

(4, 16)

into

y = a^x

16 = a^4

Solve for

a

by taking the fourth root. Take the positive value since

a > 0

a = \sqrt[4]{16} = \sqrt[4]{2^4} = 2

Determine whether the graph shows growth or decay.

Since

a = 2 > 1

, the graph of

y = 2^x

shows exponential growth.

Describe the effect of increasing

a

from 2 to 3.

Both

y = 2^x

and

y = 3^x

are exponential growth functions with y-intercept 1 and horizontal asymptote

y = 0

. For

x > 0

3^x > 2^x

, so the graph of

y = 3^x

rises more steeply. For

x < 0

3^x < 2^x

, so

y = 3^x

decreases more quickly toward 0. The effect is that the graph becomes steeper on the right and flatter on the left.

Answer

a = 2

. The graph shows exponential growth (since

a = 2 > 1

). Increasing

a

to 3 makes the graph steeper for positive

x

and closer to the asymptote for negative

x

, while keeping the y-intercept at 1.

Practise

Q1·Straightforward

Evaluate

y = 2^x

when

x = 3

Explanation

y = 2^3 = 2 \times 2 \times 2 = 8

Q2·Straightforward

Evaluate

y = 2^x

when

x = -2

. Give your answer as a fraction.

Explanation

y = 2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}

Q3·Straightforward

State the y-intercept of

y = 5^x

Explanation

When

x = 0

y = 5^0 = 1

. The y-intercept of

y = a^x

is always 1, regardless of the base

a

Q4·Straightforward

Evaluate

y = 3^x

when

x = 4

Explanation

y = 3^4 = 3 \times 3 \times 3 \times 3 = 81

Q5·Moderate

Which of the following equations represents exponential decay?

$y = 2^x$$y = (0.5)^x$$y = x^2$$y = 3x$

Explanation

y = a^x

, the graph shows decay when

0 < a < 1

. Here

a = 0.5

, which satisfies this condition, so

y = (0.5)^x

represents exponential decay. The other options are exponential growth (

y = 2^x

), a parabola (

y = x^2

), and a linear function (

y = 3x

Q6·Moderate

The graph of

y = 3^x

has a horizontal asymptote. What is its equation?

$y = 1$$y = 3$$y = 0$$x = 0$

Explanation

x \to -\infty

3^x \to 0

but never equals 0. The horizontal asymptote is

y = 0

(the x-axis). The graph of

y = 3^x

always stays above the x-axis.

Q7·Moderate

Does the graph of

y = 1.5^x

show growth or decay? Select the correct answer and reason.

Growth, because $1.5 > 1$Decay, because $1.5 > 1$Growth, because $1.5 < 2$Decay, because the graph approaches 0

Explanation

When

a > 1

, the exponential function

y = a^x

increases as

x

increases — this is exponential growth. Since

1.5 > 1

, the graph of

y = 1.5^x

shows growth.

Q8·Moderate

Evaluate

y = (0.5)^x

when

x = -3

Explanation

(0.5)^{-3} = \left(\dfrac{1}{2}\right)^{-3} = 2^3 = 8

. When you raise a fraction to a negative power, the result is the reciprocal raised to the positive power.

Q9·Challenging

The graph of

y = a^x

passes through the point

(2, 25)

. Find the value of

a

Explanation

Substituting

(2, 25)

25 = a^2

. Taking the positive square root (the base of an exponential must be positive):

a = \sqrt{25} = 5

Q10·Challenging

The graph of

y = a^x

passes through the point

(3, 0.125)

. Find the value of

a

Explanation

0.125 = \dfrac{1}{8}

, so

a^3 = \dfrac{1}{8}

. Taking the cube root:

a = \sqrt[3]{\dfrac{1}{8}} = \dfrac{1}{2} = 0.5

. Since

a = 0.5 < 1

, this graph shows exponential decay.

Q11·Challenging

For large positive values of

x

, which function produces the greatest y-values?

$y = 2^x$$y = 10^x$$y = (0.5)^x$$y = x^2$

Explanation

For large positive

x

, a larger base means faster exponential growth:

10^x > 2^x

for all

x > 0

. Both exponential functions eventually outgrow

y = x^2

(polynomial growth is always slower than exponential growth for large

x

). The function

y = (0.5)^x

decays toward 0. Therefore

y = 10^x

gives the greatest y-values.

Q12·Challenging

Which statement correctly describes the relationship between the graphs of

y = 2^x

and

y = (0.5)^x

The graph of $y = (0.5)^x$ is a reflection of $y = 2^x$ in the y-axis.The graph of $y = (0.5)^x$ is a reflection of $y = 2^x$ in the x-axis.The two graphs are identical.The graph of $y = (0.5)^x$ is a translation of $y = 2^x$ to the left.

Explanation

Since

0.5 = 2^{-1}

, we have

(0.5)^x = (2^{-1})^x = 2^{-x}

. Replacing

x

with

-x

y = 2^x

gives

y = 2^{-x} = (0.5)^x

. Substituting

-x

for

x

reflects a graph in the y-axis. Both graphs share the same y-intercept (

y = 1

when

x = 0

) and the same horizontal asymptote (

y = 0

), but

y = 2^x

increases while

y = (0.5)^x

decreases.