Transformations and asymptotes of exponential graphs | Year 11

01 / LEARN

Understand the idea.

Move the graph while tracking its features

In $y = a b^{x - h} + k$ , use $b > 0$ , $b \neq 1$ and $a \neq 0$ . The exponential factor is always positive. Multiplying by $a$ scales it vertically and reflects it when $a < 0$ ; adding $k$ moves the horizontal asymptote to $y = k$ .

The inside expression $x - h$ shifts the parent graph right by $h$ . It does not directly give the vertical intercept. To find that intercept, set $x = 0$ in the complete transformed rule.

y = a b^{x - h} + k

The curve g(x)=3 times two to the power x minus one, minus four. It passes through (0,-2.5), (1,-1), (2,2), (3,8) and approaches y=-4 to the left.

Explain domain and range from positivity

The domain is all real inputs for a valid positive exponential base. Positivity determines the range: $a > 0$ puts the graph above $k$ , while $a < 0$ puts it below $k$ . The asymptote itself is excluded because the exponential contribution never vanishes.

Use the asymptote, monotonic direction and a few exact points to sketch the curve. A graphing window only displays part of it; the edges of the screen do not create endpoints in its domain or range.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Shift and stretch

Describe $g (x) = 3 (2^{x - 1}) - 4$ and find its y-intercept.

Worked solution

1
Read the transformations from the parent function.
$2^{x} → 3 (2^{x - 1}) - 4$
Shift right 1, stretch vertically by factor 3, then move down 4.
2
Find the vertical intercept by substitution.
$g (0) = 3 (2^{- 1}) - 4 = \frac{3}{2} - 4 = - \frac{5}{2}$
The input is zero at every point on the vertical axis.
3
Use positivity to identify the features.
$Domain = ℝ, g (x) > - 4, asymptote y = - 4$
The term three times the exponential is always positive and can approach zero.
4
State the intercept and range.
$(0, - \frac{5}{2}), range (- 4, \infty)$
The asymptote is a strict lower bound and is never reached.

Answer

Intercept (0,-2.5); asymptote y=-4; range above -4.

EXAMPLE 02

Reflect an exponential

Find the range and x-intercept of $h (x) = 5 - 2^{x}$ .

Worked solution

1
Find the range from the positive exponential.
$2^{x} > 0 ⟹ 5 - 2^{x} < 5$
Subtracting a strictly positive term keeps the output strictly below 5.
2
Set the output to zero.
$5 - 2^{x} = 0 ⟹ 2^{x} = 5$
An x-intercept has zero vertical coordinate.
3
Solve for the exponent.
$x log 2 = log 5 ⟹ x = \frac{log 5}{log 2} \approx 2.322$
Common logarithms and the power law isolate the exponent.
4
State the graph features.
$Range = (- \infty, 5), x-intercept \approx (2.322, 0)$
The exponential can become arbitrarily large or approach zero, but cannot equal zero.

Answer

Range $(- \infty, 5)$ ; x-intercept approximately (2.322,0).

03 / YOUR TURN

Try it yourself.

Check your understanding

What is the asymptote of $y = 4 {(0.5)}^{x + 2} + 7$ ?

Show answer & reasoning

$y = 7$

The exponential contribution $4 {(0.5)}^{x + 2}$ approaches zero as $x$ grows. Adding 7 makes the horizontal asymptote $y = 7$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Find the domain and range of $y = 2^{x} + 3$ .
2 marks
Find the $y$ -intercept of $y = 4 (3^{x - 1}) - 2$ .
2 marks
Describe the transformations from $y = 3^{x}$ to $y = - 2 (3^{x + 1}) + 4$ .
2 marks
Find the asymptote and range of $y = 6 - 4 {(0.8)}^{x}$ .
2 marks
Find the $x$ -intercept of $y = 2^{x + 1} - 8$ .
2 marks
Explain why $y = 2^{x} + 1$ is not the same graph as $y = 2^{x + 1}$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the domain and range of $y = 2^{x} + 3$ .
2 marks
$2^{x}$ is defined and positive for all real $x$ .
Domain $ℝ$ ; range $y > 3$ .
Find the $y$ -intercept of $y = 4 (3^{x - 1}) - 2$ .
2 marks
At $x = 0$ , $\frac{4}{3} - 2 = - \frac{2}{3}$ .
$(0, - \frac{2}{3})$
Describe the transformations from $y = 3^{x}$ to $y = - 2 (3^{x + 1}) + 4$ .
2 marks
The inner $x + 1$ shifts left; the outer -2 stretches and reflects.
Left 1; vertical stretch factor 2; reflection in the $x$ -axis; up 4.
Find the asymptote and range of $y = 6 - 4 {(0.8)}^{x}$ .
2 marks
Four times the exponential is positive, so subtracting it stays below 6.
Asymptote $y = 6$ ; range $y < 6$ .
Find the $x$ -intercept of $y = 2^{x + 1} - 8$ .
2 marks
$2^{x + 1} = 8 = 2^{3}$ , so $x + 1 = 3$ and $x = 2$ .
$(2, 0)$
Explain why $y = 2^{x} + 1$ is not the same graph as $y = 2^{x + 1}$ .
3 marks
The first adds 1 to every output; the second multiplies $2^{x}$ by 2.
They also have different values at $x = 1$ : 3 and 4.
The first has asymptote $y = 1$ ; the second has asymptote $y = 0$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: exponential and logarithmic functions, growth models and numerical/algebraic prerequisites.

Official curriculum source ↗

In this lesson

Understand the idea.

Move the graph while tracking its features

Explain domain and range from positivity

See the method in action.

Shift and stretch

Read the transformations from the parent function.

Find the vertical intercept by substitution.

Use positivity to identify the features.

State the intercept and range.

Reflect an exponential

Find the range from the positive exponential.

Set the output to zero.

Solve for the exponent.

State the graph features.

Try it yourself.

Check your understanding

Keep practising