Graph transformations and parameter experiments | Year 10

01 / LEARN

Understand the idea.

Track each parameter

In $y = a b^{x - h} + k$ , take $b > 0$ , $b \neq 1$ and $a \neq 0$ . Relative to $y = b^{x}$ , the term $x - h$ shifts the graph right by $h$ and the final $+ k$ shifts it vertically by $k$ .

The horizontal asymptote becomes $y = k$ . The factor $∣ a ∣$ stretches distances from that line, while negative $a$ places outputs below it. Calculate the $y$ -intercept with $x = 0$ ; it need not equal either $a$ or $k$ .

Make a controlled comparison

For a controlled parameter experiment, vary one parameter and keep the others fixed. Compare the same inputs, intercepts and asymptotes so that a change in the graph can be linked to that single parameter.

For example, $2^{x} + 3$ is always $3$ above $2^{x}$ . Its outputs no longer double directly, but its distances above the new asymptote do: $y - 3 = 2^{x}$ . This separates vertical shifting from a change of base.

Graphs of y equals two to the x and y equals two to the x plus three for x from minus two to three. A dashed horizontal line marks the shifted asymptote y equals three. The y scale is numbered from zero to ten.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Shift right and down

For $y = 2^{x - 1} - 3$ , find the horizontal asymptote and $y$ -intercept.

Worked solution

1
Identify the vertical translation.
$k = - 3 ⟹ asymptote y = - 3$
The basic exponential approaches $0$ ; subtracting $3$ shifts that line down.
2
Set the input to zero.
$y = 2^{0 - 1} - 3 = 2^{- 1} - 3$
The $y$ -intercept occurs on the vertical axis where $x = 0$ .
3
Evaluate and state the intercept.
$y = \frac{1}{2} - 3 = - \frac{5}{2}, (0, - \frac{5}{2})$
The negative exponent gives a reciprocal, followed by the vertical shift.

Answer

$y = - 3$ ; intercept $(0, - 5 / 2)$

EXAMPLE 02

Reflect and stretch

Compare $y = - 2 (3^{x}) + 1$ with $y = 3^{x}$ .

Worked solution

1
Apply the outside multiplier.
$3^{x} ⟼ - 2 (3^{x})$
The negative sign reflects across the $x$ -axis and the factor $2$ doubles vertical distances.
2
Apply the vertical shift.
$- 2 (3^{x}) ⟼ - 2 (3^{x}) + 1$
Adding $1$ moves every transformed point up one unit, so the asymptote is $y = 1$ .
3
Determine which side of the asymptote contains the curve.
$3^{x} > 0 ⟹ - 2 (3^{x}) + 1 < 1$
The transformed range is $y < 1$ because the exponential factor is always positive.
4
Check the transformed intercept.
$y = - 2 (3^{0}) + 1 = - 1, (0, - 1)$
Substitution confirms the reflection, stretch and shift at a known input.

Answer

Reflection, vertical stretch by 2, then shift up 1.

03 / YOUR TURN

Try it yourself.

Check your understanding

State the horizontal asymptote of $y = 4 (2^{x}) + 5$ .

Show answer & reasoning

$y = 5$

In $y = 4 (2^{x}) + 5$ , the exponential term approaches $0$ in one direction. Adding $5$ moves the horizontal asymptote to $y = 5$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find the $y$ -intercept of $y = 2^{x} + 4$ .
2 marks
State the shift from $y = 3^{x}$ to $y = 3^{x + 2}$ .
2 marks
State the range of $y = 2^{x} - 7$ .
2 marks
Find the intercept and asymptote of $y = - 3 (2^{x}) + 6$ .
3 marks
Compare $y = 2^{x}$ with $y = 2^{x - 1}$ at $x = 0, 1, 2$ . Give the outputs of the second graph.
2 marks
Explain why $y = 2^{x} + 3$ does not have consecutive output ratio 2, even though its exponential part does.
3 marks

Read the worked solutions

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

Find the $y$ -intercept of $y = 2^{x} + 4$ .
2 marks
$2^{0} + 4 = 5$ .
$(0, 5)$
State the shift from $y = 3^{x}$ to $y = 3^{x + 2}$ .
2 marks
The inside change $x + 2$ gives an earlier input.
Left 2 units
State the range of $y = 2^{x} - 7$ .
2 marks
The exponential part is positive, so the result exceeds $- 7$ .
$y > - 7$
Find the intercept and asymptote of $y = - 3 (2^{x}) + 6$ .
3 marks
At zero, $- 3 + 6 = 3$ ; the vertical shift sets the asymptote.
$(0, 3)$ and $y = 6$
Compare $y = 2^{x}$ with $y = 2^{x - 1}$ at $x = 0, 1, 2$ . Give the outputs of the second graph.
2 marks
Substitute the stated inputs; each is half the original output.
$\frac{1}{2}, 1, 2$
Explain why $y = 2^{x} + 3$ does not have consecutive output ratio 2, even though its exponential part does.
3 marks
At $x = 0, 1, 2$ , outputs are $4, 5, 7$ , whose ratios are not constant. Distances above the asymptote $y = 3$ are $1, 2, 4$ .
The constant shift changes output ratios.

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Integer and fractional index laws; connecting fractional powers with roots; simplifying surds; operations with surds; rationalising denominators; exact and approximate values; exponential tables and graphs; growth/decay factors; comparing linear/quadratic/exponential patterns; exponential equations; logarithms as inverses; logarithm laws; interpreting logarithmic scales; graph transformations and parameter experiments; choosing and evaluating models.

Official curriculum source ↗

In this lesson

Understand the idea.

Track each parameter

Make a controlled comparison

See the method in action.

Shift right and down

Identify the vertical translation.

Set the input to zero.

Evaluate and state the intercept.

Reflect and stretch

Apply the outside multiplier.

Apply the vertical shift.

Determine which side of the asymptote contains the curve.

Check the transformed intercept.

Try it yourself.

Check your understanding

Keep practising