B2.3 Sketch transformed exponential functions and state domain and range

Ontario Grade 11 Mathematics

Exponential Functions

Domain, Range, and the Effect of Transformations — MCR3U Expectation B2.3

You already know how to graph linear and quadratic functions, and you have seen how transformations like shifting or flipping a parabola change its graph. In this lesson, you will apply those same transformation ideas to a new family of functions: exponential functions. By the end, you will be able to look at an equation like

f(x) = -2(3)^{x+1} + 4

, predict what its graph looks like, sketch it quickly using a few key points, and state its domain and range with confidence.

What you will learn

Identify the base exponential function and describe its key features.
Apply vertical stretches/compressions, reflections, and translations to exponential functions.
Sketch the graph of a transformed exponential function by tracking key points and the horizontal asymptote.
State the domain and range of any transformed exponential function in set notation.

Prerequisite Bridge: The Base Exponential Function

Before transforming anything, you need a solid picture of the parent function. The base exponential function has the form

f(x) = b^x

, where

b > 0

and

b \neq 1

. The value

b

is called the base. When

b > 1

, the function grows as

x

increases (exponential growth). When

0 < b < 1

, the function decays as

x

increases (exponential decay).

Consider

f(x) = 2^x

. Plugging in a few values:

f(-2) = \frac{1}{4}

,

f(-1) = \frac{1}{2}

,

f(0) = 1

,

f(1) = 2

,

f(2) = 4

. Notice that the outputs are always positive, no matter how negative

x

gets. The graph gets very close to zero on the left but never touches it — that invisible boundary is called the horizontal asymptote, located at

y = 0

for the parent function.

Two key features of

f(x) = b^x

: the domain is all real numbers, written \{x \mid x ∈

\mathbb{R}

\}, and the range is all positive real numbers, written \{y \mid y > 0, y ∈

\mathbb{R}

\}. Keep these in mind — transformations will shift or reflect these features.

f(x) = b^x

The parent form is $f(x) = b^x$ with $b > 0$ , $b \neq 1$ .
The graph never crosses the x-axis; the horizontal asymptote is $y = 0$ .
Domain of the parent function: all real numbers.
Range of the parent function: $y > 0$ (all positive real numbers).

Reading the Transformation Equation

Transformed exponential functions follow a standard template:

f(x) = a \cdot b^{x - h} + k

. Each letter controls a different transformation. Learning to read this equation before you touch the graph is the most powerful skill in this lesson.

The parameter

a

controls the vertical stretch or compression and reflections. If

|a| > 1

, the graph is stretched vertically (points move farther from the x-axis). If

0 < |a| < 1

, the graph is compressed vertically (points move closer to the x-axis). If

a

is negative, the entire graph is reflected over the horizontal asymptote — the curve now opens downward instead of upward.

The parameter

h

controls a horizontal translation. The graph shifts right by

h

units when

h > 0

, and left when

h < 0

. Be careful: in the equation, the shift appears as

x - h

inside the exponent, so

f(x) = 2^{x-3}

shifts right 3 units, while

f(x) = 2^{x+3}

(which equals

2^{x-(-3)}

) shifts left 3 units.

The parameter

k

controls a vertical translation, shifting the entire graph — including its horizontal asymptote — up by

k

units when

k > 0

, and down when

k < 0

. This is the most direct effect on the range: the horizontal asymptote moves from

y = 0

to

y = k

.

f(x) = a · b^{x-h} + k

$a$ : vertical stretch/compression; negative $a$ reflects over the asymptote.
$h$ : horizontal shift — right if $h > 0$ , left if $h < 0$ .
$k$ : vertical shift — moves the asymptote from $y = 0$ to $y = k$ .
The horizontal asymptote of $f(x) = a \cdot b^{x-h} + k$ is always $y = k$ .

How Transformations Change Domain and Range

The domain of every exponential function of this form is always all real numbers, \{x \mid x ∈

\mathbb{R}

\}. No transformation — horizontal shift, stretch, or reflection — restricts the x-values you can substitute. You can always raise a positive base to any real power.

The range, however, depends on

a

and

k

. Start with the horizontal asymptote

y = k

. If

a > 0

, the graph sits entirely above that asymptote, so the range is \{y \mid y > k, y ∈

\mathbb{R}

\}. If

a < 0

, the graph is reflected and sits entirely below that asymptote, so the range is \{y \mid y < k, y ∈

\mathbb{R}

\}. The asymptote value

k

is never actually reached — the graph approaches it forever but never touches it.

A vertical stretch or compression (changing |a|) does not change whether the output is above or below the asymptote. It only changes how quickly the function moves away from it. So |a| affects the shape of the curve but not the boundary of the range.

Domain is always \{x \mid x ∈ $\mathbb{R}$ \} for any transformed exponential function.
The horizontal asymptote is $y = k$ and is never part of the range.
If $a > 0$ : range is \{y \mid y > k, y ∈ $\mathbb{R}$ \}.
If $a < 0$ : range is \{y \mid y < k, y ∈ $\mathbb{R}$ \}.

Sketching Strategy: Five Steps That Always Work

You do not need to plot dozens of points to sketch a transformed exponential. Instead, use this five-step approach every time. First, identify

a

,

b

,

h

, and

k

from the equation. Second, draw the horizontal asymptote at

y = k

as a dashed line — this anchors your sketch. Third, choose three convenient x-values (one to the left, one at

x = h

, one to the right) and calculate the corresponding y-values. These become your key points. Fourth, draw a smooth curve through the key points that approaches the asymptote without crossing it. Fifth, label the asymptote, state the domain, and state the range.

When choosing x-values, picking

x = h - 1

,

x = h

, and

x = h + 1

is usually the most efficient strategy because the exponent becomes

-1

,

0

, and

1

respectively, which are easy to evaluate. At

x = h

, the exponent equals zero, so the function value is always

a \cdot b^0 + k = a + k

. This point is reliable and useful on every sketch.

One final check: decide which direction the curve opens. If

a > 0

and

b > 1

, the right side rises and the left side approaches the asymptote from above. If

a < 0

and

b > 1

, the right side drops and the left side approaches the asymptote from below. Visualising this before drawing prevents common sketching errors.

Draw the asymptote $y = k$ first, before any points.
The anchor point at $x = h$ always gives $y = a + k$ .
Use $x = h-1$ , $x = h$ , $x = h+1$ as your three key points.
The curve never crosses the asymptote.
Check the sign of $a$ to confirm which direction the curve opens.

Putting It All Together: Application and Interpretation

In real-world contexts, transformed exponential functions appear in population growth, cooling, and financial situations. The parameters

h

and

k

often represent a starting time or a baseline amount, while

a

scales the effect. Even without a context, being able to sketch and interpret these functions quickly is a core skill in this course.

When you are given a graph instead of an equation, you can work backwards. Read the horizontal asymptote to find

k

. Determine from the graph whether the function is above or below the asymptote to decide the sign of

a

. Pick a known point (such as the y-intercept) and substitute into

f(x) = a \cdot b^{x-h} + k

to check your reading. Stating the domain and range from a graph means identifying the asymptote and whether the curve is entirely above or below it.

Always double-check your range by substituting one point. If the point's y-value satisfies the range inequality you wrote, you can be confident your answer is correct. This habit catches sign errors quickly.

You can read $k$ directly from the horizontal asymptote on a graph.
The sign of $a$ tells you whether the graph is above or below the asymptote.
Domain is always all real numbers for this function family.
Verify your stated range by substituting a known point.

Summary of Transformation Parameters in $f(x) = a \cdot b^{x-h} + k$

Parameter	What It Controls	Effect on Graph	Effect on Asymptote / Range
$a > 1$	Vertical stretch	Points move farther from asymptote	Asymptote unchanged; range boundary unchanged
$0 < a < 1$	Vertical compression	Points move closer to asymptote	Asymptote unchanged; range boundary unchanged
$a < 0$	Reflection over asymptote	Graph flips to opposite side of asymptote	Range switches from $y > k$ to $y < k$
$h > 0$	Horizontal shift right	Entire graph moves right $h$ units	No change
$k > 0$	Vertical shift up	Entire graph moves up $k$ units	Asymptote moves to $y = k$ ; range shifts up

Worked example

Example 1: Growth Function with a Vertical Stretch and Vertical Shift

Given

f(x) = 3(2)^x - 6

, identify all transformation parameters, draw a sketch using three key points, and state the domain and range.

Identify the parameters
Match the equation to the template $f(x) = a \cdot b^{x-h} + k$ . Here $a = 3$ , $b = 2$ , $h = 0$ (no horizontal shift since the exponent is just $x$ ), and $k = -6$ .
$a = 3, b = 2, h = 0, k = -6$
Draw the horizontal asymptote
The horizontal asymptote is at $y = k$ , so draw a dashed line at $y = -6$ . The graph will approach this line but never cross it. Because $a = 3 > 0$ , the graph sits above this line.
$y = -6$
Calculate three key points
Since $h = 0$ , use $x = -1$ , $x = 0$ , and $x = 1$ . Substitute each into $f(x) = 3(2)^x - 6$ . At $x = -1$ : $f(-1) = 3(2)^{-1} - 6 = 3 \cdot \frac{1}{2} - 6 = 1.5 - 6 = -4.5$ . At $x = 0$ : $f(0) = 3(1) - 6 = -3$ . At $x = 1$ : $f(1) = 3(2) - 6 = 0$ .
$(-1, -4.5), (0, -3), (1, 0)$
Sketch the curve
Plot the three key points and the dashed asymptote at $y = -6$ . Draw a smooth curve that rises steeply to the right and approaches $y = -6$ from above as $x$ decreases toward negative infinity. The curve passes through the x-axis at $x = 1$ .
State the domain and range
The domain is all real numbers because any value of $x$ can be substituted. Since $a = 3 > 0$ , the graph is entirely above the asymptote $y = -6$ , so the range is all values greater than $-6$ .
$\text{Domain: } \{x \mid x ∈ \mathbb{R}\}, \text{Range: } \{y \mid y > -6, y ∈ \mathbb{R}\}$

Answer: Domain: \{x \mid x ∈

\mathbb{R}

\}. Range: \{y \mid y > -6, y ∈

\mathbb{R}

\}. Key points:

(-1, -4.5)

,

(0, -3)

,

(1, 0)

. Asymptote:

y = -6

.

Check: Substitute

x = 1

back in:

3(2)^1 - 6 = 6 - 6 = 0

. The point

(1, 0)

is correct. Check the range:

0 > -6

✓ and

-3 > -6

✓ — all key y-values satisfy

y > -6

.

Worked example

Example 2: Reflected and Horizontally Shifted Decay Function

Given

g(x) = -2\left(\frac{1}{3}\right)^{x+2} + 5

, identify all transformation parameters, find three key points, and state the domain and range.

Identify the parameters
Rewrite the equation as $g(x) = -2\left(\frac{1}{3}\right)^{x-(-2)} + 5$ to match the template $g(x) = a \cdot b^{x-h} + k$ . So $a = -2$ , $b = \frac{1}{3}$ , $h = -2$ , and $k = 5$ .
$a = -2, b = \frac{1}{3}, h = -2, k = 5$
Interpret each transformation
The base $b = \frac{1}{3}$ means the parent function decays (falls as $x$ increases). The value $a = -2$ means a vertical stretch by a factor of 2 and a reflection — the graph will be flipped so it sits below the asymptote. The shift $h = -2$ moves the graph 2 units to the left. The value $k = 5$ raises the asymptote to $y = 5$ .
Draw the horizontal asymptote
Draw a dashed line at $y = 5$ . Because $a = -2 < 0$ , the graph sits entirely below this line.
$y = 5$
Calculate three key points
Use $x = h - 1 = -3$ , $x = h = -2$ , and $x = h + 1 = -1$ . Substitute each into $g(x) = -2\left(\frac{1}{3}\right)^{x+2} + 5$ . At $x = -3$ : exponent is $-3 + 2 = -1$ , so $g(-3) = -2\left(\frac{1}{3}\right)^{-1} + 5 = -2(3) + 5 = -6 + 5 = -1$ . At $x = -2$ : exponent is $0$ , so $g(-2) = -2(1) + 5 = 3$ . At $x = -1$ : exponent is $1$ , so $g(-1) = -2\left(\frac{1}{3}\right) + 5 = -\frac{2}{3} + 5 = \frac{13}{3} \approx 4.33$ .
$(-3, -1), (-2, 3), (-1, \frac{13}{3})$
Sketch the curve
Plot the three key points and the dashed asymptote at $y = 5$ . Because the base is less than 1 and $a$ is negative, as $x$ increases the output rises toward 5 from below. As $x$ decreases toward negative infinity, the output drops further below 5. Draw a smooth curve below the asymptote approaching it from below on the right side.
State the domain and range
The domain is all real numbers. Since $a = -2 < 0$ , the graph is entirely below the asymptote $y = 5$ , so all output values are less than 5.
$\text{Domain: } \{x \mid x ∈ \mathbb{R}\}, \text{Range: } \{y \mid y < 5, y ∈ \mathbb{R}\}$

Answer: Domain: \{x \mid x ∈

\mathbb{R}

\}. Range: \{y \mid y < 5, y ∈

\mathbb{R}

\}. Key points:

(-3, -1)

,

(-2, 3)

,

(-1, \frac{13}{3})

. Asymptote:

y = 5

.

Check: At

x = -2

:

-2(\frac{1}{3})^{-2+2} + 5 = -2(\frac{1}{3})^0 + 5 = -2(1) + 5 = 3

✓. All key y-values (

-1

,

3

,

\frac{13}{3} \approx 4.33

) are less than

5

✓ — consistent with the stated range.

Common mistakes and how to avoid them

Writing the range as

y \geq k

instead of

y > k

, treating the asymptote as an achievable value.

Correction: The graph only approaches the asymptote — it never touches it. Always use a strict inequality:

y > k

(or

y < k

) in the range.

Confusing the direction of the horizontal shift: thinking

f(x) = 2^{x+3}

shifts right by 3.

Correction: Rewrite as

2^{x - (-3)}

, so

h = -3

, meaning the shift is 3 units to the left, not right.

Forgetting that a negative value of

a

flips the range from

y > k

to

y < k

.

Correction: Always check the sign of

a

first. A negative

a

reflects the graph below the asymptote, so the range becomes

\{y \mid y < k\}

.

Claiming the domain is

x > 0

or

x > h

because exponential functions look like they only start somewhere.

Correction: Exponential functions are defined for all real values of

x

. The domain is always \{x \mid x ∈

\mathbb{R}

\}.

Applying the vertical stretch |a| to the asymptote value, writing the asymptote as

y = ak

instead of

y = k

.

Correction: The asymptote comes from the

+k

term alone. The factor

a

multiplies the exponential part, not the constant

k

. The asymptote is simply

y = k

.

Lesson summary

The parent function $f(x) = b^x$ has domain all real numbers, range $y > 0$ , and a horizontal asymptote at $y = 0$ .
Transformed exponential functions take the form $f(x) = a \cdot b^{x-h} + k$ , where $a$ scales and reflects, $h$ shifts horizontally, and $k$ shifts vertically.
The horizontal asymptote always moves to $y = k$ , and this is the key to finding the range.
If $a > 0$ , the range is $\{y \mid y > k\}$ ; if $a < 0$ , the range is $\{y \mid y < k\}$ ; the asymptote is never included.
The domain of every transformed exponential function is \{x \mid x ∈ $\mathbb{R}$ \} — no transformation restricts the input values.
To sketch efficiently: identify parameters, draw the asymptote first, plot three key points using $x = h-1$ , $x = h$ , $x = h+1$ , then draw a smooth curve that approaches the asymptote without crossing it.

Check your understanding

Question 1

What is the horizontal asymptote of

f(x) = 4(2)^{x-1} - 3

?

$y = 4$
$y = -1$
$y = -3$
$y = 2$

Show answer and explanation

y = -3

The asymptote is determined by the vertical shift

k

. Here

k = -3

, so the asymptote is

y = -3

. The values of

a = 4

and

h = 1

do not affect the asymptote.

Question 2

What is the range of

g(x) = -5(3)^x + 2

?

\{y \mid y > 2, y ∈ $\mathbb{R}$ \}
\{y \mid y < -5, y ∈ $\mathbb{R}$ \}
\{y \mid y ≥ 2, y ∈ $\mathbb{R}$ \}
\{y \mid y < 2, y ∈ $\mathbb{R}$ \}

Show answer and explanation

\{y \mid y < 2, y ∈

\mathbb{R}

\}

Here

a = -5 < 0

and

k = 2

, so the graph is reflected below the asymptote

y = 2

. The range is all values strictly less than 2. The asymptote itself is not included, so a strict inequality is used.

Question 3

Which transformation does the equation

f(x) = 2^{x+4}

represent compared to

f(x) = 2^x

?

A shift of 4 units to the right
A vertical stretch by a factor of 4
A shift of 4 units to the left
A reflection over the x-axis

Show answer and explanation

A shift of 4 units to the left

Rewrite

2^{x+4}

as

2^{x-(-4)}

, so

h = -4

. A negative

h

means the graph shifts 4 units to the left, not right.

Question 4

For any transformed exponential function of the form

f(x) = a \cdot b^{x-h} + k

, what is always true about the domain?

The domain is \{x \mid x > h, x ∈ $\mathbb{R}$ \}.
The domain is \{x \mid x > 0, x ∈ $\mathbb{R}$ \}.
The domain is \{x \mid x ∈ $\mathbb{R}$ \}.
The domain depends on the value of $b$ .

Show answer and explanation

The domain is \{x \mid x ∈

\mathbb{R}

\}.

A positive base

b

can be raised to any real power, so there is no restriction on the input values. The domain is always all real numbers, regardless of

a

,

b

,

h

, or

k

.

Key terms

Exponential function: A function of the form $f(x) = b^x$ where $b > 0$ and $b \neq 1$ . The variable appears in the exponent.
Base ( $b$ ): The constant that is repeatedly multiplied in an exponential function. It must be positive and not equal to 1.
Horizontal asymptote: A horizontal line that a graph approaches but never touches or crosses. For $f(x) = a \cdot b^{x-h} + k$ , the asymptote is $y = k$ .
Vertical stretch: A transformation that multiplies all y-values by a factor $|a| > 1$ , pulling the graph away from the asymptote.
Vertical compression: A transformation that multiplies all y-values by a factor $0 < |a| < 1$ , pushing the graph toward the asymptote.
Reflection: A flip of the graph over a line. When $a < 0$ in an exponential function, the graph is reflected over its horizontal asymptote.
Domain: The set of all possible input values ( $x$ -values) for a function. For all transformed exponential functions, the domain is \{x \mid x ∈ $\mathbb{R}$ \}.
Range: The set of all possible output values ( $y$ -values) for a function. For transformed exponentials, the range is either $\{y \mid y > k\}$ or $\{y \mid y < k\}$ depending on the sign of $a$ .

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation B2.3. It is a study resource, not an official curriculum publication.

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B2.3 · Sketch transformed exponential functions and state domain and range

Ontario Grade 11 Mathematics

Domain, Range, and the Effect of Transformations — MCR3U Expectation B2.3

What you will learn

Prerequisite Bridge: The Base Exponential Function

Reading the Transformation Equation

How Transformations Change Domain and Range

Sketching Strategy: Five Steps That Always Work

Putting It All Together: Application and Interpretation

Summary of Transformation Parameters in $f(x) = a \cdot b^{x-h} + k$

Example 1: Growth Function with a Vertical Stretch and Vertical Shift

Example 2: Reflected and Horizontally Shifted Decay Function

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MCR3U

About this lesson