Graph and analyze exponential and logarithmic functions

Alberta Grade 12 Mathematics

Relations and Functions

Connect equations, key features, and points to understand these related graphs

An exponential function has a variable in its exponent. A logarithm answers the reverse question: which exponent produces a given value? This lesson connects these functions and shows how to analyze their graphs. The domain is the set of allowed inputs. The range is the set of possible outputs. An asymptote is a line that a graph approaches in a particular direction. In the functions studied here, an exponential graph does not cross its horizontal asymptote, and a logarithmic graph does not cross its vertical asymptote.

What you will learn

Recognize exponential and logarithmic parent functions and explain how they are related.
Graph transformed exponential and logarithmic functions using key features.
Determine domain, range, intercepts, and horizontal or vertical asymptotes.
Analyze equations and graphs using exact values and clearly labelled approximations.

1. Review powers, coordinates, and logarithms

A power tells how many times a base is used as a factor. For example,

2^3=8

. Here,

2

is the base and

3

is the exponent. In an exponential function, the variable appears in the exponent.

A coordinate

(x,y)

identifies a point on a graph. To find a point, substitute an input for

x

and calculate the output

y

. An

x

-intercept has output zero. A

y

-intercept has input zero, if zero is allowed.

A logarithm is another way to write a power statement. For example,

2^3=8

is equivalent to

\log_2 8=3

. The logarithm gives the exponent needed to produce its argument, the number inside the logarithm. For a real-valued logarithm, the argument must be positive.

b^x=y\iff x=\log_b y

The variable is in the exponent in an exponential function.
A logarithmic equation can be rewritten as an equivalent exponential equation.
A logarithm’s argument must be positive.

2. Parent functions and transformations

The parent exponential function is

y=b^x

, where

b>0

and

b\ne1

. If

b>1

, its graph increases from left to right. It approaches

y=0

as

x

decreases without bound. If

0<b<1

, the graph decreases from left to right and approaches

y=0

as

x

increases without bound. In both cases, the graph stays above

y=0

. Its domain is all real numbers and its range is

y>0

.

The parent logarithmic function is

y=\log_b x

. It asks which exponent on

b

gives

x

. Since

b^0=1

, its graph passes through

(1,0)

. Its domain is

x>0

, its range is all real numbers, and its vertical asymptote is

x=0

. The graph approaches that line as

x

approaches zero from the right; the line is not in the domain.

For the same base, the parent exponential and logarithmic functions are inverses: each undoes the other by exchanging input and output. Their graphs reflect across the line

y=x

. For example, the exponential point

(0,1)

corresponds to the logarithmic point

(1,0)

.

In

y=a b^{x-c}+d

, the value

c

shifts the graph horizontally,

d

shifts it vertically, and

a

changes its vertical scale and may reflect it. The horizontal asymptote is

y=d

, and the domain is all real numbers. If

a>0

, the range is

y>d

; if

a<0

, it is

y<d

. The graph stays strictly on one side of its horizontal asymptote.

In

y=a\log_b(x-c)+d

, where

a\ne0

, the values

c

and

d

shift the graph horizontally and vertically. The value

a

changes its vertical scale and may reflect it. The vertical asymptote is

x=c

, the domain is

x>c

, and the range is all real numbers. The asymptote is outside the domain, so the graph cannot cross it.

y=a b^{x-c}+d,\quad y=a\log_b(x-c)+d\quad(a\ne0)

For real exponential and logarithmic functions, $b>0$ and $b\ne1$ .
The logarithm’s argument determines its domain restriction.
A transformed exponential has a horizontal asymptote; a transformed logarithm has a vertical asymptote.

3. Use tables and features to sketch graphs

A table gives points that can guide a sketch. For

y=2^x

, inputs

-1,0,1,2

give outputs

\frac12,1,2,4

. Each time the input increases by one, the output doubles. The graph approaches

y=0

as

x

decreases without bound and remains above it.

For

y=\log_2 x

, reverse the coordinates of the corresponding exponential points. The points become

(\frac12,-1)

,

(1,0)

,

(2,1)

, and

(4,2)

. This graph accepts only positive inputs and approaches

x=0

as its input approaches zero from the right.

For a transformed graph, identify the base and shifts first. Then state the domain, range, and asymptote. Find intercepts by substituting zero where allowed. For an exponential equation, rewrite with a common base when possible. For a logarithmic equation, rewrite it as an exponential equation and check that the original logarithm’s argument is positive.

A graphing window shows only part of a graph. It does not change the function’s full domain or range. Use the equation and its restrictions to state those features.

b^x=y\iff x=\log_b y

Use calculated points together with the asymptote when sketching.
For logarithms, require the argument to be positive.
For inverse parent functions, corresponding points exchange coordinates.

4. Practice and diploma-style reasoning

Before sketching, find the domain and asymptote. These features help keep a logarithmic sketch within its allowed inputs and locate the line the graph approaches. Then calculate a few points that are easy to verify.

To compare two graphs, state a distinguishing feature, such as whether they increase or decrease, where their asymptotes lie, or where they cross an axis. Support your comparison with an equation or calculated point.

For multiple-choice questions, test proposed points by substitution. For logarithmic functions, also check that the argument is positive. Give an exact value first when possible, and label any decimal as an approximation.

A correct sketch agrees with calculated points and domain restrictions.
Check equation-derived answers in the original function.

Parent-function comparison

Feature	$y=b^x$ , $b>1$	$y=\log_b x$ , $b>1$
Domain	All real numbers	$x>0$
Range	$y>0$	All real numbers
Asymptote	$y=0$	$x=0$
Key point	$(0,1)$	$(1,0)$
Direction	Increasing	Increasing

Worked example

Analyze a shifted exponential

Analyze and sketch

f(x)=3\cdot2^{x-1}-2

. State its domain, range, asymptote, intercepts, and two points.

Shifted exponential

The graph shows only a window of the full domain.

Read the transformations
The parent function is $2^x$ . The graph shifts right one unit, is stretched vertically by a factor of three, and shifts down two units. The horizontal asymptote therefore shifts from $y=0$ to $y=-2$ . Since the exponential part is positive and its coefficient is positive, the graph stays above this asymptote.
$y=-2$
State domain and range
An exponential expression accepts every real input. The power of two is always positive, so multiplying it by three keeps it positive. Subtracting two gives outputs greater than $-2$ . The domain is all real numbers. The range is $y>-2$ .
Find intercepts and points
For the $y$ -intercept, substitute $x=0$ . For the $x$ -intercept, set the function equal to zero and isolate the exponential. At $x=1$ and $x=2$ , the exponents are zero and one, so these inputs give convenient points.
$f(0)=-\frac12,\quad x=1+\log_2\left(\frac23\right),\quad f(1)=1,\quad f(2)=4$

Answer: The domain is all real numbers, the range is

y>-2

, and the horizontal asymptote is

y=-2

. The

y

-intercept is

(0,-\frac12)

. The

x

-intercept is

(1+\log_2(\frac23),0)

, approximately

(0.415,0)

. Two convenient points are

(1,1)

and

(2,4)

. The graph increases because its base is greater than one and its vertical factor is positive.

Check: At

x=1

,

3\cdot2^0-2=1

. At

x=2

,

3\cdot2-2=4

. Both outputs are greater than the asymptote value

-2

. Also,

1+\log_2(\frac23)\approx0.415

, which verifies the stated approximate intercept.

Worked example

Analyze a shifted logarithm

For

g(x)=\log_3(x+2)-1

, determine the domain, range, asymptote, intercepts, and two points.

Shifted logarithm

The graph shows only a window of the full domain.

Find the restriction and asymptote
The logarithm’s argument must be positive. The argument becomes zero at $x=-2$ , which is the vertical asymptote.
$x+2>0\Longrightarrow x>-2,\quad x=-2$
Find the intercepts
For the $x$ -intercept, set the output equal to zero and convert the logarithm to exponential form. For the $y$ -intercept, substitute $x=0$ .
$x=1,\quad g(0)=\log_3 2-1$
Choose points
Arguments of one and three give logarithms with known values. Solve $x+2=1$ and $x+2=3$ to find the corresponding inputs.
$(x,g(x))=(-1,-1),\quad(1,0)$

Answer: The domain is

x>-2

, the range is all real numbers, and the vertical asymptote is

x=-2

. The

x

-intercept is

(1,0)

. The

y

-intercept is

(0,\log_3 2-1)

, approximately

(0,-0.369)

. Two points are

(-1,-1)

and

(1,0)

. The graph increases because its base is greater than one.

Check: At

x=1

, the argument is three, so

g(1)=\log_3 3-1=0

. This confirms the

x

-intercept. The input must remain greater than

-2

, so the graph cannot cross its vertical asymptote.

Worked example

Solve an exponential equation and connect it to a logarithm

Solve

5^x=\frac{1}{25}

, then express the result as a logarithmic statement.

Inverse parent functions

The graph shows only a window of the full domains.

Write a common base
Since $25=5^2$ , its reciprocal is a negative power of five. This rewrites both sides using base five.
$\frac{1}{25}=5^{-2}$
Determine the exponent
The bases are equal and greater than zero, so the exponents must be equal.
$x=-2$
Connect to a logarithm
The equivalent logarithmic statement asks which exponent on five produces $\frac{1}{25}$ . It gives the same value of $x$ .
$\log_5\left(\frac{1}{25}\right)=-2$

Answer: The solution is

x=-2

. The equivalent logarithmic statement is

\log_5(\frac{1}{25})=-2

.

Check: Substitution gives

5^{-2}=\frac{1}{25}

, so the solution satisfies the original equation.

Common mistakes and how to avoid them

Allowing a logarithm’s argument to be zero or negative.

Correction: Set the argument greater than zero and solve the resulting inequality.

Calling an exponential asymptote vertical or a logarithmic asymptote horizontal.

Correction: The exponential asymptote is horizontal. The logarithmic asymptote is vertical.

Assuming an exponential or logarithmic graph can cross its asymptote.

Correction: An exponential’s output stays strictly above or below its horizontal asymptote. A logarithm’s vertical asymptote is outside its domain.

Treating an approximate intercept as an exact value.

Correction: Give the exact logarithmic expression first, then label a decimal as an approximation.

Assuming the visible part of a graph is its full domain.

Correction: Use the equation and its restrictions to state the full domain and range.

Lesson summary

For $b>0$ and $b\ne1$ , exponential and logarithmic functions with the same base are inverses.
The parent exponential has domain all real numbers, positive range, and horizontal asymptote $y=0$ .
The parent logarithm has domain $x>0$ , range all real numbers, and vertical asymptote $x=0$ .
Transformations shift or change a graph’s scale; logarithmic arguments must remain positive.
Use substitution to check points and solutions.

Check your understanding

Question 1

What is the domain of

h(x)=\log_2(4-x)

?

$x<4$
$x>4$
All real numbers
$x\ne4$

Show answer and explanation

x<4

The argument must be positive:

4-x>0

, so

x<4

.

Question 2

What is the horizontal asymptote of

p(x)=2^x+3

?

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

Show answer and explanation

y=3

The graph of

2^x

has horizontal asymptote

y=0

. Shifting it up three units gives asymptote

y=3

.

Question 3

Which value solves

3^x=\frac{1}{27}

?

$x=3$
$x=-3$
$x=\frac13$
$x=-\frac13$

Show answer and explanation

x=-3

Since

\frac{1}{27}=3^{-3}

, the exponents are equal and

x=-3

.

Key terms

Base: The fixed number raised to a power in an exponential expression.
Domain: The set of input values for which a function is defined.
Range: The set of output values a function can produce.
Asymptote: A line that a graph approaches in a particular direction.
Inverse functions: Functions that undo each other by exchanging input and output values.
Transformation: A change that shifts, stretches, or reflects a graph.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Mathematics (Math 30-1), expectation RF9. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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RF9 · Graph and analyze exponential and logarithmic functions

Alberta Grade 12 Mathematics

Connect equations, key features, and points to understand these related graphs

What you will learn

1. Review powers, coordinates, and logarithms

2. Parent functions and transformations

3. Use tables and features to sketch graphs

4. Practice and diploma-style reasoning

Parent-function comparison

Analyze a shifted exponential

Analyze a shifted logarithm

Solve an exponential equation and connect it to a logarithm

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through Math 30-1

About this lesson and its review