3.3 Functions, graphs, and set relationships

Achievable Praxis Core: Math (5733)

3. Algebra and geometry

Our Praxis Core: Math course is currently in development and is a work-in-progress.

Functions, graphs, and set relationships

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This section covers functions and their relationship to relations, including how to evaluate and graph functions, determine domain and range, and read key features of a line, such as slope and intercepts. It also covers common linear equation forms - slope-intercept, point-slope, and standard - and uses the intercept method to graph equations in standard form.

Defining functions

A relation is any collection of ordered pairs $(x, y)$ . A function is a relation with one key rule: each input $x$ is paired with exactly one output $y$ .

So, one input can’t produce two different outputs. However, different inputs can share the same output.

For example:

The pairs $(1, 2)$ and $(2, 2)$ can both belong to a function because the inputs are different.
The pairs $(1, 2)$ and $(1, 3)$ cannot belong to a function because the same input $1$ is paired with two different outputs.

Vertical-line test

The vertical-line test checks whether a graphed relation is a function.

Draw (or imagine) vertical lines through the graph.
If any vertical line intersects the graph more than once, the relation is not a function.
If every vertical line intersects at most once, the relation is a function.

A graph is a picture of a relation or function made by plotting points $(x, y)$ on the $x y$ -plane.

Example: Determine if a relation is a function

Given the ordered pairs $(- 2, 1)$ , $(0, 3)$ , $(1, 4)$ , $(2, 3)$

List the inputs: $- 2, 0, 1, 2$ .

Each input appears only once, so no input is paired with two different outputs.

Answer: Function

Evaluating functions

Evaluating a function means finding the output for a specific input. The notation $f (a)$ means “the value of the function when $x = a$ .”

The main skill is substitution:

Replace every $x$ in the formula with the given input value.
Use parentheses around the substituted value.
Simplify using the order of operations.

Parentheses matter most when the input is negative or when the expression includes exponents or fractions.

Watch out with negative inputs: always wrap a negative input in parentheses before applying exponents. For example, $(- 2)^{2} = 4$ , not $- 4$ . Dropping the parentheses is one of the most common errors on function evaluation problems.

Example: Evaluate a quadratic function with a negative input

Let $f (x) = 2 x^{2} - 3 x + 5$ . Find $f (- 2)$ .

Substitute, keeping parentheses around $- 2$ : $f (- 2) = 2 (- 2)^{2} - 3 (- 2) + 5$ .

Compute each term: $2 (4) + 6 + 5 = 8 + 6 + 5 = 19$ .

Answer: $19$

Domain and range

The domain of a function is the set of all input values $x$ for which the function is defined. In practice, you look for inputs that make the expression invalid and exclude them.

Common restrictions include:

Division by zero: denominators cannot equal $0$ .
Even roots (such as square roots): the expression inside the root must be greater than or equal to $0$ .

The range is the set of all possible output values $f (x)$ produced by inputs in the domain. On the Praxis exam, range questions are typically limited to cases you can read from the form or the graph. For example, if a graph shows the lowest point at $y = 1$ and extends upward forever, the range is $y \geq 1$ .

Example: Find the domain

For $h (x) = \frac{3}{x - 4}$ : the denominator $x - 4$ cannot equal $0$ , so solve $x - 4 = 0 \Rightarrow x = 4$ and exclude it. Domain: $x \neq = 4$ .

For $k (x) = x + 5$ : the radicand must be nonnegative, so $x + 5 \geq 0 \Rightarrow x \geq - 5$ . Domain: $[- 5, \infty)$ .

Graphing linear functions

Graphing a linear function shows all solutions to the equation. Each point on the graph is an ordered pair $(x, y)$ that satisfies the equation, and all of those points lie on a straight line.

Because a linear function has a constant rate of change, its graph extends infinitely in both directions unless the problem states a restriction.

The method you use depends on the form of the equation:

Some forms make slope and intercepts easy to read.
Other forms are more convenient when you’re given a point on the line.

Being able to move between equations, tables, and graphs makes these questions much faster.

Slope-intercept form

Use slope-intercept form when you can read the slope $m$ and $y$ -intercept $b$ directly from the equation. It’s the most convenient form for quickly sketching a line.

For $f (x) = m x + b$ :

Plot the $y$ -intercept $(0, b)$ .
Use slope $m = \frac{Δ y}{Δ x}$ to find a second point.
Draw the line through both points.

Example: Graph $y = 2 x + 1$

$y$ -intercept: $(0, 1)$ .

Slope $2 = \frac{2}{1}$ , so from $(0, 1)$ go right $1$ and up $2$ to $(1, 3)$ .

Linear function slope of two

Answer: See graph

Horizontal and vertical lines

Two special cases come up often on the Praxis and are easy to confuse.

Horizontal lines have the equation $y = b$ , where $b$ is a constant. Every point on the line has the same $y$ -value, so the slope is $0$ (zero rise, any run). For example, $y = 3$ is a horizontal line crossing the $y$ -axis at $(0, 3)$ .

Vertical lines have the equation $x = a$ , where $a$ is a constant. Every point on the line has the same $x$ -value. Moving from one point to another involves zero run, so slope $= \frac{rise}{0}$ , which is undefined. A vertical line itself is intersected by a vertical line at infinitely many points, so it fails the vertical-line test and is not a function.

Quick reference:

$y = b$ → horizontal line, slope $= 0$ , is a function
$x = a$ → vertical line, slope is undefined, is not a function

Point-slope form

Use point-slope form when you know a point on the line and the slope, but the point isn’t the $y$ -intercept. It’s especially handy when you derive a line from two given points (see the next section).

The form

$y - y_{1} = m (x - x_{1})$

builds the line around the anchor point $(x_{1}, y_{1})$ .

When you know $(x_{1}, y_{1})$ and slope $m$ :

Write $y - y_{1} = m (x - x_{1})$ .
Use the slope to plot additional points, or rearrange into slope-intercept form if needed.

Example: Graph using point-slope form

Graph slope $3$ through $(2, - 1)$ .

Start with the point-slope formula and substitute the given values: $y - (- 1) = 3 (x - 2)$ .

Simplify the equation: $y + 1 = 3 x - 6$ , so $y = 3 x - 7$ .

Plot the given point $(2, - 1)$ .

Use the slope $3 = \frac{3}{1}$ to move right $1$ and up $3$ to $(3, 2)$ .

Linear function with slope of three

Answer: See graph

Finding a line from two points

When a problem gives you two points but no slope, you need to calculate the slope first, then use point-slope form to write the equation.

Step 1: Compute the slope using the two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$

Keep the order consistent - use the same point as “point 2” in both the numerator and denominator, or you’ll flip the sign of the slope.

Step 2: Substitute $m$ and either point into point-slope form:

$y - y_{1} = m (x - x_{1})$

Step 3: Simplify into slope-intercept form if needed.

Example: Find the equation of a line through two points

A line passes through $(1, 3)$ and $(4, 9)$ .

Compute the slope: $m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2$ .

Substitute into point-slope form using $(1, 3)$ : $y - 3 = 2 (x - 1)$ .

Simplify: $y = 2 x + 1$ .

Answer: $y = 2 x + 1$

Standard form and intercept method

Use standard form when the equation is already written as $A x + B y = C$ and you want to graph it quickly without rearranging - the intercept method gets you two points with minimal algebra.

The idea is to find where the line crosses each axis:

The $x$ -intercept occurs where the graph crosses the $x$ -axis, which happens when $y = 0$ .
The $y$ -intercept occurs where the graph crosses the $y$ -axis, which happens when $x = 0$ .

Once you have both intercepts, plot them and draw the line through the two points.

Example: Find the intercepts of $2 x + 3 y = 6$

Set $y = 0$ to find the $x$ -intercept: $2 x = 6 \Rightarrow x = 3$ , so $(3, 0)$ .

Set $x = 0$ to find the $y$ -intercept: $3 y = 6 \Rightarrow y = 2$ , so $(0, 2)$ .

Answer: $(3, 0)$ and $(0, 2)$

A function assigns exactly one output to each input; use the vertical-line test to verify.
To evaluate $f (a)$ , substitute $x = a$ and simplify.
The domain excludes inputs that cause division by zero or invalid roots; the range is the set of possible outputs.
Choose a graph form:
- Slope-intercept form $y = m x + b$ : plot $(0, b)$ and use slope $m = \frac{Δ y}{Δ x}$ .
- Point-slope form $y - y_{1} = m (x - x_{1})$ : use when a point $(x_{1}, y_{1})$ and slope $m$ are known.
- Standard form $A x + B y = C$ : use intercepts $(\frac{C}{A}, 0)$ and $(0, \frac{C}{B})$ .
Read intercepts by setting $x = 0$ or $y = 0$ .
Plot at least two accurate points before drawing the line.

Functions, graphs, and set relationships

Defining functions

A relation is any collection of ordered pairs $(x, y)$ . A function is a relation with one key rule: each input $x$ is paired with exactly one output $y$ .

So, one input can’t produce two different outputs. However, different inputs can share the same output.

For example:

The pairs $(1, 2)$ and $(2, 2)$ can both belong to a function because the inputs are different.
The pairs $(1, 2)$ and $(1, 3)$ cannot belong to a function because the same input $1$ is paired with two different outputs.

Vertical-line test

The vertical-line test checks whether a graphed relation is a function.

Draw (or imagine) vertical lines through the graph.
If any vertical line intersects the graph more than once, the relation is not a function.
If every vertical line intersects at most once, the relation is a function.

A graph is a picture of a relation or function made by plotting points $(x, y)$ on the $x y$ -plane.

Example: Determine if a relation is a function

Given the ordered pairs $(- 2, 1)$ , $(0, 3)$ , $(1, 4)$ , $(2, 3)$

List the inputs: $- 2, 0, 1, 2$ .

Each input appears only once, so no input is paired with two different outputs.

Answer: Function

Evaluating functions

Evaluating a function means finding the output for a specific input. The notation $f (a)$ means “the value of the function when $x = a$ .”

The main skill is substitution:

Replace every $x$ in the formula with the given input value.
Use parentheses around the substituted value.
Simplify using the order of operations.

Parentheses matter most when the input is negative or when the expression includes exponents or fractions.

Example: Evaluate a quadratic function with a negative input

Let $f (x) = 2 x^{2} - 3 x + 5$ . Find $f (- 2)$ .

Substitute, keeping parentheses around $- 2$ : $f (- 2) = 2 (- 2)^{2} - 3 (- 2) + 5$ .

Compute each term: $2 (4) + 6 + 5 = 8 + 6 + 5 = 19$ .

Answer: $19$

Domain and range

The domain of a function is the set of all input values $x$ for which the function is defined. In practice, you look for inputs that make the expression invalid and exclude them.

Common restrictions include:

Division by zero: denominators cannot equal $0$ .
Even roots (such as square roots): the expression inside the root must be greater than or equal to $0$ .

Example: Find the domain

For $h (x) = \frac{3}{x - 4}$ : the denominator $x - 4$ cannot equal $0$ , so solve $x - 4 = 0 \Rightarrow x = 4$ and exclude it. Domain: $x \neq = 4$ .

For $k (x) = x + 5$ : the radicand must be nonnegative, so $x + 5 \geq 0 \Rightarrow x \geq - 5$ . Domain: $[- 5, \infty)$ .

Graphing linear functions

Graphing a linear function shows all solutions to the equation. Each point on the graph is an ordered pair $(x, y)$ that satisfies the equation, and all of those points lie on a straight line.

Because a linear function has a constant rate of change, its graph extends infinitely in both directions unless the problem states a restriction.

The method you use depends on the form of the equation:

Some forms make slope and intercepts easy to read.
Other forms are more convenient when you’re given a point on the line.

Being able to move between equations, tables, and graphs makes these questions much faster.

Slope-intercept form

Use slope-intercept form when you can read the slope $m$ and $y$ -intercept $b$ directly from the equation. It’s the most convenient form for quickly sketching a line.

For $f (x) = m x + b$ :

Plot the $y$ -intercept $(0, b)$ .
Use slope $m = \frac{Δ y}{Δ x}$ to find a second point.
Draw the line through both points.

Example: Graph $y = 2 x + 1$

$y$ -intercept: $(0, 1)$ .

Slope $2 = \frac{2}{1}$ , so from $(0, 1)$ go right $1$ and up $2$ to $(1, 3)$ .

Answer: See graph

Horizontal and vertical lines

Two special cases come up often on the Praxis and are easy to confuse.

Quick reference:

$y = b$ → horizontal line, slope $= 0$ , is a function
$x = a$ → vertical line, slope is undefined, is not a function

Point-slope form

The form

$y - y_{1} = m (x - x_{1})$

builds the line around the anchor point $(x_{1}, y_{1})$ .

When you know $(x_{1}, y_{1})$ and slope $m$ :

Write $y - y_{1} = m (x - x_{1})$ .
Use the slope to plot additional points, or rearrange into slope-intercept form if needed.

Example: Graph using point-slope form

Graph slope $3$ through $(2, - 1)$ .

Start with the point-slope formula and substitute the given values: $y - (- 1) = 3 (x - 2)$ .

Simplify the equation: $y + 1 = 3 x - 6$ , so $y = 3 x - 7$ .

Plot the given point $(2, - 1)$ .

Use the slope $3 = \frac{3}{1}$ to move right $1$ and up $3$ to $(3, 2)$ .

Answer: See graph

Finding a line from two points

When a problem gives you two points but no slope, you need to calculate the slope first, then use point-slope form to write the equation.

Step 1: Compute the slope using the two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$

Keep the order consistent - use the same point as “point 2” in both the numerator and denominator, or you’ll flip the sign of the slope.

Step 2: Substitute $m$ and either point into point-slope form:

$y - y_{1} = m (x - x_{1})$

Step 3: Simplify into slope-intercept form if needed.

Example: Find the equation of a line through two points

A line passes through $(1, 3)$ and $(4, 9)$ .

Compute the slope: $m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2$ .

Substitute into point-slope form using $(1, 3)$ : $y - 3 = 2 (x - 1)$ .

Simplify: $y = 2 x + 1$ .

Answer: $y = 2 x + 1$

Standard form and intercept method

Use standard form when the equation is already written as $A x + B y = C$ and you want to graph it quickly without rearranging - the intercept method gets you two points with minimal algebra.

The idea is to find where the line crosses each axis:

The $x$ -intercept occurs where the graph crosses the $x$ -axis, which happens when $y = 0$ .
The $y$ -intercept occurs where the graph crosses the $y$ -axis, which happens when $x = 0$ .

Once you have both intercepts, plot them and draw the line through the two points.

Example: Find the intercepts of $2 x + 3 y = 6$

Set $y = 0$ to find the $x$ -intercept: $2 x = 6 \Rightarrow x = 3$ , so $(3, 0)$ .

Set $x = 0$ to find the $y$ -intercept: $3 y = 6 \Rightarrow y = 2$ , so $(0, 2)$ .

Answer: $(3, 0)$ and $(0, 2)$

Achievable Praxis Core: Math (5733)

3. Algebra and geometry

Our Praxis Core: Math course is currently in development and is a work-in-progress.

Functions, graphs, and set relationships

9 min read

Font

Discuss

Feedback

Defining functions

A relation is any collection of ordered pairs $(x, y)$ . A function is a relation with one key rule: each input $x$ is paired with exactly one output $y$ .

So, one input can’t produce two different outputs. However, different inputs can share the same output.

For example:

The pairs $(1, 2)$ and $(2, 2)$ can both belong to a function because the inputs are different.
The pairs $(1, 2)$ and $(1, 3)$ cannot belong to a function because the same input $1$ is paired with two different outputs.

Vertical-line test

The vertical-line test checks whether a graphed relation is a function.

Draw (or imagine) vertical lines through the graph.
If any vertical line intersects the graph more than once, the relation is not a function.
If every vertical line intersects at most once, the relation is a function.

A graph is a picture of a relation or function made by plotting points $(x, y)$ on the $x y$ -plane.

Example: Determine if a relation is a function

Given the ordered pairs $(- 2, 1)$ , $(0, 3)$ , $(1, 4)$ , $(2, 3)$

List the inputs: $- 2, 0, 1, 2$ .

Each input appears only once, so no input is paired with two different outputs.

Answer: Function

Evaluating functions

Evaluating a function means finding the output for a specific input. The notation $f (a)$ means “the value of the function when $x = a$ .”

The main skill is substitution:

Replace every $x$ in the formula with the given input value.
Use parentheses around the substituted value.
Simplify using the order of operations.

Parentheses matter most when the input is negative or when the expression includes exponents or fractions.

Example: Evaluate a quadratic function with a negative input

Let $f (x) = 2 x^{2} - 3 x + 5$ . Find $f (- 2)$ .

Substitute, keeping parentheses around $- 2$ : $f (- 2) = 2 (- 2)^{2} - 3 (- 2) + 5$ .

Compute each term: $2 (4) + 6 + 5 = 8 + 6 + 5 = 19$ .

Answer: $19$

Domain and range

The domain of a function is the set of all input values $x$ for which the function is defined. In practice, you look for inputs that make the expression invalid and exclude them.

Common restrictions include:

Division by zero: denominators cannot equal $0$ .
Even roots (such as square roots): the expression inside the root must be greater than or equal to $0$ .

Example: Find the domain

For $h (x) = \frac{3}{x - 4}$ : the denominator $x - 4$ cannot equal $0$ , so solve $x - 4 = 0 \Rightarrow x = 4$ and exclude it. Domain: $x \neq = 4$ .

For $k (x) = x + 5$ : the radicand must be nonnegative, so $x + 5 \geq 0 \Rightarrow x \geq - 5$ . Domain: $[- 5, \infty)$ .

Graphing linear functions

Graphing a linear function shows all solutions to the equation. Each point on the graph is an ordered pair $(x, y)$ that satisfies the equation, and all of those points lie on a straight line.

Because a linear function has a constant rate of change, its graph extends infinitely in both directions unless the problem states a restriction.

The method you use depends on the form of the equation:

Some forms make slope and intercepts easy to read.
Other forms are more convenient when you’re given a point on the line.

Being able to move between equations, tables, and graphs makes these questions much faster.

Slope-intercept form

Use slope-intercept form when you can read the slope $m$ and $y$ -intercept $b$ directly from the equation. It’s the most convenient form for quickly sketching a line.

For $f (x) = m x + b$ :

Plot the $y$ -intercept $(0, b)$ .
Use slope $m = \frac{Δ y}{Δ x}$ to find a second point.
Draw the line through both points.

Example: Graph $y = 2 x + 1$

$y$ -intercept: $(0, 1)$ .

Slope $2 = \frac{2}{1}$ , so from $(0, 1)$ go right $1$ and up $2$ to $(1, 3)$ .

Linear function slope of two

Answer: See graph

Horizontal and vertical lines

Two special cases come up often on the Praxis and are easy to confuse.

Quick reference:

$y = b$ → horizontal line, slope $= 0$ , is a function
$x = a$ → vertical line, slope is undefined, is not a function

Point-slope form

The form

$y - y_{1} = m (x - x_{1})$

builds the line around the anchor point $(x_{1}, y_{1})$ .

When you know $(x_{1}, y_{1})$ and slope $m$ :

Write $y - y_{1} = m (x - x_{1})$ .
Use the slope to plot additional points, or rearrange into slope-intercept form if needed.

Example: Graph using point-slope form

Graph slope $3$ through $(2, - 1)$ .

Start with the point-slope formula and substitute the given values: $y - (- 1) = 3 (x - 2)$ .

Simplify the equation: $y + 1 = 3 x - 6$ , so $y = 3 x - 7$ .

Plot the given point $(2, - 1)$ .

Use the slope $3 = \frac{3}{1}$ to move right $1$ and up $3$ to $(3, 2)$ .

Linear function with slope of three

Answer: See graph

Finding a line from two points

When a problem gives you two points but no slope, you need to calculate the slope first, then use point-slope form to write the equation.

Step 1: Compute the slope using the two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$

Keep the order consistent - use the same point as “point 2” in both the numerator and denominator, or you’ll flip the sign of the slope.

Step 2: Substitute $m$ and either point into point-slope form:

$y - y_{1} = m (x - x_{1})$

Step 3: Simplify into slope-intercept form if needed.

Example: Find the equation of a line through two points

A line passes through $(1, 3)$ and $(4, 9)$ .

Compute the slope: $m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2$ .

Substitute into point-slope form using $(1, 3)$ : $y - 3 = 2 (x - 1)$ .

Simplify: $y = 2 x + 1$ .

Answer: $y = 2 x + 1$

Standard form and intercept method

Use standard form when the equation is already written as $A x + B y = C$ and you want to graph it quickly without rearranging - the intercept method gets you two points with minimal algebra.

The idea is to find where the line crosses each axis:

The $x$ -intercept occurs where the graph crosses the $x$ -axis, which happens when $y = 0$ .
The $y$ -intercept occurs where the graph crosses the $y$ -axis, which happens when $x = 0$ .

Once you have both intercepts, plot them and draw the line through the two points.

Example: Find the intercepts of $2 x + 3 y = 6$

Set $y = 0$ to find the $x$ -intercept: $2 x = 6 \Rightarrow x = 3$ , so $(3, 0)$ .

Set $x = 0$ to find the $y$ -intercept: $3 y = 6 \Rightarrow y = 2$ , so $(0, 2)$ .

Answer: $(3, 0)$ and $(0, 2)$

A function assigns exactly one output to each input; use the vertical-line test to verify.
To evaluate $f (a)$ , substitute $x = a$ and simplify.
The domain excludes inputs that cause division by zero or invalid roots; the range is the set of possible outputs.
Choose a graph form:
- Slope-intercept form $y = m x + b$ : plot $(0, b)$ and use slope $m = \frac{Δ y}{Δ x}$ .
- Point-slope form $y - y_{1} = m (x - x_{1})$ : use when a point $(x_{1}, y_{1})$ and slope $m$ are known.
- Standard form $A x + B y = C$ : use intercepts $(\frac{C}{A}, 0)$ and $(0, \frac{C}{B})$ .
Read intercepts by setting $x = 0$ or $y = 0$ .
Plot at least two accurate points before drawing the line.

Functions, graphs, and set relationships

Defining functions

A relation is any collection of ordered pairs $(x, y)$ . A function is a relation with one key rule: each input $x$ is paired with exactly one output $y$ .

So, one input can’t produce two different outputs. However, different inputs can share the same output.

For example:

The pairs $(1, 2)$ and $(2, 2)$ can both belong to a function because the inputs are different.
The pairs $(1, 2)$ and $(1, 3)$ cannot belong to a function because the same input $1$ is paired with two different outputs.

Vertical-line test

The vertical-line test checks whether a graphed relation is a function.

Draw (or imagine) vertical lines through the graph.
If any vertical line intersects the graph more than once, the relation is not a function.
If every vertical line intersects at most once, the relation is a function.

A graph is a picture of a relation or function made by plotting points $(x, y)$ on the $x y$ -plane.

Example: Determine if a relation is a function

Given the ordered pairs $(- 2, 1)$ , $(0, 3)$ , $(1, 4)$ , $(2, 3)$

List the inputs: $- 2, 0, 1, 2$ .

Each input appears only once, so no input is paired with two different outputs.

Answer: Function

Evaluating functions

Evaluating a function means finding the output for a specific input. The notation $f (a)$ means “the value of the function when $x = a$ .”

The main skill is substitution:

Replace every $x$ in the formula with the given input value.
Use parentheses around the substituted value.
Simplify using the order of operations.

Parentheses matter most when the input is negative or when the expression includes exponents or fractions.

Example: Evaluate a quadratic function with a negative input

Let $f (x) = 2 x^{2} - 3 x + 5$ . Find $f (- 2)$ .

Substitute, keeping parentheses around $- 2$ : $f (- 2) = 2 (- 2)^{2} - 3 (- 2) + 5$ .

Compute each term: $2 (4) + 6 + 5 = 8 + 6 + 5 = 19$ .

Answer: $19$

Domain and range

The domain of a function is the set of all input values $x$ for which the function is defined. In practice, you look for inputs that make the expression invalid and exclude them.

Common restrictions include:

Division by zero: denominators cannot equal $0$ .
Even roots (such as square roots): the expression inside the root must be greater than or equal to $0$ .

Example: Find the domain

For $h (x) = \frac{3}{x - 4}$ : the denominator $x - 4$ cannot equal $0$ , so solve $x - 4 = 0 \Rightarrow x = 4$ and exclude it. Domain: $x \neq = 4$ .

For $k (x) = x + 5$ : the radicand must be nonnegative, so $x + 5 \geq 0 \Rightarrow x \geq - 5$ . Domain: $[- 5, \infty)$ .

Graphing linear functions

Graphing a linear function shows all solutions to the equation. Each point on the graph is an ordered pair $(x, y)$ that satisfies the equation, and all of those points lie on a straight line.

Because a linear function has a constant rate of change, its graph extends infinitely in both directions unless the problem states a restriction.

The method you use depends on the form of the equation:

Some forms make slope and intercepts easy to read.
Other forms are more convenient when you’re given a point on the line.

Being able to move between equations, tables, and graphs makes these questions much faster.

Slope-intercept form

Use slope-intercept form when you can read the slope $m$ and $y$ -intercept $b$ directly from the equation. It’s the most convenient form for quickly sketching a line.

For $f (x) = m x + b$ :

Plot the $y$ -intercept $(0, b)$ .
Use slope $m = \frac{Δ y}{Δ x}$ to find a second point.
Draw the line through both points.

Example: Graph $y = 2 x + 1$

$y$ -intercept: $(0, 1)$ .

Slope $2 = \frac{2}{1}$ , so from $(0, 1)$ go right $1$ and up $2$ to $(1, 3)$ .

Answer: See graph

Horizontal and vertical lines

Two special cases come up often on the Praxis and are easy to confuse.

Quick reference:

$y = b$ → horizontal line, slope $= 0$ , is a function
$x = a$ → vertical line, slope is undefined, is not a function

Point-slope form

The form

$y - y_{1} = m (x - x_{1})$

builds the line around the anchor point $(x_{1}, y_{1})$ .

When you know $(x_{1}, y_{1})$ and slope $m$ :

Write $y - y_{1} = m (x - x_{1})$ .
Use the slope to plot additional points, or rearrange into slope-intercept form if needed.

Example: Graph using point-slope form

Graph slope $3$ through $(2, - 1)$ .

Start with the point-slope formula and substitute the given values: $y - (- 1) = 3 (x - 2)$ .

Simplify the equation: $y + 1 = 3 x - 6$ , so $y = 3 x - 7$ .

Plot the given point $(2, - 1)$ .

Use the slope $3 = \frac{3}{1}$ to move right $1$ and up $3$ to $(3, 2)$ .

Answer: See graph

Finding a line from two points

When a problem gives you two points but no slope, you need to calculate the slope first, then use point-slope form to write the equation.

Step 1: Compute the slope using the two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ :

$m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$

Keep the order consistent - use the same point as “point 2” in both the numerator and denominator, or you’ll flip the sign of the slope.

Step 2: Substitute $m$ and either point into point-slope form:

$y - y_{1} = m (x - x_{1})$

Step 3: Simplify into slope-intercept form if needed.

Example: Find the equation of a line through two points

A line passes through $(1, 3)$ and $(4, 9)$ .

Compute the slope: $m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2$ .

Substitute into point-slope form using $(1, 3)$ : $y - 3 = 2 (x - 1)$ .

Simplify: $y = 2 x + 1$ .

Answer: $y = 2 x + 1$

Standard form and intercept method

Use standard form when the equation is already written as $A x + B y = C$ and you want to graph it quickly without rearranging - the intercept method gets you two points with minimal algebra.

The idea is to find where the line crosses each axis:

The $x$ -intercept occurs where the graph crosses the $x$ -axis, which happens when $y = 0$ .
The $y$ -intercept occurs where the graph crosses the $y$ -axis, which happens when $x = 0$ .

Once you have both intercepts, plot them and draw the line through the two points.

Example: Find the intercepts of $2 x + 3 y = 6$

Set $y = 0$ to find the $x$ -intercept: $2 x = 6 \Rightarrow x = 3$ , so $(3, 0)$ .

Set $x = 0$ to find the $y$ -intercept: $3 y = 6 \Rightarrow y = 2$ , so $(0, 2)$ .

Answer: $(3, 0)$ and $(0, 2)$