# 4.4: Graphs of Lines from Equations

**At Grade**Created by: CK-12

Can you graph the linear function \begin{align*}4y-5x=16\end{align*} on a Cartesian grid?

### Watch This

Khan Academy Slope and y-Intercept Intuition

### Guidance

The graph of any linear function can be plotted using the slope-intercept form of the equation.

- Step 1: Solve the equation for \begin{align*}y\end{align*} if it is not already in the form \begin{align*}y=mx+b\end{align*}.
- Step 2: To graph the function, start by plotting the \begin{align*}y\end{align*}-intercept.
- Step 3: Use the slope to find another point on the line. From the \begin{align*}y\end{align*}-intercept, move to the right the number of units equal to the denominator of the slope and then up or down the number of units equal to the numerator of the slope. Plot the point.
- Step 4: Connect these two points to form a line and extend the line.

*Note: You can repeat Step 3 multiple times in order to find more points on the line if you wish.*

Because the equations of horizontal and vertical lines are special, these types of lines can be graphed differently:

- The graph of a horizontal line will have an equation of the form \begin{align*}y=a\end{align*} where \begin{align*}a\end{align*} is the y-intercept of the line. You can simply draw a horizontal line through the y-intercept to sketch the graph.
- The graph of a vertical line will have an equation of the form \begin{align*}x=c\end{align*}, where \begin{align*}x\end{align*} is the x-intercept of the line. You can simply draw a vertical line through the x-intercept to sketch the graph.

#### Example A

For the following linear function, state the \begin{align*}y\end{align*}-intercept and the slope: \begin{align*}4x-3y-9=0\end{align*}.

**Solution:**

The first step is to rewrite the equation in the form \begin{align*}y=mx+b\end{align*}. To do this, solve the equation in terms of ‘\begin{align*}y\end{align*}’.

\begin{align*}& \qquad 4x-3y-9=0 && \text{Apply the zero principle to move} \ x \ \text{to the right side of the equation.}\\ & 4x{\color{red}-4x}-3y-9=0{\color{red}-4x}\\ & \qquad \quad -3y-9=-4x && \text{Apply the zero principle to move} \ -9 \ \text{to the right of the equation.}\\ & \quad \ \ -3y-9{\color{red}+9}=-4x{\color{red}+9}\\ & \qquad \qquad \ \ -3y=-4x+9 && \text{Divide all terms by the coefficient of} \ y. \ \text{Divide by} \ -3.\\ & \qquad \qquad \quad \frac{-3y}{{\color{red}-3}}=\frac{-4x}{{\color{red}-3}}+\frac{9}{{\color{red}-3}}\\ & \qquad \qquad \quad \frac{\cancel{-3}y}{{\color{red}\cancel{-3}}}=\frac{-4x}{{\color{red}-3}}+\frac{9}{{\color{red}-3}}\\ & \qquad \qquad \qquad \boxed{y=\frac{4}{3}x-3}\end{align*}

The \begin{align*}y\end{align*}-intercept is (0, -3) and the slope is \begin{align*}\frac{4}{3}\end{align*}.

#### Example B

Graph the linear function \begin{align*}y=\frac{-3}{5}x+7\end{align*} on a Cartesian grid.

**Solution:**

The \begin{align*}y\end{align*}-intercept is (0, 7) and the slope is \begin{align*}\frac{-3}{5}\end{align*}. Begin by plotting the \begin{align*}y\end{align*}-intercept on the grid.

From the \begin{align*}y\end{align*}-intercept, move to the right (run) 5 units and then move downward (rise) 3 units. Plot a point here.

Join the points with a straight line. Use a straight edge to draw the line.

#### Example C

Plot the following linear equations on a Cartesian grid.

i) \begin{align*}x=-3\end{align*}

ii) \begin{align*}y=5\end{align*}

**Solution:**

i) A line that has \begin{align*}x=-3\end{align*} as its equation passes through all points that have -3 as the \begin{align*}x-\end{align*}coordinate. The line also has a slope that is undefined. This line is parallel to the \begin{align*}y\end{align*}-axis.

ii) A line that \begin{align*}y=5\end{align*} has as its equation passes through all points that have 5 as the \begin{align*}y-\end{align*}coordinate. The line also has a slope of zero. This line is parallel to the \begin{align*}x\end{align*}-axis.

#### Concept Problem Revisited

Plot the linear function \begin{align*}4y-5x=16\end{align*} on a Cartesian grid.

The first step is to rewrite the function in slope-intercept form.

\begin{align*}& \qquad 4y-5x=16 && \text{Apply the zero principle to move} \ 5x \ \text{to the right side of the equation.}\\ & 4y-5x{\color{red}+5x}=16{\color{red}+5x}\\ & \qquad \qquad \ 4y=16+5x && \text{Divide every term} \ 4.\\ & \qquad \qquad \frac{4y}{{\color{red}4}}=\frac{16}{{\color{red}4}}+\frac{5x}{{\color{red}4}}\\ & \qquad \qquad \ \ y=4+\frac{5}{4}x && \text{Write the equation in the form} \ y=mx+b.\\ & \qquad \qquad \ \boxed{y=\frac{5}{4}x+4}\end{align*}

The slope of the line is \begin{align*}\frac{5}{4}\end{align*} and the \begin{align*}t\end{align*}-intercept is (0, 4)

Plot the \begin{align*}y\end{align*}-intercept at (0, 4). From the \begin{align*}y\end{align*}-intercept, move to the right 4 units and then move upward 5 units. Plot the point. Using a straight edge, join the points.

### Vocabulary

- Slope-Intercept Form

The ** slope-intercept form** is one method for writing the equation of a line. The slope-intercept form is \begin{align*}y=mx+b\end{align*} where \begin{align*}m\end{align*} refers to the slope and \begin{align*}b\end{align*} identifies the \begin{align*}y\end{align*}-intercept. This form is used to plot the graph of a linear function.

### Guided Practice

1. Using the slope-intercept method, graph the linear function \begin{align*}y=-\frac{3}{2}x-1\end{align*}

2. Using the slope-intercept method, graph the linear function \begin{align*}7x-3y-15=0\end{align*}

3. Graph the following lines on the same Cartesian grid. What shape is formed by the graphs?

- (a) \begin{align*}y=-3\end{align*}
- (b) \begin{align*}x=4\end{align*}
- (c) \begin{align*}y=2\end{align*}
- (d) \begin{align*}x=-6\end{align*}

**Answers:**

1. The slope of the line is \begin{align*}-\frac{3}{2}\end{align*} and the \begin{align*}y\end{align*}-intercept is (0, -1). Plot the \begin{align*}y\end{align*}-intercept. Apply the slope to the \begin{align*}y\end{align*}-intercept. Use a straight edge to join the two points.

2. Write the equation in slope-intercept form.

- \begin{align*}& \quad \ \ 7x-3y-15=0 && \text{Solve the equation in terms of the variable} \ y.\\ & 7x{\color{red}-7x}-3y-15=0{\color{red}-7x}\\ & \qquad \ \ -3y-15=-7x\\ & \quad -3y-15{\color{red}+15}=-7x{\color{red}+15}\\ & \qquad \qquad \quad -3y=-7x+15\\ & \qquad \qquad \quad \ \frac{-3y}{{\color{red}-3}}=\frac{-7x}{{\color{red}-3}}+\frac{15}{{\color{red}-3}}\\ & \qquad \qquad \qquad \ \boxed{y =\frac{7}{3}x-5}\end{align*}
- The slope is \begin{align*}\frac{7}{3}\end{align*} and the \begin{align*}y\end{align*}-intercept is (0, -5). Plot the \begin{align*}y\end{align*}-intercept. Apply the slope to the \begin{align*}y\end{align*}-intercept. Use a straight edge to join the two points.

3. There are four lines to be graphed. The lines \begin{align*}a\end{align*} and \begin{align*}c\end{align*} are lines with a slope of zero and are parallel to the \begin{align*}x\end{align*}-axis. The lines \begin{align*}b\end{align*} and \begin{align*}d\end{align*} are lines that have a slope that is undefined and are parallel to the \begin{align*}x\end{align*}-axis. The shape is a rectangle.

### Practice

For each of the following linear functions, state the slope and the \begin{align*}y\end{align*}-intercept:

- \begin{align*}y=\frac{5}{8}x+3\end{align*}
- \begin{align*}4x+5y-3=0\end{align*}
- \begin{align*}4x-3y+21=0\end{align*}
- \begin{align*}y=-7\end{align*}
- \begin{align*}9y-8x=27\end{align*}

Using the slope-intercept method, graph the following linear functions:

- \begin{align*}3x+y=4\end{align*}
- \begin{align*}3x-2y=-4\end{align*}
- \begin{align*}2x+6y+18=0\end{align*}
- \begin{align*}3x+7y=0\end{align*}
- \begin{align*}4x-5y=-30\end{align*}

Graph the following linear equations and state the slope of the line:

- \begin{align*}x=-5\end{align*}
- \begin{align*}y=8\end{align*}
- \begin{align*}y=-4\end{align*}
- \begin{align*}x=7\end{align*}

### Image Attributions

## Description

## Learning Objectives

Here you will learn how to graph a linear function from its equation without first making a table of values.

## Difficulty Level:

At Grade## Tags:

## Subjects:

## Date Created:

Dec 19, 2012## Last Modified:

Dec 04, 2014## Vocabulary

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