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# 4.4: Graphing the Linear Functions y = mx + b; y = a; x = a

Difficulty Level: At Grade Created by: CK-12

## The Graph of y = mx + b

Introduction

In this lesson you will learn how to graph a linear function that is written in slope-intercept form. This method of graphing will produce the same graph that you would have if you were to create a table of values. You will also learn to graph the special lines that are parallel to the $x-$axis and parallel to the $y-$axis without doing mathematical calculations.

Objectives

The lesson objectives for Graphing Linear Functions are:

• Understanding the process for graphing a linear function in slope-intercept form.
• Plotting the function on a Cartesian grid.
• Graphing the special lines $y = a$ and $x = a$.

Introduction

In previous lessons you have learned to graph linear functions by plotting the $x-$ and $y-$intercepts and by creating a table of values to generate the points to be plotted to graph the function. Now you will learn to create the graph of $y = mx+b$ by plotting only two points. The points that will be plotted are the $y-$intercept--that is the ‘$b$’ value of the linear function--and the point that results by applying the value of the slope, or the ‘$m$’ value of the linear function.

Watch This

Guidance

A graph for a linear function in the form $y = mx+b$ can be created by plotting the $y-$intercept and then by applying the value of the slope to the $y-$intercept.

Example A

For the following linear function, state the $y-$intercept and the slope. $4x-3y-9=0$

The first step is to rewrite the equation in the form $y=mx+b$. To do this, solve the equation in terms of ‘$y$’.

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

The $y-$intercept is (0, –3) and the slope is $\frac{4}{3}$.

Example B

Plot the linear function $y=\frac{-3}{5}x+7$ on a Cartesian grid.

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

From the $y-$intercept, move to the right (run) 5 units and then move downward (rise) 3 units. Plot a point here.

When the two points have been plotted, one point will be on the $y-$axis and the other will be to the right of the $y-$axis. The denominator of the slope should not be a negative number. Therefore, the value of ‘run’ will always be in a positive direction from the $y-$intercept. The second point will be lower than the $y-$intercept if the rise is a negative value and above the $y-$intercept if the rise is a positive value. In the above graph the second point (5, 4) is below the $y-$intercept (0, 7) since the rise is –3.

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

Example C

Plot the linear function $4y-5x=16$ on a Cartesian grid.

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

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

The slope of the line is $\frac{5}{4}$ and the $y-$intercept is (0, 4)

Plot the $y-$intercept at (0, 4). From the $y-$intercept, move to the right 4 units and then move upward 5 units. Plot the point. Using a straight edge, join the points.

Example D

Plot the following linear equations on a Cartesian grid.

i) $x=-3$

ii) $y=5$

i) A line that has $x=-3$ as its equation passes through all points that have –3 as the $x-$coordinate. The line also has a slope that is undefined. This line is parallel to the $y-$axis.

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

Vocabulary

Slope-Intercept Form

The slope-intercept form is one method for writing the equation of a line. The slope-intercept form is $y=mx+b$ where $m$ refers to the slope and $b$ identifies the $y-$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 $y=-\frac{3}{2}x-1$

2. Using the slope-intercept method, graph the linear function $7x-3y-15=0$

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

(a) $y=-3$

(b) $x=4$

(c) $y=2$

(d) $x=-6$

1. $y=-\frac{3}{2}x-1$

The slope of the line is $-\frac{3}{2}$ and the $y-$intercept is (0, –1).

Plot the $y-$intercept. Apply the slope to the $y-$intercept. Use a straight edge to join the two points.

2. $7x-3y-15=0$

Write the equation in slope-intercept form.

$& \quad \ \ 7x-3y-15=0 && \text{Solve the equation in terms of the variable} \ y.\\& 7x{\color{red}-7x}-3y-15=0{\color{red}-7x}\\& \qquad \ \ -3y-15=-7x\\& \quad -3y-15{\color{red}+15}=-7x{\color{red}+15}\\& \qquad \qquad \quad -3y=-7x+15\\& \qquad \qquad \quad \ \frac{-3y}{{\color{red}-3}}=\frac{-7x}{{\color{red}-3}}+\frac{15}{{\color{red}-3}}\\& \qquad \qquad \qquad \ \boxed{y =\frac{7}{3}x-5}$

The slope is $\frac{7}{3}$ and the $y-$intercept is (0, –5).

Plot the $y-$intercept. Apply the slope to the $y-$intercept. Use a straight edge to join the two points.

3. There are four lines to be graphed. The lines $a$ and $c$ are lines with a slope of zero and are parallel to the $x-$axis. The lines $b$ and $d$ are lines that have a slope that is undefined and are parallel to the $x-$axis.

(a) $y=-3$

(b) $x=4$

(c) $y=2$

(d) $x=-6$

The shape is a rectangle.

Summary

In this lesson you have learned that the graph of a linear function can be plotted by using the slope-intercept method. To use this method, the function must be written in the form $y=mx+b$. To plot the function, the $y-$intercept is plotted on the grid first. From the $y-$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. These two points are then joined to form the graph of the linear function. If the function is given in standard form, it must be written in slope-intercept form before the graph can be drawn.

You also learned that the graph of a line parallel to either the $x-$axis or the $y-$axis can be plotted very quickly and without doing any mathematical calculations.

Problem Set

For each of the following linear functions, state the slope and the $y-$intercept:

1. $y=\frac{5}{8}x+3$
2. $4x+5y-3=0$
3. $4x-3y+21=0$
4. $y=-7$
5. $9y-8x=27$

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

1. $3x+y=4$
2. $3x-2y=-4$
3. $2x+6y+18=0$
4. $3x+7y=0$
5. $4x-5y=-30$

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

1. $x=-5$
2. $y=8$
3. $y=-3$
4. $x=7$

For each of the following linear...

1. $y=\frac{5}{8}x+3$ The equation is written in the form $y=mx+b$. The slope of the line is $\frac{5}{8}$ and the $y-$intercept is (0, 3).
1. $4x-3y+21=0$ The equation must be written in slope-intercept form.

$& \qquad \ 4x-3y+21=0\\& 4x-4x-3y+21=0-4x\\& \qquad \quad \ -3y+21=-4x\\& \quad -3y+21-21=-4x-21\\& \qquad \qquad \quad \ -3y=-4x-21\\& \qquad \qquad \quad \ \ \frac{-3y}{-3}=\frac{-4x}{-3}-\frac{21}{-3}\\& \qquad \qquad \qquad \quad \boxed{y=\frac{4}{3}x+7}$

The slope of the line is $\frac{4}{3}$ and the $y-$intercept is (0, 7)

1. $9y-8x=27$ The equation must be written in slope-intercept form.

$& \qquad \ 9y-8x=27\\& 9y-8x+8x=27+8x\\& \qquad \qquad \ \ 9y=27+8x\\& \qquad \qquad \ \frac{9y}{9}=\frac{27}{9}+\frac{8x}{9}\\& \qquad \qquad \quad y=3+\frac{8}{9}x\\& \qquad \qquad \ \ \boxed{y=\frac{8}{9}x+3}$

The slope of the line is $\frac{8}{9}$ and the $y-$intercept is (0, 3)

Using the slope-intercept method...

$3x+y&=4\\3x-3x+y&=4-3x\\y&=4-3x\\y&=-\frac{3}{{\color{blue}1}}x+4\\m&=-\frac{3}{1} \ and \ b=(0,4)$

1. The equation must be written in slope-intercept form.

$& \qquad \ 2x+6y+18=0\\& 2x-2x+6y+18=0-2x\\& \qquad \qquad \ \ 6y+18=-2x\\& \qquad \ 6y+18-18=-2x-18\\& \qquad \qquad \qquad \ \ 6y=-2x-18\\& \qquad \qquad \qquad \ \frac{6y}{6}=\frac{-2x}{6}-\frac{18}{6}\\& \qquad \qquad \qquad \quad y=-\frac{2}{6}x-3\\& \qquad \qquad \qquad \quad y=\boxed{-\frac{1}{3}x-3}$

1. The equation must be written in slope-intercept form.

$&\qquad \ 4x-5y=-30\\& 4x-4x-5y=-30-4x\\& \qquad \quad \ -5y=-30-4x\\& \qquad \quad \ \ \frac{-5y}{-5}=\frac{-30}{-5}-\frac{4x}{-5}\\& \qquad \qquad \quad y=6+\frac{4}{5}x\\& \qquad \qquad \quad \boxed{y=\frac{4}{5}x+6}$

Graph the following linear equations...

1. $x=-5$ The slope of this line is undefined.
1. $y=-3$ The slope of this line is Zero.

Jan 16, 2013

Jan 14, 2015