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# 3.5: Graphing a Linear Function Using a Table of Values

Difficulty Level: At Grade Created by: CK-12

## Graphing a Linear Function Using a Table of Values

Objectives

The lesson objectives for Graphing a Linear Function using a table of Values are:

• Representing a problem with a linear function
• Creating a table of values from the linear function
• Using the table of values to draw the graph.

Introduction

In this concept you will learn to graph a linear function by using the function to create a table of values. These values will be the coordinates of the points that will be plotted to draw the graph of the function. A linear function will result in a graph that is a straight line.

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Guidance

Bonita will be celebrating her sixteenth birthday next month. Her parents would like to give her a surprise party at the local pool. To rent the pool for a private party costs $100 plus$55.00 for each hour the pool is rented. Write a linear function to represent the cost of the pool party and list five prices from which her parents can choose.

The cost of renting the pool is $100. This amount is a fee that must be paid to rent the pool. In addition, Bonita’s parents will also have to pay$55.00 for each hour the pool is rented. Therefore, the linear function to represent this situation is $y=55x+100$ where ‘$y$’ represents the cost in dollars and ‘$x$’ represents the time, in hours, that the pool is rented.

$y=55x+100$ - To determine five options for her parents, replace ‘$x$’ with the values 1 to 5 and calculate the cost for each of these hours.

$& \ y=55x+100 && \ y=55x+100 && \ y=55x+100 && \ y=55x+100 && \ y=55x+100\\& \ y=55(1)+100 && \ y=55(2)+100 && \ y=55(3)+100 && \ y=55(4)+100 && \ y=55(5)+100\\& \boxed{y=\155} && \boxed{y=\210} && \boxed{y=\265} && \boxed{y=\320} && \boxed{y=\375}$

These results can now be represented in a table of values:

$& X(hours) \qquad 1 \qquad \quad \ \ 2 \qquad \quad \ 3 \qquad \quad \ 4 \qquad \quad \ \ 5\\& Y(Cost) \qquad \155 \qquad \210 \qquad \265 \qquad \320 \qquad \375$

The values in the table represent the coordinates of points that are located on the graph of $y=55x+100$.

$(1,155);(2,210);(3,265);(4,320);(5,375)$

Bonita’s parents can use the table of values and/or the graph to make their decision.

Example A

Complete the table of values for the linear function $3x+2y=-6$.

Before completing the table of values, solve the given function in terms of ‘$y$’. This step is not necessary, but it does simplify the calculations.

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

$& y=\frac{-3x}{2}-3 && y=\frac{12}{2}-3 && y=\frac{-3x}{2}-3 && y=0-3\\& y=\frac{-3({\color{red}-4})}{2}-3 && y=6-3 && y=\frac{-3({\color{red}0})}{2}-3 && \boxed{y=-3}\\& && \boxed{y=3}\\& y=\frac{-3x}{2}-3 && y=\frac{-6}{2}-3 && y=\frac{-3x}{2}-3 && y=\frac{-18}{2}-3\\& y=\frac{-3({\color{red}2})}{2}-3 && y=-3-3 && y=\frac{-3({\color{red}6})}{2}-3 && y=-9-3\\& && \boxed{y=-6} && && \boxed{y=-12}$

$y=-\frac{3}{2}x-3$
$X$ $Y$
${\color{red}-4}$ $3$
${\color{red}0}$ $-3$
${\color{red}2}$ $-6$
${\color{red}6}$ $-12$

Example B

Use technology to create a table of values for the linear function $f(x)=-\frac{1}{2}x+4$.

When the table is set up, you choose the beginning number as well as the pattern for the numbers in the table. In this table, the beginning value for ‘$x$’ is -2 and the difference between each number is +2. The table is consecutive, even numbers. When consecutive numbers are used as the input numbers ($x-$values), there is a definite pattern in the output numbers ($y-$values). This will be expanded upon in a later lesson.

Example C

Complete the table of values for the following linear function, and use those values to graph the function.

$x-2y=4 && x-x-2y=-x+4 && -2y=-x+4 && \frac{-2y}{-2}=\frac{-x}{-2}+\frac{4}{-2} && \boxed{y=\frac{1}{2}x-2}$

$& \ y=\frac{1}{2}x-2 && \ y=\frac{1}{2}x-2 && \ y=\frac{1}{2}x-2 && \ y=\frac{1}{2}x-2\\& \ y=\frac{1}{2}({\color{red}-4})-2 && \ y=\frac{1}{2}({\color{red}0})-2 && \ y=\frac{1}{2}({\color{red}2})-2 && \ y=\frac{1}{2}({\color{red}6})-2\\& \ y=-2-2 && \ y=0-2 && \ y=1-2 && \ y=3-2\\& \boxed{y=-4} && \boxed{y=-2} && \boxed{y=-1} && \boxed{y=1}$

$y=\frac{1}{2}x-2$
$X$ $Y$
${\color{red}-4}$ $-4$
${\color{red}0}$ $-2$
${\color{red}2}$ $-1$
${\color{red}6}$ $1$

Vocabulary

Linear Function
The linear function is a relation between two variables, usually $x$ and $y$, in which each value of the independent variable $(x)$ is mapped to one and only one value of the dependent variable $(y)$.

Guided Practice

1. Complete the following table of values for the linear function $3x-2y=-12$

$3x-2y=-12$
$X$ $Y$
${\color{red}-6}$
${\color{red}-4}$
${\color{red}0}$
${\color{red}6}$

2. Use technology to complete a table of values for the linear function $2x-y=-8$ and use the coordinates to draw the graph.

3. A local telephone company charges a monthly fee of $25.00 plus$0.09 per minute for calls within the United States. If Sam talks for 200 minutes in one month, calculate the cost of his telephone bill.

1. $3x-2y=-12$ Solve the equation in terms of the variable ‘$y$’.

$3x-3x-2y=-3x-12 && -2y=-3x-12 && \frac{-2y}{-2}=\frac{-3x}{-2}-\frac{12}{-2}$

$\boxed{y=\frac{3}{2}x+6}$ Substitute the given values for ‘$x$’ into the function.

$& \ y=\frac{3}{2}x+6 && \ y=\frac{3}{2}x+6 && \ y=\frac{3}{2}x+6 && \ y=\frac{3}{2}x+6\\& \ y=\frac{3}{2}({\color{red}-6})+6 && \ y=\frac{3}{2}({\color{red}-4})+6 && \ y=\frac{3}{2}({\color{red}0})+6 && \ y=\frac{3}{2}({\color{red}6})+6\\& \ y=-9+6 && \ y=-6+6 && \ y=0+6 && \ y=9+6\\& \boxed{y=-3} && \boxed{y=0} && \boxed{y=6} && \boxed{y=15}$

$3x-2y=-12$
$X$ $Y$
${\color{red}-6}$ $-3$
${\color{red}-4}$ $0$
${\color{red}0}$ $6$
${\color{red}6}$ $15$

2. $2x-y=-8$ To enter the function into the calculator, it must be in the form $y= \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}$.

Solve the function in terms of the letter ‘$y$’.

$&2x-y=-8 && 2x-2x-y=-2x-8 && -y=-2x-8 && \frac{-y}{-1}=\frac{-2x}{-1}\frac{-8}{-1}\\& \boxed{y=2x+8}$

The graph can also be done using technology. The table can be used to set the window.

3. $y=.09x+25$ Write a linear function to represent the word problem.

$y &=.09(200)+25 && \text{Substitute the time of} \ 200 \ \text{minutes for the variable} \ x'.\\y&=\43.00$

The cost of Sam’s telephone bill is $43.00. Summary In this lesson, you have learned how to evaluate a linear function with given values. These values were given in table form. The table was completed by entering the values obtained by substituting the given values for ‘$x$’ into the linear function. These values were the coordinates of points that were located on the graph of the linear function. The points were used to draw the graph on a Cartesian plane. The lesson then extended into the world of technology. The graphing calculator was used to create a table of values as well as to create the graph of the function. The more you practice the keystrokes for performing these tasks on the calculator, the more efficient you will become. Problem Set Solve each of the following linear functions in terms of the variable ‘$y$’. 1. $2x-3y=18$ 2. $4x-2y=10$ 3. $3x-y=8$ 4. $5x+3y=-12$ 5. $3x-2y-2=0$ For each of the following linear functions, create a table of values that contains four coordinates: 1. $y=-4x+5$ 2. $5x+3y=15$ 3. $4x-3y=6$ 4. $2x-2y+2=0$ 5. $2x-3y=9$ For each of the linear functions, complete the table of values and use the values to draw the graph. 1. $y=-2x+1$ $& x \qquad -3 \qquad 0 \qquad 1 \qquad 5\\& y$ 1. $x=2y-3$ $& x \qquad -4 \qquad 0 \qquad 2 \qquad 6\\& y$ 1. $3x+2y=8$ $& x \qquad -6 \qquad -2 \qquad 0 \qquad 4\\& y$ 1. $4(y-1)=12x-7$ $& x \qquad -2 \qquad 0 \qquad 3 \qquad 7\\& y$ 1. $\frac{1}{2}x+\frac{1}{3}y=6$ $& x \qquad 0 \qquad 4 \qquad 6 \qquad 10\\& y$ Using technology, create a table of values for each of the following linear functions. Using technology, graph each of the linear functions. 1. $y=-2x+3$ 2. $y=-\frac{1}{2}x-3$ 3. $y=\frac{4}{3}x-2$ Mr. Red is trying to estimate the cost of renting a car to go on vacation. He has contacted a rental agency and has obtained the following information. The agency charges a daily rate of$78.00 for the vehicle plus 45 cents per mile. If Mr. Red has $350 set aside for travel, create a table of values that will give him approximate distances that he can travel with this rental car. Answers Solve each of the following linear functions... $& \qquad \ 2x-3y=18\\& 2x-2x-3y=-2x+18\\& \qquad \quad \ -3y=-2x+18\\& \qquad \quad \ \ \frac{-3y}{-3}=\frac{-2x}{-3}+\frac{18}{-3}\\& \qquad \qquad \quad \boxed{y=\frac{2}{3}x-6}$ $& \qquad \ 3x-y=8\\&3x-3x-y=-3x+8\\& \qquad \quad \ -y=-3x+8\\& \qquad \quad \ \frac{-y}{-1}=\frac{-3x}{-1}+\frac{8}{-1}\\& \qquad \qquad \boxed{y=3x-8}$ $& \quad \ \ 3x-2y-2=0\\& 3x-2y-2+2=0+2\\& \qquad \quad \ 3x-2y=2\\& \quad 3x-3x-2y=-3x+2\\& \qquad \qquad \ -2y=-3x+2\\& \qquad \qquad \ \frac{-2y}{-2}=\frac{-3x}{-2}+\frac{2}{-2}\\& \qquad \qquad \quad \ \boxed{y=\frac{3}{2}x-1}$ For each of the following linear functions... 1. $y=-4x+5$ $X$ $Y$ ${\color{red}-3}$ $17$ ${\color{red}0}$ $5$ ${\color{red}2}$ $-3$ ${\color{red}6}$ $-19$ $4x-3y&=6\\4x-4x-3y&=-4x+6\\-3y&=-4x+6\\\frac{-3y}{-3}&=\frac{-4x}{-3}+\frac{6}{-3}\\y&=\frac{4}{3}x-2$ $X$ $Y$ ${\color{red}-3}$ $-6$ ${\color{red}0}$ $-2$ ${\color{red}6}$ $6$ ${\color{red}9}$ $10$ $2x-3y&=9\\2x-2x-3y&=-2x+9\\-3y&=-2x+9\\\frac{-3y}{-3}&=\frac{-2x}{-3}+\frac{9}{-3}\\y&=\frac{2}{3}x-3$ $X$ $Y$ ${\color{red}-3}$ $-5$ ${\color{red}0}$ $-3$ ${\color{red}6}$ $1$ ${\color{red}9}$ $3$ For each of the linear functions... 1. $y=-2x+1$ $& x \qquad {\color{red}-3} \qquad {\color{red}0} \qquad \quad \ {\color{red}1} \qquad \quad \ {\color{red}5}\\& y \qquad \ \ 7 \qquad \ 1 \qquad -1 \qquad -9$ $3x+2y&=8\\3-3x+2y&=-3x+8\\2y&=-3x+8\\\frac{2y}{2}&=\frac{-3x}{2}+\frac{8}{2}\\y &= \frac{-3}{2}x+4$ $& x \qquad {\color{red}-6} \qquad {\color{red}-2} \qquad {\color{red}0} \qquad \quad {\color{red}4}\\& y \qquad \ 13 \qquad \ \ 7 \qquad 4 \qquad -2$ $\frac{1}{2}x+\frac{1}{3}y&=6\\\frac{1}{2}x-\frac{1}{2}x+\frac{1}{3}y&=-\frac{1}{2}x+6\\\frac{1}{3}y&=-\frac{1}{2}x+6\\\frac{\frac{1}{3}y}{\frac{1}{3}}&=-\frac{1}{2}\div \left(\frac{1}{3}\right)x+6 \div \left(\frac{1}{3}\right)\\y &= -\frac{1}{2}\left(\frac{3}{1}\right)x+6\left(\frac{3}{1}\right)\\y &= -\frac{3}{2}x+18$ $& x \qquad \ {\color{red}0} \qquad \ {\color{red}4} \qquad \ {\color{red}6} \qquad {\color{red}10}\\& y \qquad 18 \qquad 12 \qquad 9 \qquad \ 3$ Using technology, create a table of values... 1. $y=-2x+3$ 1. $y=\frac{4}{3}x-2$ Mr. Red is trying to estimate... $y &=0.45x+78 && y =0.45x+78 && y =0.45x+78\\y &=0.45({\color{red}100})+78 && y =0.45({\color{red}200})+78 && y =0.45({\color{red}300})+78\\y&=\123 && y=\168 && y=\213\\\\y &=0.45x+78 && y=0.45x+78 && y=0.45x+78\\y &=0.45({\color{red}400})+78 && y =0.45({\color{red}500})+78 && y =0.45({\color{red}600})+78\\y&=\258 && y=\303 && y=\348$ $& x \qquad 100 \qquad 200 \qquad 300 \qquad 400 \qquad 500 \qquad 600\\& y \qquad 123 \qquad 168 \qquad 213 \qquad 258 \qquad 303 \qquad 348$ If Mr. Red has$350 set aside for travel for his vacation, he can drive approximately 600 miles with the rental vehicle.

## Graphing a Linear Function Using the X and Y-Intercepts

Objectives

The lesson objectives for Graphing a Linear Function using a table of Values are:

• Understanding the $x$ and $y-$intercepts
• Determining the $x$ and $y-$intercepts for a given linear function
• Using the $x$ and $y-$intercepts to graph the linear function

Introduction

To graph a linear function, you need to plot only two points. These points can then be lined up with a straight edge and joined to graph the straight line. Two points that can be used to graph a linear function are the $x-$intercept and the $y-$intercept. The $x-$intercept is simply a point that is located on the $x-$axis. Its coordinates are $(x, 0)$. A $y-$intercept is a point located on the $y-$axis. Its coordinates are $(0, y)$. Graphing a linear function by plotting the $x-$ and $y-$ intercepts is often referred to as the intercept method.

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Guidance

The linear function $4x+2y=8$ can be graphed by using the intercept method.

$& \text{To determine the }x-\text{intercept, let } y=0. && \text{To determine the } y-\text{intercept, let } x=0.\\& \text{Solve for} \ x'. && \text{Solve for} \ y'.\\& 4x+2y=8 && 4x+2y=8\\& 4x+2({\color{red}0})=8 && 4({\color{red}0})+2y=8\\& 4x+{\color{red}0}=8 && {\color{red}0}+2y=8\\ & 4x=8 && 2y=8\\& \frac{4x}{4}=\frac{8}{4} && \frac{2y}{2}=\frac{8}{2}\\& x=2 && y=4\\& \text{The} \ x-\text{intercept is} \ (2, 0) && \text{The} \ y- \text{intercept is} \ (0, 4)$

Plot the $x-$intercept on the $x-$axis and the $y-$intercept on the $y-$axis. Join the two points with a straight line.

Example A

Identify the $x-$ and $y-$intercepts for each line.

(a) $2x+y-6=0$

(b) $\frac{1}{2}x-4y=4$

(a) $&\text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 2x+y-6=0 && 2x+y-6=0\\& 2x+({\color{red}0})-6=0 && 2({\color{red}0})+y-6=0\\& 2x-6=0 && y-6=0\\& 2x-6+6=0+6 && y-6+6=0+6\\& 2x=6 && y=6\\& \frac{2x}{2}=\frac{6}{2} && \text{The} \ y- \text{intercept is} \ (0, 6)\\& x=3\\& \text{The} \ x-\text{intercept is} \ (3, 0)$

(b) $& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& \frac{1}{2}x-4y=4 && \frac{1}{2}x-4y=4\\& \frac{1}{2}x-4({\color{red}0})=4 && \frac{1}{2}({\color{red}0})-4y=4\\& \frac{1}{2}x-0=4 && 0-4y=4\\& \frac{1}{2}x=4 && 4y=4\\& \overset{1}{\cancel{2}}\left(\frac{1}{\cancel{2}}\right)x=2(4) && \frac{4y}{4}=\frac{4}{4}\\& x=8 && y=1\\& \text{The} \ x-\text{intercept is} \ (8, 0) && \text{The} \ y- \text{intercept is} \ (0, 1)$

Example B

Use the intercept method to graph $2x-3y=-12$

$& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 2x-3y=-12 && 2x-3y=-12\\& 2x-3({\color{red}0})=-12 && 2({\color{red}0})-3y=-12\\& 2x-0=-12 && 0-3y=-12\\& 2x=-12 && -3y=-12\\& \frac{2x}{2}=\frac{-12}{2} && \frac{-3y}{-3}=\frac{-12}{-3}\\& x=-6 && y=4\\& \text{The} \ x-\text{intercept is} \ (-6, 0) && \text{The} \ y- \text{intercept is} \ (0, 4)$

Example C

The $x-$intercept is (-8, 0)

The $y-$intercept is (0, 4)

Use the $x-$ and $y-$intercepts of the graph to identify the linear function that matches the graph.

$& y=2x-8 && 2x+y-8=0 && x-2y+8=0\\& ({\color{red}0})=2x-8 && 2x+({\color{red}0})-8=0 && x-2({\color{red}0})+8=0\\& 0=2x-8 &&2x-8=0 && x-0+8=0\\& 0-2x=2x-2x-8 && 2x-8+8=0+8 && x+8=0\\& -2x=-8 && 2x=8 && x+8-8=0-8\\& \frac{-2x}{-2}=\frac{-8}{-2} && \frac{2x}{2}=\frac{8}{2} && x=-8\\& x=4 && x=4 && \text{The} \ x-\text{intercept is} \ (-8, 0)\\& \text{The} \ x-\text{intercept is} \ (4, 0) && \text{The} \ x- \text{intercept is} \ (4, 0) && \text{This matches the graph.}$

This does not match the graph. This does not match the graph. Confirm the $y-$intercept.

$x-2y+8&=0 && \quad \ \ -2y+8=0 && -2y=-8 && y=4\\({\color{red}0})-2y+8&=0 && -2y+8-8=0-8 && \ \frac{-2y}{-2}=\frac{-8}{-2}$

The $y-$intercept is (0, 4). This matches the graph.

The linear function that matches the graph is

$\boxed{x-2y+8=0}$

Vocabulary

Intercept Method
The intercept method is a way of graphing a linear function by using the coordinates of the $x-$ and $y-$intercepts. The graph is drawn by plotting these coordinates on the Cartesian plane and joining them with a straight line.
$X-$intercept
A $x-$intercept of a relation is the $x-$coordinate of the point where the relation intersects the $x-$axis.
$Y-$intercept

A $y-$intercept of a relation is the $y-$coordinate of the point where the relation intersects the $y-$axis.

Guided Practice

1. Identify the $x-$ and $y-$intercepts of the following linear functions:

(i) $2(x-3)=y+4$

(ii) $3x+\frac{2}{3}y-3=0$

2. Use the intercept method to graph the following relation:

(i) $5x+2y=-10$

3. Use the $x-$ and $y-$intercepts of the graph, to match the graph to its function.

(i) $2x+y=6$

(ii) $4x-3y-12=0$

(iii) $5x+3y=15$

1. (i) $2(x-3)&=y+4 && \text{Simplify the equation}\\2(x-3)&=y+4\\2x-6&=y+4\\2x-6+6&=y+4+6\\2x&=y+10 && \text{You may leave the function in this form.}\\2x-y&=y-y+10\\2x-y&=10$

If you prefer to have both variables on the same side of the equation, this form may also be used. The choice is your preference.

$& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 2x-y=10 && 2x-y=10\\& 2x-({\color{red}0})=10 && 2({\color{red}0})-y=10\\& 2x=10 && 0-y=10\\& \frac{2x}{2}=\frac{10}{2} && \frac{-y}{-1}=\frac{10}{-1}\\& x=5 && y=-10\\& \text{The} \ x-\text{intercept is} \ (5, 0) && \text{The} \ y- \text{intercept is} \ (0, -10)$

(ii) $3x+\frac{2}{3}y-3&=0 && \text{Simplify the equation.}\\3(3x)+3\left(\frac{2}{3}\right)y-3(3)&=3(0) && \text{Multiply each term by 3.}\\3(3x)+\cancel{3}\left(\frac{2}{\cancel{3}}\right)y-3(3)&=3(0)\\9x+2y-9&=0\\9x+2y-9+9&=0+9\\9x+2y&=9$

$& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 9x+2y=9 && 9x+2y=9\\& 9x+2({\color{red}0})=9 && 9({\color{red}0})+2y=9\\& 9x+0=9 && 0+2y=9\\& \frac{9x}{9}=\frac{9}{9} && \frac{2y}{2}=\frac{9}{2}\\& x=1 && y=4.5\\& \text{The} \ x-\text{intercept is} \ (1, 0) && \text{The} \ y- \text{intercept is} \ (0, 4.5)$

2. $5x+2y=-10$ Determine the $x-$ and $y-$intercepts.

$& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 5x+2y=-10 && 5x+2y=-10\\& 5x+2({\color{red}0})=-10 && 5({\color{red}0})+2y=-10\\& 5x+0=-10 && 0+2y=-10\\& \frac{5x}{5}=\frac{-10}{5} && \frac{2y}{2}=\frac{-10}{2}\\& x=-2 && y=-5\\& \text{The} \ x-\text{intercept is} \ (-2, 0) && \text{The} \ y- \text{intercept is} \ (0, -5)$

3. Identify the $x-$ and $y-$intercepts from the graph.

The $x-$intercept is (3, 0)

The $y-$intercept is (0, -4)

Determine the $x-$ and $y-$intercept for each of the functions. If the intercepts match those of the graph, then the linear function will be the one that matches the graph.

(i) $& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 2x+y=6 && 2x+y=6\\& 2x+({\color{red}0})=6 && 2({\color{red}0})+y=6\\& 2x=6 && 0+y=6\\& \frac{2x}{2}=\frac{6}{2} && y=6\\& x=3\\& \text{The} \ x-\text{intercept is} \ (3, 0) && \text{The} \ y- \text{intercept is} \ (0, 6)\\& \text{This matches the graph.} && \text{This does not match the graph.}\\& \text{Find the} \ y- \text{intercept.} && 2x+y=6 \ \text{is not the linear function for the graph.}$

(ii) $& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 4x-3y-12=0 && 4x-3y-12=0\\& 4x-3y-12+12=0+12 && 4x-3y-12+12=0+12\\& 4x-3y=12 && 4x-3y=12\\& 4x-3({\color{red}0})=12 && 4({\color{red}0})-3y=12\\& 4x-0=12 && 0-3y=12\\& 4x=12 && -3y=12\\& \frac{4x}{4}=\frac{12}{4} && \frac{-3y}{-3}=\frac{12}{-3}\\& x=3 && y=-4\\& \text{The} \ x-\text{intercept is} \ (3, 0) && \text{The} \ y- \text{intercept is} \ (0, -4)\\& \text{This matches the graph.} && \text{This matches the graph.}\\& \text{Find the} \ y-\text{intercept.} && 4x-3y-12=0 \ \text{is the linear function for the graph.}$

(iii) $& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 5x+3y=15 && 5x+3y=15\\& 5x+3({\color{red}0})=15 && 5({\color{red}0})+3y=15\\& 5x+0=15 && 0+3y=15\\& 5x=15 && 3y=15\\& \frac{5x}{5}=\frac{15}{5} && \frac{3y}{3}=\frac{15}{3}\\& x=3 && y=5\\& \text{The} \ x-\text{intercept is} \ (3, 0) && \text{The} \ y- \text{intercept is} \ (0, 5)\\& \text{This matches the graph.} && \text{This does not match the graph.}\\& \text{Find the} \ y- \text{intercept.} && 5x+3y=15 \ \text{is not the linear function for the graph.}$

Summary

In this lesson you have learned that an x-intercept is a point on the $x-$axis that has the coordinates $(\#, 0)$. You also learned that a $y-$intercept is a point on the $y-$axis that has the coordinates $(0, \#)$. When you were given a function, you learned to algebraically determine the coordinates of both the $x-$ and $y-$intercepts.

The $x-$ and $y-$intercepts were then used to draw the graph of the linear function on a Cartesian plane. This method of graphing the function is called the intercept method. You also learned to use the $x-$ and $y-$intercepts of a given graph to match a graph with its function.

Problem Set

Complete the following table:

Function $x-$intercept $y-$intercept
$7x-3y=21$
$8x-3y+24=0$
$\frac{x}{4}-\frac{y}{2}=3$
$7x+2y-14=0$
$\frac{2}{3}x-\frac{1}{4}y=-2$

Use the intercept method to graph each of the linear functions in the above table.

1. $7x-3y=21$
2. $8x-3y+24=0$
3. $\frac{x}{4}-\frac{y}{2}=3$
4. $7x+2y-14=0$
5. $\frac{2}{3}x-\frac{1}{4}y=-2$

Use the $x-$ and $y-$intercepts to match each graph to its function...

1. $7x+5y-35=0$
2. $y=5x+10$
3. $2x+4y+8=0$
4. $2x+y=2$

Complete the following table...

Function $x-$intercept $y-$intercept
$7x-3y=21$ (3, 0) (0, -7)
$8x-3y+24=0$
$\frac{x}{4}-\frac{y}{2}=3$ (12, 0) (0, -6)
$7x+2y-14=0$
$\frac{2}{3}x-\frac{1}{4}y=-2$ (-3, 0) (0, 8)

$& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 7x-3y=21 && 7x-3y=21\\& 7x-3({\color{red}0})=21 && 7({\color{red}0})-3y=21\\& 7x+0=21 && 0-3y=21\\& 7x=21 && -3y=21\\& \frac{7x}{7}=\frac{21}{7} && \frac{-3y}{-3}=\frac{21}{-3}\\& x=3 && y=-7\\& \text{The} \ x-\text{intercept is} \ (3, 0) && \text{The} \ y- \text{intercept is} \ (0, -7)$

$\frac{x}{4}-\frac{y}{2}=3 && 4\left(\frac{x}{4}\right)-4\left(\frac{y}{2}\right)=4(3) && ^{\overset{1}{\cancel{4}}}\left(\frac{x}{\cancel{4}}\right)-^{\overset{2}{\cancel{4}}}\left(\frac{y}{\cancel{2}}\right)=4(3)\\x-2y=12$

$& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& x-2y=12 && x-2y=12\\& x-2({\color{red}0})=12 && ({\color{red}0})-2y=12\\& x+0=12 && 0-2y=12\\& x=12 && -2y=12\\& && \frac{-2y}{-2}=\frac{12}{-2}\\& && y=-6\\& \text{The} \ x-\text{intercept is} \ (12, 0) && \text{The} \ y- \text{intercept is} \ (0, -6)$

$\frac{2}{3}x-\frac{1}{4}y&=-2 && 12\left(\frac{2}{3}\right)x-12\left(\frac{1}{4}\right)y=12(-2)\\^{\overset{4}{\cancel{12}}}\left(\frac{2}{\cancel{3}}\right)x-^{\overset{3}{\cancel{12}}}\left(\frac{1}{\cancel{4}}\right)y&=12(-2) && 8x-3y=-24$

$& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 8x-3y=-24 && 8x-3y=-24\\& 8x-3({\color{red}0})=-24 && 8({\color{red}0})-3y=-24\\& 8x+0=-24 && 0-3y=-24\\& 8x=-24 && -3y=-24\\& \frac{8x}{8}=\frac{-24}{8} && \frac{-3y}{-3}=\frac{-24}{-3}\\& x=-3 && y=8\\& \text{The} \ x-\text{intercept is} \ (-3, 0) && \text{The} \ y- \text{intercept is} \ (0, 8)$

Use the intercept method to graph...

1. $7x-3y=21$ The $x-$intercept is (3, 0). The $y-$intercept is (0, -7).
1. $\frac{x}{4}-\frac{y}{2}=3$ The $x-$intercept is (12, 0). The $y-$intercept is (0, -6).
1. $\frac{2}{3}x-\frac{1}{4}y=-2$ The $x-$intercept is (-3, 0). The $y-$intercept is (0, 8).

Use the $x-$ and $y-$intercepts to match...

(a) $& 7x+5y-35=0\\& 7x+5y-35+35=0+35\\& 7x+5y=35\\& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 7x+5y=35 && 7x+5y=35\\& 7x+5({\color{red}0})=35 && 7({\color{red}0})+5y=35\\& 7x+0=35 && 0+5y=35\\& 7x=35 && 5y=35\\& \frac{7x}{7}=\frac{35}{7} && \frac{5y}{5}=\frac{35}{5}\\& x=5 && y=7\\& \text{The} \ x-\text{intercept is} \ (5, 0) && \text{The} \ y- \text{intercept is} \ (0, 7)$

(b) $& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& y=5x+10 && y=5x+10\\& ({\color{red}0})=5x+10 && y=5({\color{red}0})+10\\& 0=5x+10 && y=0+10\\& -5x=5x-5x+10 && y=10\\& \frac{5x}{-5}=\frac{10}{-5} && \text{The} \ y- \text{intercept is} \ (0, 10)\\& x=-2\\& \text{The} \ x-\text{intercept is} \ (-2, 0)$

(c) $& 2x+4y+8=0\\& 2x+4y+8-8=0-8\\& 2x+4y=-8\\& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ y'.\\& 2x+4y=-8 && 2x+4y=-8\\& 2x+4({\color{red}0})=-8 && 2({\color{red}0})+4y=-8\\& 2x+0=-8 && 0+4y=-8\\& 2x=-8 && 4y=-8\\& \frac{2x}{2}=\frac{-8}{2} && \frac{4y}{4}=\frac{-8}{4}\\& x=-4 && y=-2\\& \text{The} \ x-\text{intercept is} \ (-4, 0) && \text{The} \ x-\text{intercept is} \ (0, -2)$

(d) $& 2x+y=2\\& \text{Let} \ y = 0. \ \text{Solve for} \ x'. && \text{Let} \ x = 0. \ \text{Solve for} \ `y'.\\& 2x+y=2 && 2x+y=2\\& 2x+({\color{red}0})=2 && 2({\color{red}0})+y=2\\& 2x+0=2 && 0+y=2\\& 2x=2 && y=2\\& \frac{2x}{2}=\frac{2}{2} && \text{The} \ y-\text{intercept is} \ (0, 2)\\& x=1\\& \text{The} \ x-\text{intercept is} \ (1, 0)$

(i)

(d) $2x+y=2$

(ii)

(d) $y=5x+10$

(iii)

(c) $2x+4y+8=0$

(iv)

(a) $7x+5y-35=0$

## Summary

In this lesson you have learned to create a table of values for a given linear function. The values for the dependent variable, $x$, were given in a table and you substituted the value into the function to determine the value of the dependent variable ‘$y$’. You also learned to create a table of values by using technology.

The values in the table represented coordinates of points that were located on the graph of the linear function. The coordinates were then plotted to draw the graph. In addition to drawing the graph on the Cartesian plane, you also learned to use technology to draw the graphs of the functions.

In the second lesson, you learned to algebraically calculate the values of both the $x-$ and $y-$intercepts from a given function. The $x-$ and $y-$intercepts are coordinates that intersect the axes of the Cartesian plane. By using the $x-$ and $y-$intercepts of a graph, you were able to apply this knowledge to match a graph to its linear function.

Jan 16, 2013

Jan 14, 2015