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3.4: Graphs of Linear Functions from Tables

Created by: CK-12

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.

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Khan Academy Graphing Lines 1

Guidance

One way to graph a linear function is to first create a table of points that work with the function and therefore are on the graph. A linear function will always result in a graph that is a straight line.

To create a table, substitute values for x into the function (you can choose values for x ) and use the function to calculate the corresponding value for y . Each pair of values is one point on the graph. It is easier to create the table if you first solve the equation for y

You can also use a graphing calculator to create a table of values and graph of the function. This will be explored in Example B.

Example A

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

Solution:

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 .

Solution:

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 x-2y=4 , and use those values to graph the function.

Solution:

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

Concept Problem Revisited

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.

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.

Answers:

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.

Practice

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
  4. 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.

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Date Created:

Dec 19, 2012

Last Modified:

Apr 29, 2014
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