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# 3.4: Graphs of Linear Functions from Tables

Difficulty Level: At Grade 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.

### 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 \begin{align*}x\end{align*} into the function (you can choose values for \begin{align*}x\end{align*}) and use the function to calculate the corresponding value for \begin{align*}y\end{align*}. Each pair of values is one point on the graph. It is easier to create the table if you first solve the equation for \begin{align*}y\end{align*}

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 \begin{align*}3x+2y=-6\end{align*}.

Solution:

Before completing the table of values, solve the given function in terms of ‘\begin{align*}y\end{align*}’. This step is not necessary, but it does simplify the calculations.

\begin{align*}& \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}\end{align*}

\begin{align*}& 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}\end{align*}

\begin{align*}y=-\frac{3}{2}x-3\end{align*}
\begin{align*}X\end{align*} \begin{align*}Y\end{align*}
\begin{align*}{\color{red}-4}\end{align*} \begin{align*}3\end{align*}
\begin{align*}{\color{red}0}\end{align*} \begin{align*}-3\end{align*}
\begin{align*}{\color{red}2}\end{align*} \begin{align*}-6\end{align*}
\begin{align*}{\color{red}6}\end{align*} \begin{align*}-12\end{align*}

#### Example B

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

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 ‘\begin{align*}x\end{align*}’ 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 (\begin{align*}x\end{align*}-values), there is a definite pattern in the output numbers (\begin{align*}y\end{align*}-values). This will be expanded upon in a later lesson.

#### Example C

Complete the table of values for \begin{align*}x-2y=4\end{align*}, and use those values to graph the function.

Solution:

\begin{align*}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}\end{align*}

\begin{align*}& \ 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}\end{align*}

\begin{align*}y=\frac{1}{2}x-2\end{align*}
\begin{align*}X\end{align*} \begin{align*}Y\end{align*}
\begin{align*}{\color{red}-4}\end{align*} \begin{align*}-4\end{align*}
\begin{align*}{\color{red}0}\end{align*} \begin{align*}-2\end{align*}
\begin{align*}{\color{red}2}\end{align*} \begin{align*}-1\end{align*}
\begin{align*}{\color{red}6}\end{align*} \begin{align*}1\end{align*}

#### 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 \begin{align*}y=55x+100\end{align*} where ‘\begin{align*}y\end{align*}’ represents the cost in dollars and ‘\begin{align*}x\end{align*}’ represents the time, in hours, that the pool is rented.

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

\begin{align*}& \ 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}\end{align*}

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

\begin{align*}& 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\end{align*}

The values in the table represent the coordinates of points that are located on the graph of \begin{align*}y=55x+100\end{align*}.

\begin{align*}(1,155);(2,210);(3,265);(4,320);(5,375)\end{align*}

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 \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, in which each value of the independent variable \begin{align*}(x)\end{align*} is mapped to one and only one value of the dependent variable \begin{align*}(y)\end{align*}.

### Guided Practice

1. Complete the following table of values for the linear function \begin{align*}3x-2y=-12\end{align*}

\begin{align*}3x-2y=-12\end{align*}
\begin{align*}X\end{align*} \begin{align*}Y\end{align*}
\begin{align*}{\color{red}-6}\end{align*}
\begin{align*}{\color{red}-4}\end{align*}
\begin{align*}{\color{red}0}\end{align*}
\begin{align*}{\color{red}6}\end{align*}

2. Use technology to complete a table of values for the linear function \begin{align*}2x-y=-8\end{align*} 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. \begin{align*}3x-2y=-12\end{align*} Solve the equation in terms of the variable ‘\begin{align*}y\end{align*}’.

\begin{align*}3x-3x-2y=-3x-12 && -2y=-3x-12 && \frac{-2y}{-2}=\frac{-3x}{-2}-\frac{12}{-2}\end{align*}

\begin{align*}\boxed{y=\frac{3}{2}x+6}\end{align*}

Substitute the given values for ‘\begin{align*}x\end{align*}’ into the function.

\begin{align*}& \ 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}\end{align*}

\begin{align*}3x-2y=-12\end{align*}
\begin{align*}X\end{align*} \begin{align*}Y\end{align*}
\begin{align*}{\color{red}-6}\end{align*} \begin{align*}-3\end{align*}
\begin{align*}{\color{red}-4}\end{align*}
\begin{align*}{\color{red}0}\end{align*} \begin{align*}6\end{align*}
\begin{align*}{\color{red}6}\end{align*} \begin{align*}15\end{align*}

2. \begin{align*}2x-y=-8\end{align*} To enter the function into the calculator, it must be in the form \begin{align*}y= \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}. Solve the function in terms of the letter ‘\begin{align*}y\end{align*}’.

\begin{align*}&2x-y=-8 && 2x-2x-y=-2x-8 && -y=-2x-8 && \frac{-y}{-1}=\frac{-2x}{-1}\frac{-8}{-1}\\ & \boxed{y=2x+8}\end{align*}

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

3. \begin{align*}y=.09x+25\end{align*} Write a linear function to represent the word problem.

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

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Date Created:
Dec 19, 2012