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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 x\begin{align*}x\end{align*} into the function (you can choose values for x\begin{align*}x\end{align*}) and use the function to calculate the corresponding value for y\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 y\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 3x+2y=6\begin{align*}3x+2y=-6\end{align*}.

Solution:

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

3x+2y=63x3x+2y=3x6  2y=3x6 2y2=3x262  y=3x23\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*}

y=3x23y=3(4)23y=3x23y=3(2)23y=1223y=63y=3y=623y=33y=6y=3x23y=3(0)23y=3x23y=3(6)23y=03y=3y=1823y=93y=12\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*}

y=32x3\begin{align*}y=-\frac{3}{2}x-3\end{align*}
X\begin{align*}X\end{align*} Y\begin{align*}Y\end{align*}
4\begin{align*}{\color{red}-4}\end{align*} 3\begin{align*}3\end{align*}
0\begin{align*}{\color{red}0}\end{align*} 3\begin{align*}-3\end{align*}
2\begin{align*}{\color{red}2}\end{align*} 6\begin{align*}-6\end{align*}
6\begin{align*}{\color{red}6}\end{align*} 12\begin{align*}-12\end{align*}

#### Example B

Use technology to create a table of values for the linear function f(x)=12x+4\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 ‘x\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 (x\begin{align*}x\end{align*}-values), there is a definite pattern in the output numbers (y\begin{align*}y\end{align*}-values). This will be expanded upon in a later lesson.

#### Example C

Complete the table of values for x2y=4\begin{align*}x-2y=4\end{align*}, and use those values to graph the function.

Solution:

x2y=4xx2y=x+42y=x+42y2=x2+42y=12x2\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*}

y=12x2 y=12(4)2 y=22y=4 y=12x2 y=12(0)2 y=02y=2 y=12x2 y=12(2)2 y=12y=1 y=12x2 y=12(6)2 y=32y=1\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*}

y=12x2\begin{align*}y=\frac{1}{2}x-2\end{align*}
X\begin{align*}X\end{align*} Y\begin{align*}Y\end{align*}
4\begin{align*}{\color{red}-4}\end{align*} 4\begin{align*}-4\end{align*}
0\begin{align*}{\color{red}0}\end{align*} 2\begin{align*}-2\end{align*}
2\begin{align*}{\color{red}2}\end{align*} 1\begin{align*}-1\end{align*}
6\begin{align*}{\color{red}6}\end{align*} 1\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 y=55x+100\begin{align*}y=55x+100\end{align*} where ‘y\begin{align*}y\end{align*}’ represents the cost in dollars and ‘x\begin{align*}x\end{align*}’ represents the time, in hours, that the pool is rented.

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

### Practice

Solve each of the following linear functions in terms of the variable ‘\begin{align*}y\end{align*}’.

1. \begin{align*}2x-3y=18\end{align*}
2. \begin{align*}4x-2y=10\end{align*}
3. \begin{align*}3x-y=8\end{align*}
4. \begin{align*}5x+3y=-12\end{align*}
5. \begin{align*}3x-2y-2=0\end{align*}

For each of the following linear functions, create a table of values that contains four coordinates:

1. \begin{align*}y=-4x+5\end{align*}
2. \begin{align*}5x+3y=15\end{align*}
3. \begin{align*}4x-3y=6\end{align*}
4. \begin{align*}2x-2y+2=0\end{align*}
5. \begin{align*}2x-3y=9\end{align*}

For each of the linear functions, complete the table of values and use the values to draw the graph.

1. \begin{align*}y=-2x+1\end{align*}

\begin{align*}& x \qquad -3 \qquad 0 \qquad 1 \qquad 5\\ & y\end{align*}

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

\begin{align*}& x \qquad -4 \qquad 0 \qquad 2 \qquad 6\\ & y\end{align*}

1. \begin{align*}3x+2y=8\end{align*}

\begin{align*}& x \qquad -6 \qquad -2 \qquad 0 \qquad 4\\ & y\end{align*}

1. \begin{align*}4(y-1)=12x-7\end{align*}

\begin{align*}& x \qquad -2 \qquad 0 \qquad 3 \qquad 7\\ & y\end{align*}

1. \begin{align*}\frac{1}{2}x+\frac{1}{3}y=6\end{align*}

\begin{align*}& x \qquad 0 \qquad 4 \qquad 6 \qquad 10\\ & y\end{align*}

Using technology, create a table of values for each of the following linear functions. Using technology, graph each of the linear functions.

1. \begin{align*}y=-2x+3\end{align*}
2. \begin{align*}y=-\frac{1}{2}x-3\end{align*}
3. \begin{align*}y=\frac{4}{3}x-2\end{align*}
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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Dec 19, 2012
Mar 04, 2015

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