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# 3.5: Graphs of Linear Functions from Intercepts

Difficulty Level: At Grade Created by: CK-12

What are the intercepts of 4x+2y=8\begin{align*}4x+2y=8\end{align*}? How could you use the intercepts to quickly graph the function?

### Guidance

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. While any two points can be used to graph a linear function, two points in particular that can be used are the x\begin{align*}x\end{align*}-intercept and the y\begin{align*}y\end{align*}-intercept. Graphing a linear function by plotting the x\begin{align*}x-\end{align*} and y\begin{align*}y-\end{align*} intercepts is often referred to as the intercept method.

The x\begin{align*}x\end{align*}-intercept is where the graph crosses the x\begin{align*}x\end{align*}-axis. Its coordinates are (x,0)\begin{align*}(x, 0)\end{align*}. Because all x\begin{align*}x\end{align*}-intercepts have a y\begin{align*}y\end{align*} coordinate equal to 0, you can find an x\begin{align*}x\end{align*}-intercept by substituting 0 for y\begin{align*}y\end{align*} in the equation and solving for x\begin{align*}x\end{align*}.

The y\begin{align*}y\end{align*}-intercept is where the graph crosses the y\begin{align*}y\end{align*}-axis. Its coordinates are (0,y)\begin{align*}(0, y)\end{align*}. Because all y\begin{align*}y\end{align*}-intercepts have a x\begin{align*}x\end{align*} coordinate equal to 0, you can find an y\begin{align*}y\end{align*}-intercept by substituting 0 for x\begin{align*}x\end{align*} in the equation and solving for y\begin{align*}y\end{align*}.

#### Example A

Identify the x\begin{align*}x-\end{align*} and y\begin{align*}y\end{align*}-intercepts for each line.

(a) 2x+y6=0\begin{align*}2x+y-6=0\end{align*}

(b) 12x4y=4\begin{align*}\frac{1}{2}x-4y=4\end{align*}

Solution:

(a)

Let y=0. Solve for x'.2x+y6=02x+(0)6=02x6=02x6+6=0+62x=62x2=62x=3The x-intercept is (3,0)Let x=0. Solve for y'.2x+y6=02(0)+y6=0y6=0y6+6=0+6y=6The y-intercept is (0,6)

(b)

#### Example B

Use the intercept method to graph \begin{align*}2x-3y=-12\end{align*}

Solution:

#### Example C

Use the \begin{align*}x-\end{align*} and \begin{align*}y\end{align*}-intercepts of the graph to identify the linear function that matches the graph.

The \begin{align*}x\end{align*}-intercept is (-8, 0) and the \begin{align*}y\end{align*}-intercept is (0, 4).

Solution:

This does not match the graph. This does not match the graph. Confirm the \begin{align*}y\end{align*}-intercept.

The \begin{align*}y\end{align*}-intercept is (0, 4). This matches the graph.

The linear function that matches the graph is

#### Concept Problem Revisited

The linear function \begin{align*}4x+2y=8\end{align*} can be graphed by using the intercept method.

Plot the \begin{align*}x\end{align*}-intercept on the \begin{align*}x\end{align*}-axis and the \begin{align*}y\end{align*}-intercept on the \begin{align*}y\end{align*}-axis. Join the two points with a straight line.

### Vocabulary

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

A \begin{align*}y\end{align*}-intercept of a relation is the \begin{align*}y-\end{align*}coordinate of the point where the relation intersects the \begin{align*}y\end{align*}-axis.

### Guided Practice

1. Identify the \begin{align*}x-\end{align*} and \begin{align*}y\end{align*}-intercepts of the following linear functions:

(i) \begin{align*}2(x-3)=y+4\end{align*}
(ii) \begin{align*}3x+\frac{2}{3}y-3=0\end{align*}

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

(i) \begin{align*}5x+2y=-10\end{align*}

3. Use the \begin{align*}x-\end{align*} and \begin{align*}y\end{align*}-intercepts of the graph, to match the graph to its function.

(i) \begin{align*}2x+y=6\end{align*}
(ii) \begin{align*}4x-3y-12=0\end{align*}
(iii) \begin{align*}5x+3y=15\end{align*}

1. (i)

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.

(ii)

2. \begin{align*}5x+2y=-10\end{align*} Determine the \begin{align*}x-\end{align*} and \begin{align*}y\end{align*}-intercepts.

3. Identify the \begin{align*}x-\end{align*} and \begin{align*}y\end{align*}-intercepts from the graph.

The \begin{align*}x\end{align*}-intercept is (3, 0)

The \begin{align*}y\end{align*}-intercept is (0, -4)

Determine the \begin{align*}x-\end{align*} and \begin{align*}y\end{align*}-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)

(ii)

(iii)

### Practice

For 1-10, complete the following table:

Function \begin{align*}x\end{align*}-intercept \begin{align*}y\end{align*}-intercept
\begin{align*}7x-3y=21\end{align*} 1. 2.
\begin{align*}8x-3y+24=0\end{align*} 3. 4.
\begin{align*}\frac{x}{4}-\frac{y}{2}=3\end{align*} 5. 6.
\begin{align*}7x+2y-14=0\end{align*} 7. 8.
\begin{align*}\frac{2}{3}x-\frac{1}{4}y=-2\end{align*} 9. 10.

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

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

Use the \begin{align*}x-\end{align*} and \begin{align*}y\end{align*}-intercepts to match each graph to its function.

a.\begin{align*}7x+5y-35=0\end{align*}
b.\begin{align*}y=5x+10\end{align*}
c.\begin{align*}2x+4y+8=0\end{align*}
d.\begin{align*}2x+y=2\end{align*}

## Date Created:

Dec 19, 2012

Apr 29, 2014
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