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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)\begin{align*}&\text{Let} \ y = 0. \ \text{Solve for} \ x\text{'}. && \text{Let} \ x = 0. \ \text{Solve for} \ y\text{'}.\\ & 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)\end{align*}

(b)

Let y=0. Solve for x'.12x4y=412x4(0)=412x0=412x=421(12)x=2(4)x=8The x-intercept is (8,0)Let x=0. Solve for y'.12x4y=412(0)4y=404y=44y=44y4=44y=1The y-intercept is (0,1)\begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x\text{'}. && \text{Let} \ x = 0. \ \text{Solve for} \ y\text{'}.\\ & \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)\end{align*}

#### Example B

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

Solution:

\begin{align*}& \text{Let} \ y = 0. \ \text{Solve for} \ x\text{'}. && \text{Let} \ x = 0. \ \text{Solve for} \ y\text{'}.\\ & 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)\end{align*}

#### 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:

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

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

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

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

The linear function that matches the graph is \begin{align*}\boxed{x-2y+8=0.}\end{align*}

#### Concept Problem Revisited

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

\begin{align*}& \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{'}. && \text{Solve for} \ y\text{'} .\\ & 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)\end{align*}

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)

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

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.

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

(ii)

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

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

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

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

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)

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

(ii)

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

(iii)

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

### 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*}

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Dec 19, 2012
Apr 29, 2014

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