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# Intercepts by Substitution

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Practice Intercepts by Substitution
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Graphs of Linear Functions from Intercepts

What are the intercepts of $4x+2y=8$ ? 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 $x$ -intercept and the $y$ -intercept. Graphing a linear function by plotting the $x-$ and $y-$ intercepts is often referred to as the intercept method.

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

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

#### Example A

Identify the $x-$ and $y$ -intercepts for each line.

(a) $2x+y-6=0$

(b) $\frac{1}{2}x-4y=4$

Solution:

(a)

$&\text{Let} \ y = 0. \ \text{Solve for} \ x\text{'}. && \text{Let} \ x = 0. \ \text{Solve for} \ y\text{'}.\\& 2x+y-6=0 && 2x+y-6=0\\& 2x+({\color{red}0})-6=0 && 2({\color{red}0})+y-6=0\\& 2x-6=0 && y-6=0\\& 2x-6+6=0+6 && y-6+6=0+6\\& 2x=6 && y=6\\& \frac{2x}{2}=\frac{6}{2} && \text{The} \ y \text{-intercept is} \ (0, 6)\\& x=3\\& \text{The} \ x \text{-intercept is} \ (3, 0)$

(b)

$& \text{Let} \ y = 0. \ \text{Solve for} \ x\text{'}. && \text{Let} \ x = 0. \ \text{Solve for} \ y\text{'}.\\& \frac{1}{2}x-4y=4 && \frac{1}{2}x-4y=4\\& \frac{1}{2}x-4({\color{red}0})=4 && \frac{1}{2}({\color{red}0})-4y=4\\& \frac{1}{2}x-0=4 && 0-4y=4\\& \frac{1}{2}x=4 && -4y=4\\& \overset{1}{\cancel{2}}\left(\frac{1}{\cancel{2}}\right)x=2(4) && \frac{-4y}{-4}=\frac{4}{-4}\\& x=8 && y=-1\\& \text{The} \ x \text{-intercept is} \ (8, 0) && \text{The} \ y \text{-intercept is} \ (0, -1)$

#### Example B

Use the intercept method to graph $2x-3y=-12$ .

Solution:

$& \text{Let} \ y = 0. \ \text{Solve for} \ x\text{'}. && \text{Let} \ x = 0. \ \text{Solve for} \ y\text{'}.\\& 2x-3y=-12 && 2x-3y=-12\\& 2x-3({\color{red}0})=-12 && 2({\color{red}0})-3y=-12\\& 2x-0=-12 && 0-3y=-12\\& 2x=-12 && -3y=-12\\& \frac{2x}{2}=\frac{-12}{2} && \frac{-3y}{-3}=\frac{-12}{-3}\\& x=-6 && y=4\\& \text{The} \ x \text{-intercept is} \ (-6, 0) && \text{The} \ y \text{-intercept is} \ (0, 4)$

#### Example C

Use the $x-$ and $y$ -intercepts of the graph to identify the linear function that matches the graph.

a) $y=2x-8$

b) $x-2y+8=0$

c) $2x+y-8=0$

The $x$ -intercept is (–8, 0) and the $y$ -intercept is (0, 4).

Solution: Find the $x$ and $y$ intercepts for each equation and see which matches the graph.

a) $x$ intercept: $0=2x-8 \rightarrow x=4$

$y$ intercept: $y=2(0)-8 \rightarrow y=-8$

b) $x$ intercept: $x-2(0)+8=0 \rightarrow x=-8$

$y$ intercept: $0-2y+8=0 \rightarrow y=4$

c) $x$ intercept: $2x+0-8=0 \rightarrow x=4$

$y$ intercept: $2(0)+y-8=0 \rightarrow y=8$

The $x$ and $y$ intercepts match for $x-2y+8=0$ so this is the equation of the line.

#### Concept Problem Revisited

The linear function $4x+2y=8$ can be graphed by using the intercept method.

$& \text{To determine the }x \text{-intercept, let } y=0. && \text{To determine the } y \text{-intercept, let } x=0.\\& \text{Solve for} \ x\text{'}. && \text{Solve for} \ y\text{'} .\\& 4x+2y=8 && 4x+2y=8\\& 4x+2({\color{red}0})=8 && 4({\color{red}0})+2y=8\\& 4x+{\color{red}0}=8 && {\color{red}0}+2y=8\\ & 4x=8 && 2y=8\\& \frac{4x}{4}=\frac{8}{4} && \frac{2y}{2}=\frac{8}{2}\\& x=2 && y=4\\& \text{The} \ x \text{-intercept is} \ (2, 0) && \text{The} \ y \text{-intercept is} \ (0, 4)$

Plot the $x$ -intercept on the $x$ -axis and the $y$ -intercept on the $y$ -axis. Join the two points with a straight line.

### Guided Practice

1. Identify the $x-$ and $y$ -intercepts of the following linear functions:

(i) $2(x-3)=y+4$
(ii) $3x+\frac{2}{3}y-3=0$

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

(i) $5x+2y=-10$

3. Use the $x-$ and $y$ -intercepts of the graph, to match the graph to its function.

(i) $2x+y=6$
(ii) $4x-3y-12=0$
(iii) $5x+3y=15$

1. (i)

$2(x-3)&=y+4 && \text{Simplify the equation}\\2(x-3)&=y+4\\2x-6&=y+4\\2x-6+6&=y+4+6\\2x&=y+10 && \text{You may leave the function in this form.}\\2x-y&=y-y+10\\2x-y&=10$

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.

$& \text{Let} \ y = 0. \ \text{Solve for} \ x. && \text{Let} \ x = 0. \ \text{Solve for} \ y.\\& 2x-y=10 && 2x-y=10\\& 2x-({\color{red}0})=10 && 2({\color{red}0})-y=10\\& 2x=10 && 0-y=10\\& \frac{2x}{2}=\frac{10}{2} && \frac{-y}{-1}=\frac{10}{-1}\\& x=5 && y=-10\\& \text{The} \ x \text{-intercept is} \ (5, 0) && \text{The} \ y \text{-intercept is} \ (0, -10)$

(ii)

$3x+\frac{2}{3}y-3&=0 && \text{Simplify the equation.}\\3(3x)+3\left(\frac{2}{3}\right)y-3(3)&=3(0) && \text{Multiply each term by 3.}\\3(3x)+\cancel{3}\left(\frac{2}{\cancel{3}}\right)y-3(3)&=3(0)\\9x+2y-9&=0\\9x+2y-9+9&=0+9\\9x+2y&=9$

$& \text{Let} \ y = 0. \ \text{Solve for} \ x. && \text{Let} \ x = 0. \ \text{Solve for} \ y.\\& 9x+2y=9 && 9x+2y=9\\& 9x+2({\color{red}0})=9 && 9({\color{red}0})+2y=9\\& 9x+0=9 && 0+2y=9\\& \frac{9x}{9}=\frac{9}{9} && \frac{2y}{2}=\frac{9}{2}\\& x=1 && y=4.5\\& \text{The} \ x \text{-intercept is} \ (1, 0) && \text{The} \ y \text{-intercept is} \ (0, 4.5)$

2.

$& \text{Let} \ y = 0. \ \text{Solve for} \ x. && \text{Let} \ x = 0. \ \text{Solve for} \ y.\\& 5x+2y=-10 && 5x+2y=-10\\& 5x+2({\color{red}0})=-10 && 5({\color{red}0})+2y=-10\\& 5x+0=-10 && 0+2y=-10\\& \frac{5x}{5}=\frac{-10}{5} && \frac{2y}{2}=\frac{-10}{2}\\& x=-2 && y=-5\\& \text{The} \ x \text{-intercept is} \ (-2, 0) && \text{The} \ y \text{-intercept is} \ (0, -5)$

3. Identify the $x-$ and $y$ -intercepts from the graph.

The $x$ -intercept is (3, 0)

The $y$ -intercept is (0, -4)

Determine the $x-$ and $y$ -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)

$& \text{Let} \ y = 0. \ \text{Solve for} \ x. && \text{Let} \ x = 0. \ \text{Solve for} \ y.\\& 2x+y=6 && 2x+y=6\\& 2x+({\color{red}0})=6 && 2({\color{red}0})+y=6\\& 2x=6 && 0+y=6\\& \frac{2x}{2}=\frac{6}{2} && y=6\\& x=3\\& \text{The} \ x \text{-intercept is} \ (3, 0) && \text{The} \ y \text{-intercept is} \ (0, 6)\\& \text{This matches the graph.} && \text{This does not match the graph.}$

$2x+y=6$ is not the linear function for the graph.

(ii)

$& \text{Let} \ y = 0. \ \text{Solve for} \ x. && \text{Let} \ x = 0. \ \text{Solve for} \ y.\\& 4x-3y-12=0 && 4x-3y-12=0\\& 4x-3y-12+12=0+12 && 4x-3y-12+12=0+12\\& 4x-3y=12 && 4x-3y=12\\& 4x-3({\color{red}0})=12 && 4({\color{red}0})-3y=12\\& 4x-0=12 && 0-3y=12\\& 4x=12 && -3y=12\\& \frac{4x}{4}=\frac{12}{4} && \frac{-3y}{-3}=\frac{12}{-3}\\& x=3 && y=-4\\& \text{The} \ x \text{-intercept is} \ (3, 0) && \text{The} \ y \text{-intercept is} \ (0, -4)\\& \text{This matches the graph.} && \text{This matches the graph.}$

$4x-3y-12=0$ is the linear function for the graph.

(iii)

$& \text{Let} \ y = 0. \ \text{Solve for} \ x. && \text{Let} \ x = 0. \ \text{Solve for} \ y.\\& 5x+3y=15 && 5x+3y=15\\& 5x+3({\color{red}0})=15 && 5({\color{red}0})+3y=15\\& 5x+0=15 && 0+3y=15\\& 5x=15 && 3y=15\\& \frac{5x}{5}=\frac{15}{5} && \frac{3y}{3}=\frac{15}{3}\\& x=3 && y=5\\& \text{The} \ x \text{-intercept is} \ (3, 0) && \text{The} \ y \text{-intercept is} \ (0, 5)\\& \text{This matches the graph.} && \text{This does not match the graph.}$

$5x+3y=15$ is not the linear function for the graph.

### Explore More

For 1-10, complete the following table:

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

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

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

Use the $x-$ and $y$ -intercepts to match each graph to its function.

a. $7x+5y-35=0$
b. $y=5x+10$
c. $2x+4y+8=0$
d. $2x+y=2$
1. .

1. .

1. .

1. .

### Vocabulary Language: English

$x-$intercept

$x-$intercept

An $x-$intercept is a location where a graph crosses the $x-$axis. As a coordinate pair, this point will always have the form $(x, 0)$. $x-$intercepts are also called solutions, roots or zeros.
$y$-intercept

$y$-intercept

A $y-$intercept is a location where a graph crosses the $y-$axis. As a coordinate pair, this point will always have the form $(0, y)$.
Intercept

Intercept

The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis.
Intercept Method

Intercept Method

The intercept method is a way of graphing a linear function by using the coordinates of the $x$ and $y$-intercepts. The graph is drawn by plotting these coordinates on the Cartesian plane and then joining them with a straight line.