일차함수와 일차방정식

x-intercept (2,0): y-intercept (0,−2)

x-intercept (3,0); y-intercept (0,2)

First, find the $x\text{-intercept}.$ Let $y=0,$

$\begin{array}{c} \\ \quad -x+2y=6 \\ -x+2(0)=6 \\ \quad -x=6 \\ \quad x=-6\end{array}$

The $x\text{-intercept}$ is $(-6,0).$

Now find the $y\text{-intercept}.$ Let $x=0.$

$\begin{array}{c} \\ -x+2y=6 \\ -0+2y=6 \\ \\ \\ \quad 2y=6 \\ \quad y=3\end{array}$

The $y\text{-intercept}$ is $(0,3).$

Find a third point. We’ll use $x=2,$

$\begin{array}{c} \\ -x+2y=6 \\ -2+2y=6 \\ \\ \\ \quad 2y=8 \\ \quad y=4\end{array}$

A third solution to the equation is $(2,4).$

Summarize the three points in a table and then plot them on a graph.

Do the points line up? Yes, so draw line through the points.

Find the intercepts and a third point.

We list the points and show the graph.

This line has only one intercept! It is the point $(0,0).$

To ensure accuracy, we need to plot three points. Since the intercepts are the same point, we need two more points to graph the line. As always, we can choose any values for $x,$ so we’ll let $x$ be $1$ and $-1.$

Organize the points in a table.

Plot the three points, check that they line up, and draw the line.

ⓐ $\,y=-3$

This equation has only one variable, $y.$ Its graph is a horizontal line crossing the $y\text{-axis}$ at $-3.$

ⓑ $\,4x-6y=12$

This equation is of the form $Ax+By=C.$ Find the intercepts and one more point.

ⓒ $\,x=2$

There is only one variable, $x.$ The graph is a vertical line crossing the $x\text{-axis}$ at $2.$

ⓓ $\,y=\frac{2}{5}x-1$

Since $y$ is isolated on the left side of the equation, it will be easiest to graph this line by plotting three points.

• ⓐ intercepts
• ⓑ horizontal line
• ⓒ plotting points
• ⓓ vertical line

• ⓐ vertical line
• ⓑ plotting points
• ⓒ horizontal line
• ⓓ intercepts

(3,0),(0,3)

(5,0),(0,−5)

(−2,0),(0,−2)