###### Example 4.9.8 Using intercepts

If we use the \(x\)- and \(y\)-intercepts to plot \(y=-2x+1\text{,}\) we have some calculation to do. While it is apparent that the \(y\)-intercept is at \((0,1)\text{,}\) where is the \(x\)-intercept? Here are two methods to find it.

Set \(y=0\text{.}\)

\begin{align*}
y\amp=-2x+1\\
\substitute{0}\amp=-2x+1\\
0\subtractright{1}\amp=-2x\\
-1\amp=-2x\\
\divideunder{-1}{-2}\amp=x\\
\frac{1}{2}\amp=x
\end{align*}

Factor out the coefficient of \(x\text{.}\)

\begin{align*}
y\amp=-2x+1\\
y\amp=-2x+(\highlight{-2})\left(\highlight{-\frac{1}{2}}\right)1\\
y\amp=-2\left(x+\left(-\frac{1}{2}\right)1\right)\\
y\amp=-2\left(\highlight{x}-\frac{1}{2}\right)
\end{align*}

And now it is easy to see that substituting \(x=\frac{1}{2}\) would make \(y=0\text{.}\)

So the \(x\)-intercept is at \(\left(\frac{1}{2},0\right)\text{.}\) Plotting both intercepts:

This worked, but here are some observations about why this method is not the greatest.

- We had to plot a point with fractional coordinates.
- We only plotted two points and they turned out very close to each other, so even the slightest inaccuracy in our drawing skills could result in a line that is way off.