###### Example4.9.8Using 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.

###### Factor out the coefficient of $$x\text{.}$$
\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*}
\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.
in-context