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:

Figure 4.9.9 Marking intercepts
Figure 4.9.10 Using slope triangles

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

in-context