#### Example 7.14.

Graph \(f(x)=\sin (x)\) and \(g(x)= \sin \left(x - \dfrac{\pi}{4}\right)\) for \(-2\pi \le x \le 2\pi\text{.}\) How is the graph of \(g\) different from the graph of \(f\text{?}\)

## Solution.

The graphs are shown below. The graph of \(g(x)= \sin \left(x - \dfrac{\pi}{4}\right)\) has the same amplitude, midline, and period as the graph of \(f(x)=\sin (x)\text{,}\) but the graph of \(g\) is shifted to the right by \(\dfrac{\pi}{4}\) units, compared to the graph of \(f\text{.}\)

We can see why this shift occurs by studying a table of values for the two functions.

\(x\) | \(0\) | \(\dfrac{\pi}{4}\) | \(\dfrac{\pi}{2}\) | \(\dfrac{3\pi}{4}\) | \(\pi\) | \(\dfrac{5\pi}{4}\) | \(\dfrac{3\pi}{2}\) | \(\dfrac{7\pi}{4}\) | \(2\pi\) |

\(f(x)\) | \(0\) | \(\dfrac{\sqrt{2}}{2}\) | \(1\) | \(\dfrac{\sqrt{2}}{2}\) | \(0\) | \(\dfrac{-\sqrt{2}}{2}\) | \(-1\) | \(\dfrac{-\sqrt{2}}{2}\) | \(0\) |

\(g(x)\) | \(\dfrac{-\sqrt{2}}{2}\) | \(0\) | \(\dfrac{\sqrt{2}}{2}\) | \(1\) | \(\dfrac{\sqrt{2}}{2}\) | \(0\) | \(\dfrac{-\sqrt{2}}{2}\) | \(-1\) | \(\dfrac{-\sqrt{2}}{2}\) |

Notice that in the table, \(g\) has the same function values as \(f\text{,}\) but each one is shifted \(\dfrac{\pi}{4}\) units to the right. The same thing happens in the graph: each \(y\)-value appears \(\dfrac{\pi}{4}\) units farther to the right on \(g\) than it does on \(f\text{.}\)