Question

(a) How is the graph of \(y = {\bf{1}} + \sqrt x \) related to the graph of \(y = \sqrt x \)? Use the answer and Figure 4(a) to sketch the graph of \(y = {\bf{1}} + \sqrt x \).

(b) How is the graph of \(y = {\bf{5sin}}\pi x\) related to the graph of \(y = {\bf{sin}}x\)? Use your answer and Figure 6 to sketch the graph of \(y = {\bf{5sin}}\pi x\).

Short answer

a. The graph of a function \(y = 1 + \sqrt x \) is obtained by shifting the graph of \(y = \sqrt x \) by 1 unit upward.

b. The graph of \(y = \sin \pi x\) is obtained from the graph of \(y = \sin x\) by compressing horizontally by a factor of \(\pi \). And the period of the function \(y = \sin \pi x\) is 2.

Similarly, the graph of \(y = 5\sin \pi x\) is obtained from the graph of \(y = \sin \pi x\) by stretching it vertically \(y = \sin \pi x\) by a factor of 5.

Step-by-step solution

Find the transformation for \(y = {\bf{1}} + \sqrt x \)

The graph of a function \(y = 1 + \sqrt x \) is obtained by shifting the graph of \(y = \sqrt x \) by 1 unit upward.

Sketch the graph of \(y = {\bf{1}} + \sqrt x \)

The sketch is shown below:

Find the transformation for \(y = {\bf{5sin}}\pi x\)

The graph of \(y = \sin \pi x\) is obtained from the graph of \(y = \sin x\) by compressing horizontally by a factor of \(\pi \). And the period of the function \(y = \sin \pi x\) becomes \(\frac{{2\pi }}{\pi } = 2\).

Similarly, the graph of \(y = 5\sin \pi x\) is obtained from the graph of \(y = \sin \pi x\) by stretching it vertically \(y = \sin \pi x\) by a factor of 5.

Sketch the graph of \(y = {\bf{5sin}}\pi x\)

The sketch is shown below: