6.1 Graphs of the sine and cosine functions  (Page 5/13)

Let’s begin by comparing the equation to the form $y=A\sin (Bx).$

• Step 1. We can see from the equation that $A=-2,$ so the amplitude is 2.
  $\left|A\right|=2$
• Step 2. The equation shows that $B=\frac{\pi }{2},$ so the period is
  $$\begin{array}{l}P=\frac{2\pi }{\frac{\pi }{2}}\hfill \\ =2\pi \cdot \frac{2}{\pi }\hfill \\ =4\hfill \end{array}$$
• Step 3. Because $A$ is negative, the graph descends as we move to the right of the origin.
• Step 4–7. The x -intercepts are at the beginning of one period, $x=0,$ the horizontal midpoints are at $x=2$ and at the end of one period at $x=4.$

The quarter points include the minimum at $x=1$ and the maximum at $x=3.$ A local minimum will occur 2 units below the midline, at $x=1,$ and a local maximum will occur at 2 units above the midline, at $x=3.$ [link] shows the graph of the function.

• Step 1. The function is already written in general form: $f(x)=3\sin \left(\frac{\pi }{4}x-\frac{\pi }{4}\right).$ This graph will have the shape of a sine function    , starting at the midline and increasing to the right.
• Step 2. $\left|A\right|=\left|3\right|=3.$ The amplitude is 3.
• Step 3. Since $\left|B\right|=\left|\frac{\pi }{4}\right|=\frac{\pi }{4},$ we determine the period as follows.
  $P=\frac{2\pi }{\left|B\right|}=\frac{2\pi }{\frac{\pi }{4}}=2\pi \cdot \frac{4}{\pi }=8$

  The period is 8.

• Step 4. Since $C=\frac{\pi }{4},$ the phase shift is
  $\frac{C}{B}=\frac{\frac{\pi }{4}}{\frac{\pi }{4}}=1.$

  The phase shift is 1 unit.

• Step 5. [link] shows the graph of the function.