sketch a graph of the function f(x)=3sin(π/4 x)-2

Explanation:

Step1: Identify the amplitude

The general form of a sinusoidal function is $y = A\sin(Bx - C)+D$. For $f(x)=3\sin(\frac{\pi}{4}x)-2$, the amplitude $|A| = 3$. This means the graph oscillates 3 units above and below the mid - line.

Step2: Identify the period

The period of a sine function $y=\sin(Bx)$ is given by $T=\frac{2\pi}{|B|}$. Here, $B = \frac{\pi}{4}$, so $T=\frac{2\pi}{\frac{\pi}{4}}=8$.

Step3: Identify the vertical shift

The value of $D=-2$, so the mid - line of the graph is $y = - 2$.

Step4: Find key points

• Start with the standard sine - function key points and adjust for the transformations. For a standard sine function $y=\sin(x)$, key points in one period are $(0,0),(\frac{\pi}{2},1),(\pi,0),(\frac{3\pi}{2}, - 1),(2\pi,0)$.
• For $y = 3\sin(\frac{\pi}{4}x)-2$, when $x = 0$, $y=3\sin(0)-2=-2$.
• When $\frac{\pi}{4}x=\frac{\pi}{2}\Rightarrow x = 2$, $y=3\sin(\frac{\pi}{2})-2=3 - 2=1$.
• When $\frac{\pi}{4}x=\pi\Rightarrow x = 4$, $y=3\sin(\pi)-2=-2$.
• When $\frac{\pi}{4}x=\frac{3\pi}{2}\Rightarrow x = 6$, $y=3\sin(\frac{3\pi}{2})-2=-3 - 2=-5$.
• When $\frac{\pi}{4}x = 2\pi\Rightarrow x = 8$, $y=3\sin(2\pi)-2=-2$.

Then plot these key points and draw a smooth sinusoidal curve with amplitude 3, period 8, and mid - line $y=-2$ repeating every 8 units.

Answer:

The graph is a sinusoidal curve with amplitude 3, period 8, and mid - line $y=-2$. Key points in one period are $(0, - 2),(2,1),(4, - 2),(6, - 5),(8, - 2)$ and the curve repeats every 8 units.