graph $y = 5sinleft(\frac{pi}{2}x\right)-4$ in the interactive widget. note that one moveable point always…

Explanation:

Step1: Identify the general form of sine - function

The general form of a sine - function is $y = A\sin(Bx - C)+D$. For the function $y = 5\sin(\frac{\pi}{2}x)-4$, we have $A = 5$, $B=\frac{\pi}{2}$, $C = 0$, and $D=-4$.

Step2: Find the amplitude

The amplitude is given by $|A|$. Here, $|A|=|5| = 5$.

Step3: Find the period

The period of a sine - function $y = A\sin(Bx - C)+D$ is $T=\frac{2\pi}{|B|}$. Substituting $B = \frac{\pi}{2}$, we get $T=\frac{2\pi}{\frac{\pi}{2}}=4$.

Step4: Find the mid - line

The mid - line of the function is $y = D$. So, the mid - line is $y=-4$.

Step5: Find the maximum and minimum values

The maximum value of the function occurs when $\sin(\frac{\pi}{2}x)=1$. Then $y_{max}=5\times1 - 4=1$. The minimum value occurs when $\sin(\frac{\pi}{2}x)=-1$. Then $y_{min}=5\times(-1)-4=-9$.

Step6: Plot key points

We know that for $y = \sin x$, key points are $(0,0),(\frac{\pi}{2},1),(\pi,0),(\frac{3\pi}{2},-1),(2\pi,0)$. For $y = 5\sin(\frac{\pi}{2}x)-4$, when $x = 0$, $y=5\sin(0)-4=-4$; when $x = 1$, $y = 5\sin(\frac{\pi}{2})-4=1$; when $x = 2$, $y=5\sin(\pi)-4=-4$; when $x = 3$, $y = 5\sin(\frac{3\pi}{2})-4=-9$; when $x = 4$, $y=5\sin(2\pi)-4=-4$.

Answer:

The graph of $y = 5\sin(\frac{\pi}{2}x)-4$ is a sine - wave with amplitude 5, period 4, mid - line $y=-4$, maximum value $y = 1$ and minimum value $y=-9$. Key points include $(0, - 4),(1,1),(2,-4),(3,-9),(4,-4)$.