determine the amplitude, range, and period of the function y = - 5/4 sin(4x) and sketch its graph. the…

Explanation:

Step1: Find the amplitude

For a sine - function of the form $y = A\sin(Bx)$, the amplitude is $|A|$. Given $y=-\frac{5}{4}\sin(4x)$, $A =-\frac{5}{4}$, so the amplitude is $\left|-\frac{5}{4}\right|=\frac{5}{4}$.

Step2: Find the range

The range of the basic sine function $y = \sin(x)$ is $[- 1,1]$. For the function $y = A\sin(Bx)$, the range is $[ - |A|,|A|]$. Since $A=-\frac{5}{4}$, the range is $\left[-\frac{5}{4},\frac{5}{4}\right]$.

Step3: Find the period

For a sine - function of the form $y=\sin(Bx)$, the period is given by $T=\frac{2\pi}{|B|}$. Given $y =-\frac{5}{4}\sin(4x)$, $B = 4$, so the period is $T=\frac{2\pi}{4}=\frac{\pi}{2}$.

Step4: Find the y - coordinates of the quarter - points

The quarter - points of $y=\sin(x)$ are at $x = 0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$. For $y =-\frac{5}{4}\sin(4x)$:

• When $x = 0$, $y=-\frac{5}{4}\sin(0)=0$, so the point is $(0,0)$.
• When $x=\frac{\pi}{8}$, $y =-\frac{5}{4}\sin\left(4\times\frac{\pi}{8}\right)=-\frac{5}{4}\sin\left(\frac{\pi}{2}\right)=-\frac{5}{4}$, so the point is $\left(\frac{\pi}{8},-\frac{5}{4}\right)$.
• When $x=\frac{\pi}{4}$, $y=-\frac{5}{4}\sin\left(4\times\frac{\pi}{4}\right)=-\frac{5}{4}\sin(\pi)=0$, so the point is $\left(\frac{\pi}{4},0\right)$.
• When $x=\frac{3\pi}{8}$, $y =-\frac{5}{4}\sin\left(4\times\frac{3\pi}{8}\right)=-\frac{5}{4}\sin\left(\frac{3\pi}{2}\right)=\frac{5}{4}$, so the point is $\left(\frac{3\pi}{8},\frac{5}{4}\right)$.
• When $x=\frac{\pi}{2}$, $y=-\frac{5}{4}\sin\left(4\times\frac{\pi}{2}\right)=-\frac{5}{4}\sin(2\pi)=0$, so the point is $\left(\frac{\pi}{2},0\right)$.

Answer:

The amplitude is $\frac{5}{4}$. The range is $\left[-\frac{5}{4},\frac{5}{4}\right]$. The period is $\frac{\pi}{2}$. The quarter - points are $(0,0),\left(\frac{\pi}{8},-\frac{5}{4}\right),\left(\frac{\pi}{4},0\right),\left(\frac{3\pi}{8},\frac{5}{4}\right),\left(\frac{\pi}{2},0\right)$