answer the following questions for (y = sinleft(-\frac{pi}{4}x-\frac{pi}{2}\right)-5). express your answers…

answer the following questions for (y = sinleft(-\frac{pi}{4}x-\frac{pi}{2}\right)-5). express your answers as simplified fractions if necessary and express points as ordered pairs.\n(a) identify the amplitude, period, phase shift, and vertical shift.\n(b) graph the function and identify the key points on one full period.\npart 1 of 5\n(a) identify the amplitude, period, phase shift, and vertical shift.\nthe amplitude is 1.\npart 2 of 5\nthe period is 8.\npart 3 of 5\nthe phase shift is -2.\npart 4 of 5\nthe vertical shift is -5.\npart: 4 / 5\npart 5 of 5\n(b) graph the function and identify the key points on one full period.\nto draw the graph, plot all points corresponding to the relative minima, relative maxima, and any key points within one cycle. then click on the graph icon.

answer the following questions for (y = sinleft(-\frac{pi}{4}x-\frac{pi}{2}\right)-5). express your answers as simplified fractions if necessary and express points as ordered pairs.\n(a) identify the amplitude, period, phase shift, and vertical shift.\n(b) graph the function and identify the key points on one full period.\npart 1 of 5\n(a) identify the amplitude, period, phase shift, and vertical shift.\nthe amplitude is 1.\npart 2 of 5\nthe period is 8.\npart 3 of 5\nthe phase shift is -2.\npart 4 of 5\nthe vertical shift is -5.\npart: 4 / 5\npart 5 of 5\n(b) graph the function and identify the key points on one full period.\nto draw the graph, plot all points corresponding to the relative minima, relative maxima, and any key points within one cycle. then click on the graph icon.

Answer

Explanation:

Step1: Recall sine - function form

The general form of a sine function is $y = A\sin(Bx - C)+D$, where $A$ is the amplitude, $T=\frac{2\pi}{|B|}$ is the period, $\frac{C}{B}$ is the phase - shift, and $D$ is the vertical shift. For the function $y=\sin(-\frac{\pi}{4}x-\frac{\pi}{2}) - 5$, we have $A = 1$, $B=-\frac{\pi}{4}$, $C =-\frac{\pi}{2}$, and $D=-5$.

Step2: Find key - points for one period

The period $T = 8$. We start by finding the starting point of the period using the phase - shift. Let $-\frac{\pi}{4}x-\frac{\pi}{2}=0$, then $x=-2$.

  • For the maximum point:
    • The maximum value of $\sin u$ is 1. We want to find $x$ when $\sin(-\frac{\pi}{4}x-\frac{\pi}{2}) = 1$. So $-\frac{\pi}{4}x-\frac{\pi}{2}=\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}$. Solving for $x$ gives $x=-4$. Substituting $x = - 4$ into $y=\sin(-\frac{\pi}{4}x-\frac{\pi}{2})-5$, we get $y=1 - 5=-4$. The maximum point is $(-4,-4)$.
  • For the minimum point:
    • The minimum value of $\sin u$ is - 1. We want to find $x$ when $\sin(-\frac{\pi}{4}x-\frac{\pi}{2})=-1$. So $-\frac{\pi}{4}x-\frac{\pi}{2}=-\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}$. Solving for $x$ gives $x = 0$. Substituting $x = 0$ into $y=\sin(-\frac{\pi}{4}x-\frac{\pi}{2})-5$, we get $y=-1 - 5=-6$. The minimum point is $(0,-6)$.
  • For the mid - points:
    • When $-\frac{\pi}{4}x-\frac{\pi}{2}=0$, $x=-2$ and $y=-5$.
    • When $-\frac{\pi}{4}x-\frac{\pi}{2}=\pi$, $x=-6$ and $y=-5$.

The key points for one period $[-6,2]$ are $(-6,-5),(-4,-4),(-2,-5),(0,-6),(2,-5)$.

Answer:

The key points for one full period are $(-6,-5),(-4,-4),(-2,-5),(0,-6),(2,-5)$