Answer:
We have function,
[tex]y = 3 - 6 \sin {}^{} (2x + \frac{\pi}{2} ) [/tex]
Standard Form of Sinusoid is
[tex]y = - 6 \sin(2x + \frac{\pi}{2} ) + 3[/tex]
Which corresponds to
[tex]y = a \sin(b(x + c)) + d[/tex]
where a is the amplitude
2pi/b is the period
c is phase shift
d is vertical shift or midline.
In the equation equation, we must factor out 2 so we get
[tex]y = - 6(2(x + \frac{\pi}{4} )) + 3[/tex]
Also remeber a and b is always positive
So now let answer the questions.
a. The period is
[tex] \frac{2\pi}{ |b| } [/tex]
[tex] \frac{2\pi}{ |2| } = \pi[/tex]
So the period is pi radians.
b. Amplitude is
[tex] | - 6| = 6[/tex]
Amplitude is 6.
c. Domain of a sinusoid is all reals. Here that stays the same. Range of a sinusoid is [-a+c, a-c]. Put the least number first, and the greatest next.
So using that rule, our range is [6+3, -6+3]= [9,-3] So our range is [-3,9].
D. Plug in 0 for x.
[tex]3 - 6 \sin((2(0) + \frac{\pi}{2} ) [/tex]
[tex]3 - 6 \sin( \frac{\pi}{2} ) [/tex]
[tex]3 - 6(1)[/tex]
[tex]3 - 6[/tex]
[tex] = - 3[/tex]
So the y intercept is (0,-3)
E. To find phase shift, set x-c=0 to solve for phase shift.
[tex]x + \frac{\pi}{4} = 0[/tex]
[tex]x = - \frac{\pi}{4} [/tex]
Negative means to the left, so the phase shift is pi/4 units to the left.
f. Period is PI, so use interval [0,2pi].
Look at the graph above,
