Question

100 points!!! Pre calculus. I need helpppppppppp

Answer 1

Answer:

We have function,

$y = 3 - 6 \sin {}^{} (2x + \frac{\pi}{2} )$

Standard Form of Sinusoid is

$y = - 6 \sin(2x + \frac{\pi}{2} ) + 3$

Which corresponds to

$y = a \sin(b(x + c)) + d$

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

$y = - 6(2(x + \frac{\pi}{4} )) + 3$

Also remeber a and b is always positive

So now let answer the questions.

a. The period is

$\frac{2\pi}{ |b| }$

$\frac{2\pi}{ |2| } = \pi$

So the period is pi radians.

b. Amplitude is

$| - 6| = 6$

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.

$3 - 6 \sin((2(0) + \frac{\pi}{2} )$

$3 - 6 \sin( \frac{\pi}{2} )$

$3 - 6(1)$

$3 - 6$

$= - 3$

So the y intercept is (0,-3)

E. To find phase shift, set x-c=0 to solve for phase shift.

$x + \frac{\pi}{4} = 0$

$x = - \frac{\pi}{4}$

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,