Graph a sine function whose amplitude is 4, period is pi, Midline is y= -3, and y- intercept is (0, -3). The graph is not a reflection of the parent function over the x axis. The first point Must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point

Guest Jan 28, 2019

#2**0 **

Period = 4pi

T = 4pib = 2pi/T = 2pi/(4pi) = 1/2 = 0.5

Amplitude = 2a = 2

Midline: y = 3

d = 3

y-intercept: (0,3)

The function is a reflection of its parent function over the x-axis, so 'a' is negative meaning a = -2 instead of a = 2

The function is

y = a*sin(bx-c)+d

y = -2*sin(0.5x-0)+3

y = -2*sin(0.5x)+3

Mclovin123 Jan 28, 2019

#3**+1 **

We have the form

y = A sin (Bx + C ) + D

A = the amplitude = 4

B = the number of periods in 2pi....so ...B = 2pi / period = 2pi/pi = 2

C is the phase shift....there is none....so....C = 0

D = the shift up / down = -3

So.....the equation is

y = 4sin (2x) - 3

Here is the graph : https://www.desmos.com/calculator/67jvpq3xip

CPhill Jan 28, 2019