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 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 

 Jan 28, 2019
 #1
avatar+145 
0

Peri

 Jan 28, 2019
 #2
avatar+145 
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

 Jan 28, 2019
 #3
avatar+95866 
+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

 

 

cool cool cool

 Jan 28, 2019

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