sine graph with max at 0,5 and min at pi/2, 1, what would the graph look like, and what would the equation be?

Answer 1

y = Asin(Bx+C) + D

A = Amplitude = (max-min)/2 = (5-1)/2 = 2

D = midline = (max+min)/2 = (5+1)/2 = 3

5 = 2sin(B(0) +C ) + 3

sinC = (5-3)/2 = 1

C = pi/2

1 = 2sin(B(pi/2) + pi/2) + 3

-2/2= -1 = sin(Bpi/2 + pi/2)

Bpi/2+ pi/2= arcsin-1= - pi/2

Bpi/2 = -pi

B= -2

y= 2sin(-2x + pi/2) + 3

or use B=+2,

y = 2sin(2x + pi/2) +3

Answer 2

The peak-to-peak amplitude would be 5 - 1 = 4.

From its midline, the graph would go up 2, and then down 2. So the midline is located at 1 + 2, or 3. Thus we need a vertical shift (displacement) of 3 to move the standard midline of 0 up to 3.

Half of 4, or 2, is the vertical stretch.

The minimum occurs after the maximum, so there is no x-axis reflection.

The distance from the maximum to the minimum along the axis is π/2, so the full cycle is twice this amount, or π. The normal period of a sine wave is 2π, so this is happening at twice the normal rate. So, the horizontal shrink is 1/2.

Since the maximum y-value occurs at 0, this requires a left shift of π/2.

Putting all these facts together, we determine that:

y = 2sin(2x+π/2) + 3

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