SOLUTION: Determine the amplitude, midline, and an equation for the sine graph with a max at (-5pi/12, -1) and a min at (pi/12, -5).
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Question 1178489: Determine the amplitude, midline, and an equation for the sine graph with a max at (-5pi/12, -1) and a min at (pi/12, -5). Answer by greenestamps(13209) (Show Source):
a is the amplitude -- half the difference between the minimum and maximum values of the function
b is the horizontal stretch/shrink factor -- the period is 2pi/b
c is the horizontal shift
d is the vertical shift
The minimum and maximum values of the function are -5 and -1; that makes the midline -3 and the amplitude 2: a=2; d=-3
Half the period (from where the function value is maximum to where it is minimum) is from -5pi/12 to pi/12, a difference of 6pi/12=pi/2; so the period of the function is pi. That makes b=2pi/pi=2.
In the basic sine graph, the function value is 0 and increasing at x=0. To determine the horizontal shift, we need to determine where this function value is 0 and increasing. That place is one-quarter period to the left of the maximum. The maximum is at -5pi/12 and one-quarter period is pi/4. So the function is 0 and increasing at -5pi/12-pi/4 = -5pi/12-3pi/12 = -8pi/12 = -2pi/3. So c is 2pi/3.
The function is then
A graph showing about 2 periods; maximum at (-5pi/12,-1) = (-1.3,-1) and minimum at (pi/12,-5) = (0.26,-5)
