Answer:
At ; I = 3, P is at maximum
Step-by-step explanation:
P ( I ) =
at I = 3
P" ( I ) < 0
therefore at I = 3 the value of P( I ) is at maximum
attached below is the detailed solution
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1 answer:
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Answer:
amplitude = 2
period = 4
midline = y = -1
equation = either or
depending on what format you're using
Step-by-step explanation:
the amplitude is the distance from the bottommost point to the topmost point
so if we use the points (0,1) and (2,-3), the min y value is -3 and the max y value is 1
so amplitude = ymax - ymin = 1 - - 3 = 4
amplitude = 4
the period is the distance it takes to complete one cycle
we have (0,1) and (4,1) which are two crest points
the distance between these is x2 - x1 = 4 - 0 = 4
period = 4
the midline is a horizontal line bisecting the sin line, which would be in the middle of its min and max y values
we already found that ymax = 1 and ymin = -3
so the midline would be y = (ymax + ymin) / 2 = (1 -3) /2 = -1
midline = y = -1
to find the sin equation, we will fill out the equation
asin(b(x-d))+c
a is the amplitude = 2
b = 2π/period = 2π/4 = π/2
c is the vertical shift. the midline of sine is originally at y = 0
since it is at y = -1 here, the vertical shift is -1 - 0 = -1
d is the horizontal shift. the middle of the upwards portion of the sine graph is originally at x = 0, but in the picture we see that it is at x = -1 so the shift is -1 - 0 = -1
a = 2, b = π/2, c = -1, d = -1
plug these values into the equation to get
asin(b(x-d))+c
2sin(π/2(x--1))+-1
=2sin(π/2(x+1))-1
=2sin(πx/2+π/2)-1
equation = either or
depending on what format you're using