SOLUTION: This problem has a graph that goes with it but I cannot paste it here. The graph is of a negative sin function that passes through the origin with an amplitude of 5, period 6pi and

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Question 1158754: This problem has a graph that goes with it but I cannot paste it here. The graph is of a negative sin function that passes through the origin with an amplitude of 5, period 6pi and phase shift -3pi. What I have trouble with is writing the formula in the form y=asin (bx+ or -c). I put y=5sin(1/3x-pi) which is wrong and I want to know why its wrong. Please help me! If you know how to do this please explain the steps so i understand where I went wrong. Thank you!
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!

A = -5 since the amplitude is 5 and the function is a negative sine function.
Recall that |A| is the amplitude.

T = 6pi is the period
B = 2pi/T
B = 2pi/(6pi)
B = 2/6
B = 1/3 is the coefficient of the x term

C = is the phase shift
D = 0 is the midline

Plug those A,B,C,D values into the equation below. Simplify.








is the final answer

-----------------------------------------------------

If you want to describe this as a series of transformations, then you would start with the parent function and follow the steps below
  1. Vertically stretch the graph by a factor of 5 (ie make it 5 times taller). This changes the amplitude from 1 to 5. So A = 1 becomes A = 5.
  2. Reflect the graph over the x axis. This is from the "negative sin function" portion in the instructions. So A goes from A = 5 to A = -5.
  3. Horizontally stretch the graph by a factor of 3. It is now 3 times wider. The old period 2pi has tripled to 6pi. At the same time, B changes from B = 1 to B = 1/3. As the value of B decreases toward 0, the graph will stretch out horizontally more and more.
  4. Finally, apply a phase shift of -3pi. This is another way of saying shift the graph 3pi units to the left.

There is no vertical shifting, which is why D = 0.
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