Question 1158754
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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 = {{{-3pi}}} is the phase shift
D = 0 is the midline


Plug those A,B,C,D values into the equation below. Simplify.
{{{y = A*sin(B(x-C)) + D}}}


{{{y = -5*sin(expr(1/3)(x-(-3pi))) + 0}}}


{{{y = -5*sin(expr(1/3)(x+3pi))}}}


{{{y = -5*sin(expr(1/3)x+expr(1/3)*3pi)}}}


{{{y = -5*sin(expr(1/3)x+pi)}}} <font color=red>is the final answer</font>


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If you want to describe this as a series of transformations, then you would start with the parent function {{{y = sin(x)}}} and follow the steps below
<ol><li>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.</li><li>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.</li><li>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.</li><li>Finally, apply a phase shift of -3pi. This is another way of saying shift the graph 3pi units to the left.</li></ol>
There is no vertical shifting, which is why D = 0.
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