document.write( "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! \n" ); document.write( "

Algebra.Com's Answer #781732 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "A = -5 since the amplitude is 5 and the function is a negative sine function.
\n" ); document.write( "Recall that |A| is the amplitude.\r
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\n" ); document.write( "\n" ); document.write( "T = 6pi is the period
\n" ); document.write( "B = 2pi/T
\n" ); document.write( "B = 2pi/(6pi)
\n" ); document.write( "B = 2/6
\n" ); document.write( "B = 1/3 is the coefficient of the x term\r
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\n" ); document.write( "\n" ); document.write( "C = $\"-3pi\"$ is the phase shift
\n" ); document.write( "D = 0 is the midline\r
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\n" ); document.write( "\n" ); document.write( "Plug those A,B,C,D values into the equation below. Simplify.
\n" ); document.write( " $\"y+=+A%2Asin%28B%28x-C%29%29+%2B+D\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"y+=+-5%2Asin%28expr%281%2F3%29%28x-%28-3pi%29%29%29+%2B+0\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"y+=+-5%2Asin%28expr%281%2F3%29%28x%2B3pi%29%29\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"y+=+-5%2Asin%28expr%281%2F3%29x%2Bexpr%281%2F3%29%2A3pi%29\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"y+=+-5%2Asin%28expr%281%2F3%29x%2Bpi%29\"$ is the final answer\r
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\n" ); document.write( "\n" ); document.write( "-----------------------------------------------------\r
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\n" ); document.write( "\n" ); document.write( "If you want to describe this as a series of transformations, then you would start with the parent function $\"y+=+sin%28x%29\"$ and follow the steps below
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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.
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.
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.
Finally, apply a phase shift of -3pi. This is another way of saying shift the graph 3pi units to the left.

\n" ); document.write( "There is no vertical shifting, which is why D = 0.
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