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Question: How do I find the equation for this curve in its final position? The graph of \[y=\sin x\] is shifte...

How do I find the equation for this curve in its final position? The graph of y=sinxy=\sin x is shifted a distance of π4\dfrac{\pi }{4} to the right, reflected in the X-axis, then translated one unit upward.

Explanation

Solution

We have with us the equation y=sinxy=\sin x. We will go step wise here. Firstly, we have our position shifted towards the right by a distance of π4\dfrac{\pi }{4}, that means we will have the graph with the equation as y=sin(xπ4)y=\sin \left( x-\dfrac{\pi }{4} \right). Now, it is reflected in the X-axis, that is, now the graph gets shifted to the negative Y-axis direction. Now, when we translate one unit upward, we will add positive 1 to the equation we get. Hence, we will have the equation for the given curve.

Complete step by step answer:
According to the question given to us, we are given an equation of the graph which is, y=sinxy=\sin x.
Now, we are given certain statements which we have to write in the equation form.
Firstly, we have the equation of the line as,
y=sinxy=\sin x-----(1)
Now, the question says that we have shifted the equation to the right by a distance of π4\dfrac{\pi }{4}. So, we will subtract x by that many factor, that is, we will have,
y=sin(xπ4)y=\sin \left( x-\dfrac{\pi }{4} \right)-----(2)
Next, we have reflected the given function in the X-axis, so we get the function in the negative Y-axis and so we will multiply the function by -1, and we get the function as,
y=sin(xπ4)y=-\sin \left( x-\dfrac{\pi }{4} \right)----(3)
Next, we have to translate one step or one unit upwards, so we get the function as,
y=sin(xπ4)+1y=-\sin \left( x-\dfrac{\pi }{4} \right)+1

Therefore, the equation of the curve in the final position is y=sin(xπ4)+1y=-\sin \left( x-\dfrac{\pi }{4} \right)+1.

Note: : In the equation (2), we have moved towards to the right direction yet still we wrote xx as xπ4x-\dfrac{\pi }{4}, this is because when we translated the function in the right direction we have already added bits to the xx, so in order to plug in a 0, we will naturally have to subtract the bits that we had added.