Question

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

Answer 1

Hint: We have with us the equation \[y=\sin x\]. We will go step wise here. Firstly, we have our position shifted towards the right by a distance of \[\dfrac{\pi }{4}\], that means we will have the graph with the equation as \[y=\sin \left( x-\dfrac{\pi }{4} \right)\]. 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, \[y=\sin x\].
Now, we are given certain statements which we have to write in the equation form.
Firstly, we have the equation of the line as,
\[y=\sin x\]-----(1)
Now, the question says that we have shifted the equation to the right by a distance of \[\dfrac{\pi }{4}\]. So, we will subtract x by that many factor, that is, we will have,
\[y=\sin \left( x-\dfrac{\pi }{4} \right)\]-----(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,
\[y=-\sin \left( x-\dfrac{\pi }{4} \right)\]----(3)
Next, we have to translate one step or one unit upwards, so we get the function as,
\[y=-\sin \left( x-\dfrac{\pi }{4} \right)+1\]

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

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