The function y=sinx has been transformed. It now has amplitude of 5.0, a period of 26, a phase shift of 2 units to the right, a vertical translation of 6.5 units down, and is reflected over the x-axis. Given that (π/6,1/2) is a point in the parent function, use mapping notation to determine the x-coordinate of its image point in the transformed function.
Enter the numerical value of the x-coordinate only in the box below rounded to two decimals.
Start with y = sinx
It now has amplitude of 5.0 : y = sinx ----> y = 5sinx
a period of 26: y = 5sinx ----> y = 5 sin (πx/13)
a phase shift of 2 units to the right: y = 5sin (π/13x) ---> y = 5sin (π/13(x-2))
a vertical translation of 6.5 units down: y = 5sin (π/13(x-2)) ---> y = 5sin (π/13(x-2)) - 6.5
reflected over the x-axis: y = 5sin (π/13(x-2)) - 6.5 ---->y = -5sin (π/13(x-2)) + 6.5
To find the x-coordinate of the image point in the transformed function, we need to apply the given transformations step-by-step to the x-coordinate of the point (π/6, 1/2) in the parent function.
Step 1: Phase Shift
Given a phase shift of 2 units to the right, we subtract 2 from the x-coordinate of the point.
New x-coordinate for the point: π/6 - 2 = -7π/6
Step 2: Period
Given a period of 26, the point should be scaled to fit within this new period. The period affects the x-coordinate, not the amplitude.
We divide the new x-coordinate by the current period and take the remainder.
Remainder: (-7π/6) mod 26
Calculating the remainder: -7π/6 ≈ -7.81
Remainder: (-7.81) mod 26 ≈ -7.81
Thus, the new x-coordinate remains at approximately -7.81.
Step 3: Reflection over the x-axis
Given that the function is reflected over the x-axis, we change the sign of the x-coordinate.
New x-coordinate: -(-7.81) = 7.81
Therefore, the x-coordinate of the image point in the transformed function is approximately 7.81.
To find the x-coordinate of the image point in the transformed function, we need to apply each transformation to the x-coordinate of the given point in the parent function.
Let's start with the given point (π/6, 1/2). The x-coordinate is π/6.
1. Phase Shift: The function has a phase shift of 2 units to the right. To adjust the x-coordinate for the phase shift, we subtract the phase shift value:
π/6 - 2 = -11π/6
2. Reflect over the x-axis: The function is reflected over the x-axis, meaning the y-coordinate will change its sign, but the x-coordinate remains the same. Therefore, the x-coordinate remains -11π/6.
So, the x-coordinate of the image point in the transformed function is -11π/6.
However, since the question asks for the numerical value rounded to two decimal places, we need to compute the approximate decimal value of -11π/6.
Using the approximate value of π as 3.14, we can calculate:
-11π/6 ≈ -11 * 3.14 / 6 = -5.73
Therefore, the numerical value of the x-coordinate (rounded to two decimal places) of the image point in the transformed function is -5.73.