A graph of a function, y = sin (x), can be shifted left/right, shifted up/down, squeezed, stretched, or reflected. Think of y = sin (x) as your master (or base) function. Then y = A sin (Bx + C) + CDis your transformed function. Here's what the transformations do:
A, which can be positive or negative, squeezes or stretches the function vertically. It changes the sine function's amplitude - the distance from the mid line to the top (or the bottom). When A is negative, the sine function is reflected and instead of going midline - top - midline - bottom - midline, it goes to the bottom first.
B squeezes or stretches the function horizontally. It changes the period - how long it takes for the sine function to go from midline to top to midline to bottom and back to midline again. Frequency or cycle are often used as well, especially as units of the sine curve against time.
The new period of the function will be 2Ï€ divided by the coefficient out front. The new frequency will be the coefficient in front of the x. This is because frequency and period are reciprocals.
In the case of both A and B, when A and B are more than 1, the graph stretches and when it's less than 1, it squeezes.
C is a horizontal shift. When C is negative the graph goes to the right and when C is positive the graph goes to the left.
D is a vertical shift. When D is negative we move the graph down and when D is positive we move the graph up.
Now it's possible to put all these changes together at once. When that happens, simply attack them one at a time. Use y = sin (x) with an amplitude of 1, midline of x = 0, period of 2Ï€ as your template. You can either look at the functions or look at the squeezes/stretches to do this problem. I'll explain it given the four choices below the functions.
Choice 1: amplitude 1, frequency 4, midline at 2.
Our base function is y = sin (x). We now apply the shifts to create something in the form y = A sin (Bx + C) + D. We want an amplitude of 1, so we make A = 1. We want a frequency of 4, so B is 4. The midline represents where the new place of the vertical shift will be, so D = 2.
y = 1 sin (4x + C) + 2
r(x) is sin (4x + 8) + 2 which matches our template.
Choice 2: amplitude of 4, frequency of 2, midline of 8. Again we use y = A sin (Bx + C) + D as the template. A is for amplitude, B for frequency, D for midline. We put them into the template and get y = 4 sin (2x + C) + 8. q(x) has this and also shifts the graph π units. Thus q(x) is our match.
Choice 3: amplitude of 2, frequency of 8, midline of 1. Again we use y = A sin (Bx + C) + D as the template. We substitute values of A for amplitude, B for frequency, D for midline. That gives y = 2 sin (8x + C) + 1. g(x) has those pieces and a horizontal shift too.
Choice 4: Amplitude of 8, frequency of π, midline of 4. Again we use y = A sin (Bx + C) + D and substitute for A, B, and D as before. That gives us y = 8 sin (πx + C) + 4 and p(x) has those pieces.
Thus, Choice 1 is r(x), Choice 2 is q(x), Choice 3 is g(x), Choice 4 is p(x).
