Help me plss, i need this done by today

with the template above in mind, let's check the graph of that function, it has a midline at y = 0, namely the x-axis, so it has the same midline as sine, we'll use sine, because sine graph starts at 0 and then it moves up and then down, this one starts at 0, reason why, I guess we could also use a -cosine, which has the same effect, anyhow, let's use sine.

from the midlne up, we have a peak at 1.5 or 3/2 which our Amplitude.

we can also see that from 0 to 8, is the graph, after that it simply repeats, so we can say that the period of that one is 8, thus

and since the midline is same as sin(θ) that means it doesn't have a vertical shift, so our D = 0, and since it's beginning at x = 0, it doesn't have any horizontal shift from the sin(θ) either, so C = 0, thus we can write it as

Check the picture below.

c is the answer

step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.