Question

f(x)=sin(πx4)

Use the sine tool to graph the function. The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.

Answer 1

Answer: Graph is below

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Explanation:

If we plugged in x = 0, then we get...

$y = \sin\left(\frac{\pi x}{4}\right)\\\\y = \sin\left(\frac{\pi*0}{4}\right)\\\\y = \sin\left(\frac{0}{4}\right)\\\\y = \sin\left(0\right)\\\\y = 0$

Therefore, this sine curve goes through the origin. More specifically, this point is on the midline y = 0, aka the x axis. This is point A in the diagram below, which I made with GeoGebra (free graphing software).

Now to find another point on this sine curve. We need either a min or max point. Note that the expression $\frac{\pi x}{4}$ is the same as $\frac{\pi}{4}x$. We will divide 2pi over the coefficient of pi/4 to get

(2pi) divide (pi/4) = 2pi * 4/pi = 8

This tells us the period of this sine curve is 8 units in the horizontal x axis direction. The distance from midline to the nearest peak or valley is 1/4 of this, so (1/4)*8 = 2 units. Move 2 units to the right to arrive at x = 2. Plug this into the original function to get the following:

$y = \sin\left(\frac{\pi x}{4}\right)\\\\y = \sin\left(\frac{\pi*2}{4}\right)\\\\y = \sin\left(\frac{\pi}{2}\right)\\\\y = 1$

This matches with the fact that sin(x) maxes out at 1. So we've reached a max or highest point here. The location of which is B(2,1).