Graphing Sinusoidal Functions Part 2

Question

I can't seem to figure the equation for the graph also in BLUE. Where do I start? I Know the amplitude is 2, but how do I get the period to find the rest of the equation?

Answer 1

Using the general equation

$$ y=A\sin\left(\frac{2\pi}{p}(x-h)\right)+k\tag{1} $$

we see that the vertical distance between the high and low points is $4$ which is twice the amplitude $A$, therefore $A=2$. The first peak is at $x=\frac{\pi}{4}$ and the second peak is at $x=\frac{9\pi}{4}$ so the period is $p=\frac{9\pi}{4}-\frac{\pi}{4}=\frac{8\pi}{4}=2\pi$. There is no vertical shift so $k=0$, but we see that instead of passing through the origin, the sine graph passes through $\left(-\frac{\pi}{4},0\right)$, so it has been horizontally shifted by $h=-\frac{\pi}{4}$. Substituting these values into the equation give a result of

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

Answer 2

The amplitude is $2$.

the period is $2\pi$, thus the equation is

$$y=f(x)=2\sin(\frac{2\pi}{2\pi}x+\phi)$$

$$=2\sin(x+\phi)$$

but $f(-\frac{\pi}{4})=0$, thus $\phi=\frac{\pi}{4}$ and finally,

$$f(x)=2\sin(x+\frac{\pi}{4}).$$