# How do you find the x intercepts of y=sin((pix)/2)+1?

Mar 17, 2017

$x$-intercepts for $y = \sin \left(\frac{\pi x}{2}\right) + 1$ are at $\left\{\ldots \ldots . , - 5 , - 1 , 3 , 7 , 11 , \ldots \ldots .\right\}$

#### Explanation:

To determine $x$-intercepts of any function $y = f \left(x\right)$, we put $y = 0$

(and we put $x = 0$ to find $y$-intercepts).

Hence $x$-intercepts are given by $f \left(x\right) = 0$

and here $\sin \left(\frac{\pi x}{2}\right) + 1 = 0$

or $\sin \left(\frac{\pi x}{2}\right) = - 1 = \sin \left(\frac{3 \pi}{2}\right)$

Therefore $\frac{\pi x}{2} = 2 n \pi + \left(\frac{3 \pi}{2}\right)$, where $n$ is an integer.

This happens as in the domain $0 < x < 2 \pi$, only for $\sin \left(\frac{3 \pi}{2}\right) = - 1$ and sine ratio has a cycle of $2 \pi$.

Now as $\frac{\pi x}{2} = 2 n \pi + \left(\frac{3 \pi}{2}\right)$, and dividing by $\pi$ we have

$\frac{x}{2} = 2 n + \frac{3}{2}$ and $x = 4 n + 3$

Hence, we have $x$-intercepts for $y = \sin \left(\frac{\pi x}{2}\right) + 1$

at $\left\{\ldots \ldots . , - 5 , - 1 , 3 , 7 , 11 , \ldots \ldots .\right\}$
graph{sin((pix)/2)+1 [-10, 10, -5, 5]}