# How do you graph g(x) = -2sin(pix+pi/3)?

Jul 28, 2015

#### Answer:

There are several steps.

#### Explanation:

$y = A \sin \left(B x + C\right)$

(or, in some treatments $y = A \sin \left(B x - C\right)$ for the method advocated in this answer, it doesn't matter.)

We know that sine takes on values between $- 1$ and $1$, so when we multiply after finding sine that will change the range of values.

The amplitude reflects this fact and is equal to $\left\mid A \right\mid$

For the basic sine graph, $y = \sin x$, we often think of this as starting the first cycle when we take the sine of 0 and finishing when we take the sine of $2 \pi$. The period of the basic function is $2 \pi$.

The period of $y = A \sin \left(B x + C\right)$ is $\frac{2 \pi}{B} \text{ }$ (Multiplying by $B$ changes the scale on the $x$ axis.)

Phase shift (or Horizontal Shift) tells us where we "start the first period" it tells us when we take the sine of 0.

To find the shift, solve $B x + C = 0$
(Or $B x - C = 0$ depandng on your textbook.)

So here we go:

For $g \left(x\right) = - 2 \sin \left(\pi x + \frac{\pi}{3}\right)$,

We have Amplitude = $2 \text{ }$
(The minus sign will flip the graph across the $x$ axis.)

Period is $\frac{2 \pi}{\pi} = 2$

Phase Shift is the solution to $\pi x + \frac{\pi}{3} = 0$,

it is $- \frac{1}{3}$

The graph starts on the $x$ axis at $x = - \frac{1}{3}$, then decreases to a minimum of $- 2$, back up to and through the $x$ axis, up to a maximum of $2$, before returning to the $x$ axis. Then repeat the pattern as required.

Here is one period:

graph{y = -2sin(pix+pi/3) *(sqrt(1-(x-2/3)^2))/(sqrt(1-(x-2/3)^2)) [-4.177, 5.69, -2.59, 2.343]

Note

Another way of finding the period is to find the beginning of the first cycle (phase shift) by solving $B x + C = 0$ as before, and the find the end of the cycle by solving $B x + C = 2 \pi$

The period is the difference between the end and the beginning. If you go through the algebra, you'll see that the difference is $\frac{2 \pi}{B}$