# How do you translate y=2sin (4x-pi/3) from the parent function?

Dec 1, 2015

Start from $\sin \left(x\right)$. We did the following transformations:

1. $\sin \left(x\right) \setminus \to \sin \left(4 x\right)$
2. $\sin \left(4 x\right) \setminus \to \sin \left(4 x - \frac{\pi}{3}\right)$
3. $\sin \left(4 x - \frac{\pi}{3}\right) \setminus \to 2 \sin \left(4 x - \frac{\pi}{3}\right)$

Let's see how these changes affect the graph:

1. When we change $f \left(x\right) \setminus \to f \left(k x\right)$, we change the "speed" with which the $x$ variable runs. This means that, if $k$ is positive, the $x$ values arrive earlier. For istance, if $k = 4$, we have $f \left(4\right)$ when $x = 4$, of course. But when computing $f \left(4 x\right)$, we have $f \left(4\right)$ for $x = 1$. This means that $\sin \left(4 x\right)$ is a horizontally compressed version of $\sin \left(x\right)$. Here's the graphs .

2. When we change from $f \left(x\right)$ to $f \left(x + k\right)$, we are translating horizontally the function, and the reasons are similar to those in the first point. Is $k$ is positive, the function is shifted to the left, if $k$ is negative to the right. So, in this case, the function is shifted to the right by $\frac{\pi}{3}$ units. Here's the graphs

3. When we change from $f \left(x\right)$ to $k \cdot f \left(x\right)$, we simply multiply every point in the graph by $k$, resulting in a vertical stretch (expanding if $k > 0$ or contracting if $k < 0$). Here's the graphs.