# How do you graph and list the amplitude, period, phase shift for y=3cos4x?

Jul 11, 2016

The cosine function's general form is

$y = A \cos \left(k x + \phi\right)$

and the basic form we all know and love has $A = 1 , k = 1 \mathmr{and} \phi = 0$

If we focus on functions with $\phi = 0$ first and adjust A, we see that at $x = 0$, $y = A$ because:

$y \left(0\right) = A \cos \left(0\right) = A$

Hence A adjusts the amplitude of the cosine function.

Now, lets look at what k does.

If we were to put $x = \frac{\pi}{2}$ into the basic cosine, we would obtain zero.

More generally, we get $y \left(\frac{\pi}{2}\right) = A \cos \left(\frac{k \pi}{2}\right)$

as we increase k to some value $k > 1$, we obtain values that would be further along on the basic cosine function - we are decreasing the period. The opposite is also true, decreasing k to some value $k < 1$ lengthens the period.

Finally, lets look at the phase constant.

$y = A \cos \left(x + \phi\right)$

Lets say $\phi$ is some arbitrary number $0 \le \phi \le 2 \pi$. A positive phase means that a larger x value is going into the cosine function so it will produce a value that we would expect further to the right - positive phase shifts the graph to the left!

So, lets look at our function:

$y \left(x\right) = 3 \cos \left(4 x\right)$

We can see that $A = 3 , k = 4 \mathmr{and} \phi = 0$

We know that it will start where we expect since there is no phase shift. The amplitude is now 3, so the graph will be stretched in the y direction - ranging from $- 3 \le y \le 3$.

$k = 4 \implies$period is 4 times shorter than the basic function. So there will be 4 wavelengths in the period $\left[0 , 2 \pi\right)$.

graph{3cos(4x) [-2.7, 13.32, -3.57, 4.44]}

As you can see from the graph, this is true. The amplitude is 3 and there are 4 wavelengths between 0 and $2 \pi$