# How do you graph y + 1 = 3 cos 4 (x-2)?

Dec 6, 2015

Write the equation as $y = 3 \cos \left(4 x - 8\right) + 1$. Now you're in the generic form

$A \cos \left(\omega x + \phi\right) + k$.

A function in this form has four important informations:

1. $A$ is the amplitude, which is the maximum value reached by the function. Of course, the standard amplitude is $1$, since $\cos \left(x\right)$ ranges between $- 1$ and $1$. And in fact, a function with amplitude $A$ ranges from $- A$ to $A$.

2. $\omega$ affects the period, because it changes the "speed" with which the function grows. Look at this example: if we have the standard function $\cos \left(x\right)$, if you want to go from $\cos \left(0\right)$ to $\cos \left(2 \setminus \pi\right)$, the variable $x$ must from from $0$ to $2 \setminus \pi$. Now try $\cos \left(2 x\right)$: in this case, if $x$ runs from $0$ to $\pi$, your function ranges from $\cos \left(0\right)$ to $\cos \left(2 \setminus \pi\right)$. So, we needed "half" the $x$ travel to cover a whole period. In general, the formula states that the period $T$ is $T = \frac{2 \setminus \pi}{\setminus} \omega$.

3. $\setminus \phi$ is a phase shift, and again look at this example: with the standard function $\cos \left(x\right)$, you have $\cos \left(0\right)$ for $x = 0$, of course. Now we try $\cos \left(x - 1\right)$. To have $\cos \left(0\right)$, we must input $x = 1$. So, the same value has been shifted ahead of $1$ unit. In general, if $\phi$ is positive, it shifts the function backwards (which means to the left on the $x$-axis) of $\phi$ units, and if $\phi$ is negative, the shift is to the right.

4. Finally, the $+ 1$ at the end is a vertical shift. Think of it like this: when you have $y = \cos \left(x\right)$, it means that you are associating with every $x$ the $y$ value "$\cos \left(x\right)$". Now, you change to $y = \cos \left(x\right) + 1$. This means that now you associate to the same old $x$ the new value $\cos \left(x\right) + 1$, which is one more than the old value. So, if you add one unit on the $y$ axis, you shift upwards. Of course, if $k$ is negative, the shift is downwards.

So, in the end, you start from the standard cosine function. Then, you do all the transformations: