Hello unstopabl3 Originally Posted by

**unstopabl3** Hi, I know how to draw simple graphs like sinx, cosx, sin2x, cos2x ... etc.

But I am stuck at the following:

1) 2-2cos3x

2) 3-4cos2x (for 0-180)

3) 2-2cosx+2

Could someone tell me the easiest way to draw these on a graph paper?

I know that 3x or 2x in the above means the # of cycles in a 360 range. But I am confused about what to do when 3-4 or 1-2 is added in-front of the functions like in the above examples.

OK. Here's how you do #1.

First, when the graph of $\displaystyle y= \cos3x$ is changed into $\displaystyle y=2\cos3x$, all the $\displaystyle y$-values are multiplied by $\displaystyle 2$. In other words, the graph is *stretched *up and down by a factor of $\displaystyle 2$, from the $\displaystyle x$-axis. Instead of taking values from $\displaystyle -1$ to $\displaystyle 1$, therefore, it will take values from $\displaystyle -2$ to $\displaystyle +2$.

Next, when $\displaystyle y = 2\cos3x$ is changed into $\displaystyle y = -2\cos3x$, all the $\displaystyle y$-values have their sign changed. In other words, the graph is *reflected *in the $\displaystyle x$-axis. So, for example, when $\displaystyle x = 0,\; y = -2$ instead of $\displaystyle +2$.

Finally, changing $\displaystyle y = -2\cos3x$ into $\displaystyle y = 2-2\cos3x$ adds $\displaystyle 2$ to all the $\displaystyle y$-values. In other words, the graph is *shifted* (translated) upwards by $\displaystyle 2$ units.

So:

- Can you sketch the graph for #1 now?

. - Can you do #2 in the same way?

***

I assume that for #3 you mean

$\displaystyle y = 2-2\cos(x+2)$

Clearly, this is related to the graph of

$\displaystyle y = 2-2\cos x$

which you can sketch using the same techniques as above. Then, you need to learn that when $\displaystyle x$ is changed into $\displaystyle (x+2)$, the graph is translated $\displaystyle 2$ units to the *left. *

Yes, it is to the left, not the right, because when $\displaystyle x = 0$, the value of $\displaystyle (x+2)$ is $\displaystyle 2$. So the new graph *starts *at the point where $\displaystyle x = 2$ on the original graph. So it is exactly the same as the original, but it's $\displaystyle 2$ units to the left. (You may need to think a bit about that!)

Can you complete #3 now?

Grandad