# Thread: Transforming a cosine graph.

1. ## Transforming a cosine graph.

I am supposed to write an equation for a graph that shows the starting point at (0,-2) max at (3,0) next min at (just over 6, -2)
and it continues.

18c

So it says write a sine/cosine equation.
I wrote a sine, but the answer shows a cosine, I know it can be either but my question is

for my a value I get 1
my d value is -1
and my x = 3 because the first max is at 3.

it gives y=-cosx-1 ?
so how did they get that?
I have a =+ but If you have a -a does your max become a min?

2. The amplitude is half the difference between the max and the min. Your max is 0, your min is -2, so the amplitude is 1. We're going to transform the standard cosine graph, so we have to keep in mind that the standard cosine graph "starts" (at x=0) with a maximum. Your graph "starts" with a minimum, so we have to flip the graph by making the amplitude negative (this is exactly what you were asking about when "a", the amplitude, is negative--doing this makes a max a min and vice versa). Now shift the graph down so that the maximum is 0--do this by making d=-1 (because otherwise the maximum is 1, so you need to drop the graph by 1 unit). This will give you y=-cos(x)-1.

To see the relation between this and the transformed sine graph version, you need only to know that sin(x)=cos(pi/2 -x) (or 90-x if you like degree measures).

Oh, and x is a variable--it doesn't take on any one specific value like a and d can. To generate the graph, x ranges over all real numbers!

3. so couldnt it be a = + and horizontally shifted 3 right also?

4. don't make it more difficult than it really is ... look at the graph again and think in terms of transformations.

the period is a bit over 6 ... $\displaystyle (2\pi)$ ?

looks like an inverted cosine graph ... $\displaystyle -\cos(x)$ ?

graph is shifted down 1 unit ... $\displaystyle (-1)$ vertical shift ?

$\displaystyle y = -\cos(x) - 1$