# Thread: X-intercepts of a graph

1. ## X-intercepts of a graph

Hi all!

Ive been having some trouble with the same types of questions over and over again. I cannot seem to get them right..

I have trouble finding the x-intercepts, after a series of transformations to the original graph.

Here is an example:

y = 2-3cos 2(θ - π/2)

I know that:
- amplitude: 3
- translation of π/2 units in the positive direction of the x-axis
- dilation of factor 1/2 from the y-axis
- translation of 2 units in the positive direction of the y-axis

I understand how it has been transformed from the original cos graph, but i can't get the x-intercepts after all the transformations. How do i start it? Can anyone give me a good method??

Anyones help would be greatly appreciated.

2. Originally Posted by steph_r
Hi all!

Ive been having some trouble with the same types of questions over and over again. I cannot seem to get them right..

I have trouble finding the x-intercepts, after a series of transformations to the original graph.

Here is an example:

y = 2-3cos 2(θ - π/2)

I know that:
- amplitude: 3
- translation of π/2 units in the positive direction of the x-axis
- dilation of factor 1/2 from the y-axis
- translation of 2 units in the positive direction of the y-axis

I understand how it has been transformed from the original cos graph, but i can't get the x-intercepts after all the transformations. How do i start it? Can anyone give me a good method??

Anyones help would be greatly appreciated.
For that example you could use $\displaystyle cos(A+B) = cosAcosB - sinAsinB$ to reduce $\displaystyle cos(\theta - \frac{\pi}{2})$ to $\displaystyle -sin(\theta)$ and solve that way.

The phase shift means each point will cross the x axis $\displaystyle \frac{\pi}{2}$ 'later' so your original intercepts would now be at $\displaystyle k + \frac{\pi}{2}$ rather than at $\displaystyle k$

Applying the dilation factor means each point is at half it's old value (it is "squashed" so to speak). This means your points which were at $\displaystyle k+\frac{\pi}{2}$ would be at $\displaystyle \frac{1}{2}(k+\frac{\pi}{2})$

3. Originally Posted by e^(i*pi)
For that example you could use $\displaystyle cos(A+B) = cosAcosB - sinAsinB$ to reduce $\displaystyle cos(\theta - \frac{\pi}{2})$ to $\displaystyle -sin(\theta)$ and solve that way.

The phase shift means each point will cross the x axis $\displaystyle \frac{\pi}{2}$ 'later' so your original intercepts would now be at $\displaystyle k + \frac{\pi}{2}$ rather than at $\displaystyle k$

Applying the dilation factor means each point is at half it's old value (it is "squashed" so to speak). This means your points which were at $\displaystyle k+\frac{\pi}{2}$ would be at $\displaystyle \frac{1}{2}(k+\frac{\pi}{2})$
Thank you i'll give that a try, and i hope if i have any questions u can help me out thank you

4. Originally Posted by steph_r
Thank you i'll give that a try, and i hope if i have any questions u can help me out thank you
All right, I'll do what I can before I turn into bed XD