# Thread: translations of the graphs of sine and cosine functions

1. ## translations of the graphs of sine and cosine functions

I have a test tmrw and just went over the information today so I am a little confused. Some of the graphs I get right but some of them I am getting the wrong X values.

So if i have y=-4 sin 2(x-pi/2)
so the Amplitude is 4
and the period is pi
but how in the world do i get the x axis?

I've tried everything and i'll find one way works for one type but doesn't work for another. Is it really that way? I was told to multiply pi/2 by 1/4 2/4 and 3/4 but it doesnt come out correct. Does anyone know what to add or multiply by to get the x axis? Thanks

2. Originally Posted by KoolFlair

so the Amplitude is 4
correct

Originally Posted by KoolFlair

and the period is pi
correct

Originally Posted by KoolFlair

So if i have y=-4 sin 2(x-pi/2)
but how in the world do i get the x axis?
find the x axis intercepts by making y = 0

$y=-4 \sin 2(x-\frac{\pi}{2})$

$0=-4 \sin 2(x-\frac{\pi}{2})$

$0= \sin 2(x-\frac{\pi}{2})$

can you solve from here?

3. Originally Posted by KoolFlair
I have a test tmrw and just went over the information today so I am a little confused. Some of the graphs I get right but some of them I am getting the wrong X values.

So if i have y=-4 sin 2(x-pi/2)
so the Amplitude is 4
and the period is pi
but how in the world do i get the x axis?
what do you mean by "getting the x-axis"?

I've tried everything and i'll find one way works for one type but doesn't work for another. Is it really that way? I was told to multiply pi/2 by 1/4 2/4 and 3/4 but it doesnt come out correct. Does anyone know what to add or multiply by to get the x axis? Thanks
...

4. im not sure how to finish it that way.

I was told to make an inequality. So i got

pi/2<x<9pi/2

And pi/2 starts the x
and
9pi/2 ends as the 4th so i need to find the 3 in between. Do i add pi/2 to each one?

5. Originally Posted by skeeter
...

like if you set up and x/y tree. I need to get the X then take the X and insert it into the cos x

6. Originally Posted by KoolFlair
im not sure how to finish it that way.

I was told to make an inequality. So i got

pi/2<x<9pi/2

And pi/2 starts the x
and
9pi/2 ends as the 4th so i need to find the 3 in between. Do i add pi/2 to each one?
The function

$y=-4 \sin 2(x-\frac{\pi}{2})$

has period of $\pi$ with x-intercepts at $0+\frac{\pi}{2},\frac{\pi}{2}+\frac{\pi}{2}$ and $\pi+\frac{\pi}{2}$

This gives all the solutions on $[\frac{\pi}{2},\frac{3\pi}{2}]$

If you need solutions on $[\frac{\pi}{2},\frac{9\pi}{2}]$

keep adding $\pi$ to the solutions above

7. Thanks pickslides
Thats what i thought.

But my solutions manuel says it should be

Pi/2
11pi/4
5pi
19pi/4
9pi/2

So how are they getting 11pi/4?

8. You sure about that? I graphed the function and I'm getting x-intercepts at $0,\; \frac{\pi}{2},\; \pi,\; \frac{3\pi}{2},...$.

If your original function is
$y = -4{\color{red}\cos} \left(2\left(x - \frac{\pi}{2}\right)\right)$
then the x-intercepts occur at
$\frac{\pi}{4},\; \frac{3\pi}{4},\; \frac{5\pi}{4},\; \frac{7\pi}{4},...$
but these answer still don't match yours.

01