Posted by **Jill ** on Friday, March 5, 2010 at 11:48am.

Determine the equation of the graph in the form y = acos(bx).

I will try and describe the Graph.

-It starts at the origin (0,0)

-It has a minimum value of -2

-It has a maximum value of 2.

-It has an x-intercept of 90 degrees.

-It starts moving downward below the x-axis, and then curves back upwars, and continues in this pattern.

My answer:

I think "a" should= 2.5, but I do not not what "b" should be. Whatever I put in, I can not get the graph to start at the origin. It always starts above or below y=0.

Could someone please help me?

Thanks.

- Math -
**Reiny**, Friday, March 5, 2010 at 2:03pm
A cosine curve without a phase shift cannot start at (0,0)

The curve that would fit your description would have to be

**y = 2cos 2(x + 45 degrees)**

Check:

(0,0) lies on it.

when x = 90, y = 2 cos 270 = 0

so there is an x-intercept at 90

when x=45, y = 2 cos 180 = -2

when x = 135, y = 2 cos 360 = +2

the period of the curve is 180 , shown by b=2

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