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.
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?
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)
(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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