check the wording of your question.
A cosine curve of the type you stated does not have any asymptotes.
the answer you gave of -pi/8<x<3pi/8
represents one period of the curve.
change y=3cos(4x+pi/2)-3
to y=3cos4(x+pi/8)-3
notice that the curve y = 3cos4x has been shifted pi/8 to the left or -pi/8,
the period is 2pi/4 or pi/2
if we add pi/2 to -pi/8 we get 3pi/8
Is there a formula to find the correct asymptotes on y=3cos(4x+p/2)-3
p=pie
Now I know that answer is
-p/8<x<3p/8
How do I find the asymptotes?
2 answers
Those points are two of the points where y = 0 and the function is tangent to the x axis. There are an infinite number of them.
Make a table
x ______________ -pi/8, 0, pi/8, pi/4, 3pi/8
4 x + pi/2_______ 0, pi/2, pi, 3pi/2,2pi
cos (4x+pi/2)____ 1, 0, -1, 0, 1
3cos(4x+pi/2)____ 3, 0, -3, 0, 3
3cos(4x+pi/2)-3 _ 0, -3, -6, -3, 0
that is one complete cycle of the cosine function from y = 0 to y = 0 again. It goes on and on in both directions of course
Make a table
x ______________ -pi/8, 0, pi/8, pi/4, 3pi/8
4 x + pi/2_______ 0, pi/2, pi, 3pi/2,2pi
cos (4x+pi/2)____ 1, 0, -1, 0, 1
3cos(4x+pi/2)____ 3, 0, -3, 0, 3
3cos(4x+pi/2)-3 _ 0, -3, -6, -3, 0
that is one complete cycle of the cosine function from y = 0 to y = 0 again. It goes on and on in both directions of course
1 answer