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Math Help - Finding the asymptote of a tangent...

  1. #1
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    Finding the asymptote of a tangent...

    I cant seem to figure it out. im asked to find a line that is an asymptote for y=4tan(5x)

    so I know that is = to y= (4sin(5x)/cos (5x) and I dont know where to go from there.
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  2. #2
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    Quote Originally Posted by ninobrn99 View Post
    I cant seem to figure it out. im asked to find a line that is an asymptote for y=4tan(5x)

    so I know that is = to y= (4sin(5x)/cos (5x) and I dont know where to go from there.
    Think about what values of x result in an undefined expression. Specifically, what values of x cause the denominator to be 0? (We can't divide by 0.)
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  3. #3
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    right, so I set cos(5x)=0
    =cos(3pi/2)
    =5x=(3pi/2)
    x=(3/10)PI

    I dont know how I got the (3pi/2) though.
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    Quote Originally Posted by ninobrn99 View Post
    right, so I set cos(5x)=0
    =cos(3pi/2)
    =5x=(3pi/2)
    x=(3/10)PI

    I dont know how I got the (3pi/2) though.
    Well you should be familiar with the shape of y=cos(x) in the Cartesian plane (you could also think in terms of triangles and the unit circle), and know that

    cos(0) = 1
    cos(pi/2) = cos(-pi/2) = 0
    cos(pi) = cos(-pi) = -1
    cos(2pi/3) = cos(-2pi/3) = 0
    cos(2pi) = cos(-2pi) = 1

    etc.

    So, your answer is correct, and another answer is given by

    5x = pi/2
    x = pi/10
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  5. #5
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    Thank you very much!!!!!
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