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Math Help - Find the equation of the tangent line

  1. #1
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    Find the equation of the tangent line

    Find the equation of the tangent line to the graph of the function at z = 3.

    f(z)= (4z^2)/(5z^2+2z)

    I found the slope to be (8/289)

    Do I just plug in 3 to the original equation, and use whatever y I get to be...

    y-(whatever I get)=(8/289)(x-3)?

    Thanks,

    -Tyler
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  2. #2
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    equation of line

    f(z)=(4z^2)/(5z^2+2z)
    f'(z)=8/(25*(z^2)+(20*z)+4) (through use of partial fractions)
    f'(3)=8/(25*(3^2)+(20*3)+4)
    f'(3)=8/289
    then find the y coordinate to accompany x

    y=(4z^2)/(5z^2+2z)
    y=(4*(3^2)/(5*(3^2)+(2*3))
    y=0.71(approximation)
    giving the coordinate (3,0.71)
    then use y-b=m(x-a)
    y-0.71=8/289(x-3)
    giving y=(8/289)x-0.97966
    rounding to y=(8/289)x-1
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