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Math Help - tangent line equations

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    tangent line equations

    Find an equation of the line that is tangent to the graph of f(x)=x^3 and parallel to the line: 12x-y+5=0
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    Quote Originally Posted by phalange View Post
    Find an equation of the line that is tangent to the graph of f(x)=x^3 and parallel to the line: 12x-y+5=0
    Rearrange the second equation into y=mx+c to find the gradient

    y = 12x+5 therefore the gradient is 12

    If the tangent [ie: f'(x)] is parallel to this line it must be equal.

    Find f'(x) and set it equal to 12 then get x.

    Spoiler:
    f'(x) = 3x^2 = 12
    x = \pm 2


    Put your value(s) of x into f(x) to get a co-ordinate.

    Spoiler:
    f(-2)=  -8 \: , \: f(2) = 8


    You can now use the equation of a straight line y-y_1 = m(x-x_1)

    To my knowledge I am not sure if you can get an explicit solution
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