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Thread: Find the equation of a tangent: f(x)=-x^2+3x+2

  1. #1

    Find the equation of a tangent: f(x)=-x^2+3x+2


    I have a function called: f(x)=-x^2+3x+2

    The graph of the function have two tangent, which both go through (0,0). How can I find an equation for both tangents?

  2. #2
    Elite Member
    Join Date
    Jan 2012
    Any line that goes through (0, 0) is of the form y= ax for some number a. In order to be tangent to y= f(x), at [tex](x_0, f(x_0))[/tex] we must have [tex]ax_0= f(x_0)[/tex] and [tex]a= f'(x_0)[/tex]. Solve those two equations for [tex]x_0[/tex] and a.

    (Are you sure you have copied the problem correctly? y= -x^2+ 3x+ 2 is a parabola with vertex at (3/2, 17/4) opening downward. The origin is inside the parabola. No lines through the origin can be tangent to this parabola. If you do what I suggest above you will get non-real values.)

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