
New Member
Find the equation of a tangent: f(x)=x^2+3x+2
Hi.
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?

Elite Member
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 nonreal values.)
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