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

1. ## 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?

2. 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 $(x_0, f(x_0))$ we must have $ax_0= f(x_0)$ and $a= f'(x_0)$. Solve those two equations for $x_0$ 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.)