# Find all parameters $a$ , $b$ such that a polynomial $f ( x ) = x ^ 3 + ax ^ 2 + bx$ has exactly one tangent that passes through the origin .

Determine all parameters $a$ and $b$ such that a polynomial given by $f ( x ) = x ^ 3 + ax ^ 2 + bx$ has exactly one tangent that passes through the origin .
My work:
Let use equation for tangent line:
$y-y_0=y'(x_0)(x-x_0)$. We know that $x=y=0$ because tangent line pass through $(0,0)$(origin).
$y_0=y'(x_0)*x_0$. Next we need to find derivative of $y$.
$y'(x_0)=3x_0^2+2ax_0+b$
Our eqation of tangent line now look like this.
$$y_0=(3x_0^2+2ax_0+b)x_0$$ $$x_0^3+ax_0^2+bx_0=3x_0^3+2ax_0^2+bx_0$$ $$2x_0^3+ax_0^2=0$$ $$x_0^2(2x_0+a)=0$$ Or $x_0=0$ or $a=-2x_0$
But I'am not sure how to finish my work.

• Since $f(0)=0$, you know that the tangent at $x=0$ passes through the origin. You need to make sure that none others do. You’ve derived a condition for the tangent at $(x_0,f(x_0))$ to pass through the origin so... ?
– amd
Apr 20 '16 at 8:42
• You have shown that the tangent at $x=0$ passes through the origin iff $x_0=-\frac{a}{2}$ or $x=0$. So what does that tell you about the number of tangents that pass through the origin? Apr 20 '16 at 8:42

The given function always has a tangent at ${x_0} = 0$. Since we can choose any ${x_0}$ the point on f, (${x_0},f({x_0}$) is also a point on a tangent through the origin when $a = - 2{x_0}$.
The ordinate of this point, for any value, $b$, is given by:$$x_0^3 - 2{x_0}x_0^2 + b{x_0} = - x_0^3 + b{x_0}.$$