1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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... ? $\endgroup$
    – amd
    Apr 20 '16 at 8:42
  • 1
    $\begingroup$ 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? $\endgroup$
    – almagest
    Apr 20 '16 at 8:42
2
$\begingroup$

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:$$x_0^3 - 2{x_0}x_0^2 + b{x_0} = - x_0^3 + b{x_0}.$$

A dynamic sketch of the situation can be seen at Geogebra:

Find All Parameters

$\endgroup$
1
  • $\begingroup$ @user328032 The GeoGebra example is terrific. $\endgroup$
    – dantopa
    Mar 22 '17 at 0:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.