Determine the value of K so the function have only one point of one horizontal point of tangency 
Consider the polynomial function  define by
$ f(x) = x^3 + x^2 + kx - 1 $
Determine the value of k so f has only on point of horizontal
tangency.

So I know that $   f^{'}(x) = 3x^2 + 2x + k  $
And I should consider  $   f^{'}(x) = 0 $ but how I find a k that makes the $   f^{'}(x) = 3x^2 + 2x + k  $   have only one root?
I also thought that if $ k = -x^2 $ then the function would have only one point of horizontal tangency. But this does not works because this not equal the answer.  ( Though I think it shouldwork too )
Btw the answer is  $ \frac{1}{3} $ And I still can't figure why.
 A: For $f(x) = x^3 + x^2 + kx - 1$ , when $\frac{df(x)}{dx}= 3x^2+2x+k$=0 has only one solution or its discriminant $\Delta = b^2-4ac=0$, you will  have only one horizontal tangent line. Now, a=3,b=2,c=k, solve  $2^2$-4 x 3 k=0, k=1/3.
A: Continuing where you left off:
We know that the horizontal tangent line is found when $$f'(x) = 3x^2 + 2x + k =0$$
So we need to find the value of $k$ which only gives us $1$ root for this polynomial. Let's us use the quadratic formula to solve for the roots.
$$x=\frac{-2\pm \sqrt{2^2-4(3)(k)}}{2(3)}\\
x=\frac{-2\pm \sqrt{4-4(3)(k)}}{6}\\
x=\frac{-2\pm \sqrt{4(1-3k)}}{6}\\
x=\frac{-2\pm 2\sqrt{1-3k}}{6}\\
x=\frac{-1\pm\sqrt{1-3k}}{3}$$
Now, the only way that $x$ can have one solution is if $$\sqrt{1-3k}=0\\1-3k=0\\ 3k=1\\ \therefore k=\frac{1}{3}$$
Plugging $k$ back in to solve for our single root of $x$, $$x=\frac{-1\pm\sqrt{1-3\left(\frac{1}{3}\right)}}{3}\\
x=\frac{-1\pm\sqrt{1-1}}{3}\\
x=\frac{-1}{3}$$
The only way $x$ can have a single root is if $k=\frac{1}{3}$.
