This question seems simple but I can't wrap my head around it 
Determine the two points on the graph $y = x^4 - 2x^2 - x$ which share a common tangent.

I tried plotting the graph out but I couldn't really wrap my head around how $2$ points can have the same tangent. I could be missing something completely obvious but I would really appreciate if someone could solve it and walk me through it.
 A: Let the common tangent be $y=kx+m$, whose intersection points with the curve are given by,
$$x^4-2x^2-(1+k)x-m=(x^2-1)^2-(1+k)x-(m+1)=0$$
Given that there are two points that share the tangent line, the above equation has to have exactly two roots, which requires that it has the form of a complete square, i.e. $k=m=-1$.
Thus, the two points are $(1,-2)$ and $(-1,0)$, with the tangent line $y=-x-1$.
A: It's a trick. We don't need calculus, just "incidence."
If we begin with $$   y = (x^2 - 1)^2 = x^4 - 2 x^2 + 1 $$
we see that the $x$ axis is tangent to the graph twice, at $x = \pm 1$
Next, we will subtract off a linear function, namely $\color{red}{x+1},$ as your function is
$$ (x^2 - 1)^2 - x - 1 = x^4 - 2 x^2 - x $$
Ignoring questions of slope, the line $y = 0 - x - 1$ or
$$ y = -x - 1 $$
intersects your graph exactly twice. Therefore it really is tangent twice.


One of the comments suggests looking at the (repeat) tangents to $y = \sin x,$ namely $y = 1$ and $y = -1.$  Here we use the same trick to find lines that are tangent, infinitely often, to $$y = \frac{x}{2} + \sin x $$

A: I just made a guess ( guided by the idea described in the comment by @Izaak van Dongen): the derivative is of course
$$y'=4x^3-4x-1$$
and $y'=-1$ is easy to solve: $x=0$ or $x=1$ or $x=-1$ and the latter two correspond to tangents with slope $-1$ to the points $(1|-2)$ and $(-1|0)$ and both tangents turn out to be one and the same with the equation:
$$y=-x-1$$
which is easy to check.
A: For the equation $\hat y=k\hat x+b$ of tangent at point $x$ we know that:
$$
k=4x^3-4x^2-1,\qquad\text{and}\qquad b=-3x^4+2x^2
$$
Let those points have coordinate $x=p$ and $x=q$, then
$$
4p^3-4p-1=4q^3-4q-1,\\
-3p^4+2p^2 = -3q^4+2q^2
$$
Rearragning, we get the following:
$$
(p-q)(p+q)(3p^2+3q^2+2)=0,\\
(p-q)(p^2+pq+q^2-1)=0
$$
The rest I leave for you.
