Can one find a line that is tangent to a cubic polynomial more than once? I know that any line cannot be tangent to the graph of $y=Ax^3+Bx^2+Cx+D$ at more than one point.
Question: how can one show this, or even prove it?
 A: Look at the difference between the cubic and the line, this is also a cubic, and it has a double root at the tangent point. This can only happen at one point since the cubic has only three roots (counted with multiplicity).
Let the cubic be $p(x)$, the tanget $L(x)$ (as in line) and the x value of the tangent point is $a$. Then we have $p(a) - L(a) = 0$ and $p'(a) - L'(a) = 0$ so the Taylor expansion of $p-L$ around $a$ starts with $\frac{(x-a)^2}{2}p''(a)$ showing that $p-L$ has a double root at $a$.
A: Suppose the cubic has a tangent line with slope $m$, and for the sake of contradiction assume the tangent line touches at 2 distinct points $x=a$ and $x=b$.
By the mean value theorem, there must be a $c$ such that $a < c < b$ and $f'(c) = m$.  This means that $f'(x) = m$ has three solutions $\{a, b, c\}$, but as $f'(x) = m$ is a quadratic, it can only have 2 solutions.
A: HINT...In order for the line to be tangent to the cubic, solving simultaneously will give a polynomial with a double root at the tangent point. Clearly this can happen once but not twice in the case of a cubic as a cubic only has three roots and not four (i.e. Two double roots)
