Prove that if P(x) is a quartic polynomial, then there can only be at most one line that is tangent to it at two points. Prove that if P(x) is a quartic polynomial, then there can only be at most one line that is tangent to it at two points.
From what I have seen, the only case in which I have found a tangent line at 2 points is when the tangent line is horizontal. Is there any other way to approach it?

Line $y=ax + b$ is tangent to polynomial $y = P(x)$ at $x_1$ iff $P(x_1)=ax_1+b$ and $P'(x_1)=a$ which is equivalent to $Q(x)=P(x)-ax-b$ having a double root $Q(x_1)=Q'(x_1)=0$ at $x_1$.
Given $P(x)$ is a quartic then $Q(x)$ is a quartic as well, thus it can have at most $4$ real roots (counting multiplicities). Therefore $Q(x)$ can not have more than $2$ double roots, so there are at most $2$ points of tangency between the quartic and any line.
