I think I have the answer, but it's a tricky problem, so I'm looking for verification here.
Find a number $a$ so that the quadratic polynomial with roots $a$ and $a^2$ has the maximum possible slope at the point $x = 0.$
First, a quadratic polynomial with roots $a$ and $a^2$ is $$f(x) = (x - a)(x-a^2) = x^2 - ax - a^2x + a^3.$$
Then we note that the derivative of $f(x)$ is $$f'(x) = 2x - a - a^2.$$
Then we want to find the value $a$ that gives the maximum slope at $$f'(0) = -a - a^2.$$
So now we can think about this as a function of $a,$ $$g(a) = -a - a^2.$$ The derivative of this function is $$g'(a) = -1 - 2a,$$ which has a critical point at $a = -\frac{1}{2},$ which is in fact a global maximum of $g(a).$
The max slope is then $g(-\frac{1}{2}) = \frac{1}{4},$ which isn't a very big slope.