Second derivative $= 0$ at turning point $(0,0)$ for $y = x^5-5x^4$

Question

My understanding is that at a maximum turning point of a function the first derivative is zero and the second derivative is negative. However, I'm confused that with $y = x^5-5x^4$ the turning point $(0,0)$ has a second derivative of $0$ instead of a negative value. I feel I'm missing something simple here.