AMC12 2006 12B, Problem 12

The parabola $y=ax^2+bx+c$ has vertex $(p,p)$ and $y$-intercept $(0,-p)$, where $p\ne 0$. What is $b$?

Solution: A parabola with the given equation and with vertex $(p,p)$ must have equation $y=a(x-p)^2+p$. Because the $y$-intercept is $(0,-p)$ and $p\ne 0$, it follows that $a=-2/p$. Thus\[y=-\frac{2}{p}(x^2-2px+p^2)+p=-\frac{2}{p}x^2+4x-p,\] so $\fbox{4}$.

Answer: 4