If $y = -x^2 + 5$ and $x$ is a real number, then what is the maximum value possible for $y$?
I notice that this is a quadratic function with a negative coefficient for $x^2$, which means that it opens downward and has a vertex at the top. I also know that the vertex of a quadratic function is at the point $(-b/2a, f(-b/2a))$, where $a$ and $b$ are the coefficients of $x^2$ and $x$, respectively. In this case, $a = -1$ and $b = 0$, so the vertex is at $(-0/2(-1), f(-0/2(-1))) = (0, f(0))$. To find $f(0)$, I plug in $x = 0$ into the function: $f(0) = -(0)^2 + 5 = 5$. Therefore, the vertex is at $(0, 5)$, and this is the maximum value possible for $y$, since any other value of $x$ would make $y$ smaller. 