The smallest value of $x^2+8x$ for real values of $x$ is
$\text{(A) } -16.25\quad \text{(B) } -16\quad \text{(C) } -15\quad \text{(D) } -8\quad \text{(E) None of these}$
Let's imagine this as a quadratic equation. To find the minimum or maximum value, we always need to find the vertex of the quadratic equation. The vertex of the quadratic is $\frac{-b}{2a}$ in $ax^2+bx+c=0$. Then to find the output, or the y value of the quadratic, we plug the vertex "x" value back into the equation. In this quadratic, a=1, b=8, and c=0. So the "x" value is $\frac{-b}{2a} \Rightarrow \frac{-8}{2} = -4$. Plugging it back into $x^2 + 8x$, we get $16-32 = -16 \Rightarrow$ $\fbox{B}$