Ans: D
Since the graph opens upwards, then $a>0$.
Since the graph opens upwards, then $a>0$.
Note that the $x$-intercept of the graph is lying on the negative $x$-axis, then the root must be negative. Consider the equation of the graph $y=a(x+b)^2$, the root is $-b$. Hence we have
$\begin{array}{rcl}
-b & < & 0 \\
b & > & 0
\end{array}$
Therefore, $a>0$ and $b>0$.
