2015-II-08

Ans: D
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 $b>0$.