Ans: A
I is true. Rewrite the function of the graph into general form, we have
$\begin{array}{rcl}
y & = & (h-x)(k-x) \\
y & = & x^2 -hx -kx +hk \\
y & = & x^2 -(h+k)x +hk
\end{array}$
Since the coefficient of $x^2$ is positive, then the graph opens upwards.
II is true. Sub. $y=0$ into the function of the graph, we have
$\begin{array}{rcl}
(h-x)(k-x) & = & 0 \\
(x-h)(x-k) & = & 0
\end{array}$
$\therefore x=h$ or $x=k$.
Therefore, there are two $x$-intercepts.
III is not true. Note that the $y$-intercept of the graph is $hk$, which is negative.
