Ans: D
A is not true. Rewrite the equation of the graph to the general form, we have
$\begin{array}{rcl}
y & = & 5+(x-3)^2 \\
y & = & 5+x^2-6x+9 \\
y & = & x^2-6x+14 \ldots \unicode{x2460}
\end{array}$
Since the coefficient of $x^2$ is positive, the graph opens upwards.
B is not true. The discriminant of $\unicode{x2460}$
$\begin{array}{cl}
= & (-6)^2 -4(1)(14) \\
= & -20 \\
< & 0
\end{array}$
Therefore, there is no $x$-intercept for the graph.
C is not true. From $\unicode{x2460}$, the $y$-intercept is $14$.
D is true. Sub. $x=3$ into the RS of $\unicode{x2460}$, we have
$\begin{array}{rcl}
y & = & 3^2-6(3)+14 \\
y & = & 5
\end{array}$
Therefore, the graph passes through $(3,5)$.
