Ans: A
I is true. Rewrite the equation to general form, we have
I is true. Rewrite the equation to general form, we have
$\begin{array}{rcl}
y & = & (3 – x)(x + 2) + 6 \\
y & = & -x^2 + x + 12
\end{array}$
Since the coefficient of $x^2$ is $-1$, then the graph opens downwards.
II is not true. Sub. $x = 1$ into the right side of the equation, we have
$\begin{array}{rcl}
\text{RS} & = & -(1)^2 + 1 + 12 \\
& = & 12 \\
& \neq & 10
\end{array}$
Therefore, the graph dose not passes through the point $(1, 10)$.
III is not true. Sub. $y = 0$ into the equation, we have
$\begin{array}{rcl}
0 & = & -x^2 + x + 12 \\
x^2 – x – 12 & = & 0 \\
(x – 4)(x + 3) & = & 0
\end{array}$
Therefore, $x = -3$ or $x = 4$.
Hence, the $x$-intercepts of the graph are $-3$ and $4$.
