Ans: (a) $24x-48$ (b) $4$
- Let $h = k_1 + k_2 x$, where $k_1$ and $k_2$ are non-zero constants.
Consider $h(-2) = -96$, we have
$\begin{array}{rcl}
-96 & = & k_1 + k_2(-2) \\
-96 & = & k_1 – 2k_2 \ \ldots \unicode{x2460}
\end{array}$Consider $h(5) = 72$, we have
$\begin{array}{rcl}
72 & = & k_1 + k_2(5) \\
72 & = & k_1 + 5k_2 \ \ldots \unicode{x2461}
\end{array}$$\unicode{x2461} – \unicode{x2460}$, we have
$\begin{array}{rcl}
168 & = & 5k_2 \\
k_2 & = & 24
\end{array}$Sub. $k_2 = 24$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
-96 & = & k_1 – 2(24) \\
k_1 & = & -48
\end{array}$Therefore, $h(x) = 24x – 48$.
-
$\begin{array}{rcl}
h(x) & = & 3x^2 \\
24x – 48 & = & 3x^2 \\
x^2 – 8x + 16 & = & 0 \\
(x – 4)^2 & = & 0
\end{array}$Therefore, $x = 4\ \text{(repeated)}$.
