- Let $f(x) =k_1 x^2 +k_2 x$, where $k_1, k_2 \neq 0$.
$\begin{array}{rcl}
f(4) & = & 96 \\
k_1(4)^2+k_2(4) & = & 96 \\
4k_1 +k_2 & = & 24 \ldots \unicode{x2460}
\end{array}$Also,
$\begin{array}{rcl}
f(-5) & = & 15 \\
k_1(-5)^2 +k_2(-5) & = & 15 \\
5k_1-k_2 & = & 3 \ldots \unicode{x2461}
\end{array}$$\unicode{x2461} +\unicode{x2460}$, we have
$\begin{array}{rcl}
9k_1 & = & 27 \\
k_1 & = & 3
\end{array}$Sub. $k_1=3$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
4(3) +k_2 & = & 24 \\
k_2 & = & 12
\end{array}$$\therefore f(x)=3x^2+12x$.
- Sub. $y=0$ into $y=8f(x)$, we have
$\begin{array}{rcl}
8(3x^2+12x) & = & 0 \\
x^2+4x & = & 0 \\
x(x+4) & = & 0
\end{array}$$\therefore x=0$ or $x=-4$.
Hence, the $x$-intercepts are $0$ and $-4$.
-
$\begin{array}{rcl}
f(x) & = & k \\
3x^2+12x & = & k \\
3x^2 +12x -k & = & 0 \\
\end{array}$For the equation has two distinct real roots, we have
$\begin{array}{rcl}
\Delta & > & 0 \\
(12)^2 -4(3)(-k) & > & 0 \\
144+12k & > & 0 \\
12k & > & -144 \\
k & > & -12
\end{array}$
2022-I-10
Ans: (a) $f(x)=3x^2+12x$ (b) $0$ and $-4$ (c) $k>-12$
