Range of $f(x)=ax^2-c$ The function $f(x)=ax^2-c$ satisfies $-4\le f(1) \le -1$ and $-1\le f(2) \le 5$. Which of the following statement is true ?
(1)$-7\le f(3) \le 26$
(2)$-4\le f(3) \le 15$
(3)$-1\le f(3) \le 20$
(4)$-\frac {28}{3}\le f(3) \le \frac {35}{3}$ 
My approach is as follow
$f(1)=a-c$
$f(2)=4a-c$
$f(3)=9a-c$
$-4\le a-c \le -1$ and $-1\le 4a-c \le 5$
We need to find $a'\le 9a-c \le b'$, how do I proceed.
 A: First, consider how to express $f(3)$ as a linear combination of $f(1)$ and $f(2)$ so you can effectively use their adjusted limits to determine the limits for $f(3)$. To do this, for some real constants $d$ and $e$ you have
$$\begin{equation}\begin{aligned}
9a - c & = d(a - c) + e(4a - c) \\
& = (d)a - (d)c + (4e)a - (e)c \\
& = (d + 4e)a + (-d - e)c
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Matching the coefficients of $a$ and $c$ between the left & right sides gives
$$d + 4e = 9 \tag{2}\label{eq2A}$$
$$-d - e = -1 \tag{3}\label{eq3A}$$
Adding \eqref{eq2A} and \eqref{eq3A} gives $3e = 8 \implies e = \frac{8}{3}$ which, substituting back into \eqref{eq3A} gives $-d - \frac{8}{3} = -1 \implies d = -\frac{5}{3}$. This now gives
$$\begin{equation}\begin{aligned}
f(3) & = 9a - c \\
& = - \frac{5}{3}(a - c) + \frac{8}{3}(4a - c) \\
& = - \frac{5}{3}f(1) + \frac{8}{3}f(2)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Next, you've got
$$-4\le f(1) \le -1 \implies \frac{20}{3} \ge -\frac{5}{3}f(1) \ge \frac{5}{3} \tag{5}\label{eq5A}$$
$$-1\le f(2) \le 5 \implies -\frac{8}{3} \le \frac{8}{3}f(2) \le \frac{40}{3} \tag{6}\label{eq6A}$$
Adding \eqref{eq5A} and \eqref{eq6A} gives, from using \eqref{eq4A}, that
$$\frac{5}{3} - \frac{8}{3} \le - \frac{5}{3}f(1) + \frac{8}{3}f(2) \le \frac{20}{3} + \frac{40}{3} \implies -1 \le f(3) \le 20 \tag{7}\label{eq7A}$$
This matches option ($3$).
