WAEC 2008 · Paper 2 · Q1✱✱

(a)
A function $f$ is defined on the set $\mathbb{R}$ of real numbers by $f : x \to px^2 + qx + 2$ , where $p$ and $q$ are constants. If $f(-2) = 0$ and $f(1) = -3$ , find $f(-4)$ .

Worked solution (try it first)

Put $x = -2$ into $f(x) = px^2 + qx + 2$ and use $f(-2) = 0$ : $4p - 2q + 2 = 0$ .
Divide by 2: $2p - q = -1$ .
Put $x = 1$ and use $f(1) = -3$ : $p + q + 2 = -3$ , so $p + q = -5$ .
Add the two equations to remove $q$ : $3p = -6$ , so $p = -2$ .
Then $q = -5 - p$ , which gives $q = -3$ .
So $f(x) = -2x^2 - 3x + 2$ .
Put $x = -4$ : $-2(16) - 3(-4) + 2 = -32 + 12 + 2$ .
$f(-4) = -18$ .