WAEC 2020 · Paper 1 · Q28

A function $f$ defined by $f : x \to x^2 + px + q$ is such that $f(3) = 6$ and $f'(3) = 0$ . Find the value of $q$ .

Worked solution (try it first)

Differentiate: $f'(x) = 2x + p$ .
Then $f'(3) = 6 + p = 0$ , so $p = -6$ .
Put $x = 3$ into $f$ : $f(3) = 9 + 3p + q = 9 - 18 + q = q - 9$ .
Set $q - 9 = 6$ , so $q = 15$ , option C.

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