NECO 2023 · Paper 1 · Q23

Applications of differentiation

Find the minimum value of $y = x^2 - 2x - 3$ .

Worked solution (try it first)

Differentiate: $\dfrac{dy}{dx} = 2x - 2$ , which is zero at $x = 1$ .
$\dfrac{d^2y}{dx^2} = 2$ is positive, so this turning point is a minimum.
The minimum value is $y = 1^2 - 2(1) - 3 = -4$ , option D.

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