WAEC 2013 · Paper 2 · Q12

(a)
Find the maximum and minimum points of the curve $y = 2x^3 - 3x^2 - 12x + 4$ .
Show the answer
maximum $(-1, 11)$ , minimum $(2, -16)$
(b)
Sketch the curve in 12(a) above.
Model answer
A sketch shows the shape and the key points, not an accurate plot. Mark the maximum $(-1, 11)$ and minimum $(2, -16)$ from (a), and the $y$ -intercept $(0, 4)$ . Then draw an S-shaped cubic: it rises to the maximum, falls through the minimum and rises again (positive $x^3$ term, so it goes up on the right). The curve crosses the $x$ -axis three times: $x = -2$ (check: $-16 - 12 + 24 + 4 = 0$ ), and from $2x^2 - 7x + 2 = 0$ , $x \approx 0.3$ and $3.2$ .

Plot the curves, move them, and read values off the graph.

Worked solution (try it first)

Differentiate: $\dfrac{dy}{dx} = 6x^2 - 6x - 12$
$= 6(x - 2)(x + 1)$ .
Stationary points where $\dfrac{dy}{dx} = 0$ : $x = 2$ or $x = -1$ .
The second derivative is $\dfrac{d^2y}{dx^2} = 12x - 6$ .
At $x = -1$ : $\dfrac{d^2y}{dx^2} = -18 < 0$ , a maximum, with $y = -2 - 3 + 12 + 4 = 11$ .
So the maximum point is $(-1, 11)$ .
At $x = 2$ : $\dfrac{d^2y}{dx^2} = 18 > 0$ , a minimum, with $y = 16 - 12 - 24 + 4 = -16$ .
So the minimum point is $(2, -16)$ .

The curve is a cubic with a positive $x^3$ term: it rises to the maximum $(-1, 11)$ , falls through $(0, 4)$ to the minimum $(2, -16)$ , then rises again.