WAEC 2009 · Paper 2 · Q10

(a)
Using the same axes, sketch the curves $y = 6 - x - x^2$ and $y = 3x^2 - 2x + 3$ .
Model answer
Sketch, don't plot a table. $y = 6 - x - x^2$ opens downwards and cuts the $x$ -axis at $(-3, 0)$ and $(2, 0)$ and the $y$ -axis at $(0, 6)$ ; its maximum is at $(-\frac12, 6\frac14)$ . $y = 3x^2 - 2x + 3$ opens upwards, cuts the $y$ -axis at $(0, 3)$ , has its minimum at $(\frac13, 2\frac23)$ and never meets the $x$ -axis. The curves cross at $x = -\frac34$ and $x = 1$ .
(b)
Find the $x$ -coordinates of the points of intersection of the two curves in (a).
Separate values with commas, e.g. 3, −2
(c)
Calculate the area of the finite region bounded by the two curves in (a).

Worked solution (try it first)

For $y = 6 - x - x^2 = -(x + 3)(x - 2)$ : it opens downwards, cuts the $x$ -axis at $-3$ and $2$ , and the $y$ -axis at $6$ .
Its vertex is at $x = -\frac12$ , $y = 6\frac14$ .
For $y = 3x^2 - 2x + 3$ : it opens upwards and cuts the $y$ -axis at $3$ .
Its discriminant is $4 - 36 < 0$ , so it doesn't meet the $x$ -axis.
Its vertex is at $x = \frac13$ , $y = 2\frac23$ .

Between these points the first curve is on top, so the area is $\displaystyle\int_{-3/4}^{1} [(6 - x - x^2) - (3x^2 - 2x + 3)]\,dx = \int_{-3/4}^{1} (3 + x - 4x^2)\,dx$ .
Integrate: $\left[3x + \frac{x^2}{2} - \frac{4x^3}{3}\right]_{-3/4}^{1}$ .
At $x = 1$ : $3 + \frac12 - \frac43 = \frac{13}{6}$ .
At $x = -\frac34$ : $-\frac94 + \frac{9}{32} + \frac{9}{16} = -\frac{45}{32}$ .
Subtract: $\frac{13}{6} + \frac{45}{32} = \frac{208 + 135}{96}$
$= \frac{343}{96}$ .
The area is $3\frac{55}{96}$ square units, about $3.57$ .