WAEC 2019 · Paper 2 · Q9

(a)
If $y = 2x^2 - 5x + 3$ is expressed in the form $y = p(x + q)^2 + r$ , where $p$ , $q$ and $r$ are constants, find $(q + r)$ .
(b)
Find the area enclosed by the curves $y = x^2 - 3x + 2$ and $y = -x^2 + 3x + 2$ .

Try it on a graph

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

Worked solution (try it first)

(a)

Take 2 out of the $x$ terms only: $y = 2\left(x^2 - \frac52x\right) + 3$ .
Complete the square inside: half of $-\frac52$ is $-\frac54$ , so $x^2 - \frac52x = \left(x - \frac54\right)^2 - \frac{25}{16}$ .
Multiply back by 2: $y = 2\left(x - \frac54\right)^2 - \frac{25}{8} + 3$
$= 2\left(x - \frac54\right)^2 - \frac18$ .
So $p = 2$ , $q = -\frac54$ and $r = -\frac18$ .
$q + r = -\frac{10}{8} - \frac18$
$= -\frac{11}{8}$ .

(b)

The curves meet where $x^2 - 3x + 2 = -x^2 + 3x + 2$ , so $2x^2 - 6x = 0$ .
Factorise: $2x(x - 3) = 0$ , so $x = 0$ or $x = 3$ .
Between them, $y = -x^2 + 3x + 2$ is on top (at $x = 1$ it is 4, the other is 0).
Area $= \displaystyle\int_0^3 \left[(-x^2 + 3x + 2) - (x^2 - 3x + 2)\right]dx$
$= \int_0^3 (-2x^2 + 6x)\,dx$ .
$= \left[-\frac{2x^3}{3} + 3x^2\right]_0^3$
$= -18 + 27$
$= 9$ square units.

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