JAMB 2002 · UME · Q31

Find the maximum value of $y$ in the equation $y = 1 - 2x - 3x^2$ .

Worked solution (try it first)

The turning point is at $x = -\frac{b}{2a}$ .
With $a = -3$ and $b = -2$ , that is $x = -\frac13$ .
Substitute: $y = 1 - 2\left(-\frac13\right) - 3\left(\frac19\right)$
$= 1 + \frac23 - \frac13$ .
So the maximum value is $\frac43$ , option B.