JAMB 1999 · UME · Q33

If the maximum value of $y = 1 + hx - 3x^2$ is 13, find $h$ .

Worked solution (try it first)

The turning point is at $x = -\frac{b}{2a}$ .
With $a = -3$ and $b = h$ , that is $x = \frac h6$ .
Substitute: $y = 1 + h \cdot \frac h6 - 3 \cdot \frac{h^2}{36}$ , which simplifies to $1 + \frac{h^2}{12}$ .
Set it equal to 13: $\frac{h^2}{12} = 12$ , so $h^2 = 144$ and $h = \pm 12$ .
The value in the options is $h = 12$ , option B.