JAMB 1986 · UME · Q23

Quadratics & their graphs

The curve $y = -x^2 + 3x + 4$ intersects the coordinate axes at

Worked solution (try it first)

On the $y$ -axis $x = 0$ , so $y = 4$ .
The point is $(0, 4)$ .
On the $x$ -axis $y = 0$ : $-x^2 + 3x + 4 = 0$ .
Multiply by $-1$ : $x^2 - 3x - 4 = 0$ .
Factorise: $(x - 4)(x + 1) = 0$ , so $x = 4$ or $x = -1$ .
So the points are $(0, 4)$ , $(4, 0)$ and $(-1, 0)$ , option D.

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