JAMB 1988 · UME · Q27

The solutions of $x^2 - 2x - 1 = 0$ are the points of intersection of two graphs. If one of the graphs is $y = 2 + x - x^2$ , find the second graph.

Worked solution (try it first)

At the intersections, $2 + x - x^2$ equals the second graph's $y$ .
That equation must be the same as $x^2 - 2x - 1 = 0$ .
Rewrite the given equation as $0 = -x^2 + 2x + 1$ , and subtract it from the curve: $(2 + x - x^2) - (-x^2 + 2x + 1) = 1 - x$ .
So the second graph is $y = 1 - x$ .
Check: $2 + x - x^2 = 1 - x$ rearranges to $x^2 - 2x - 1 = 0$ , option A.