JAMB 1987 · UME · Q28

Complete the square: half of $-6$ is $-3$, so $x^2 - 6x = (x - 3)^2 - 9$.

So $y = (x - 3)^2 - 9 + 8 = (x - 3)^2 - 1$.

A square is never negative, so the least value of $y$ is $-1$ (when $x = 3$), option D.