If y= (x-1)(x+2), then what is the least possible value of y

\(y= (x-1)(x+2)=x^2+x-2\).

Theory:
Quadratic expression \(ax^2+bx+c\) reaches its extreme values when \(x=-\frac{b}{2a}\).
When \(a>0\) extreme value is minimum value of \(ax^2+bx+c\) (maximum value is not limited).
When \(a<0\) extreme value is maximum value of \(ax^2+bx+c\) (minimum value is not limited).

You can look at this geometrically: \(y=ax^2+bx+c\) when graphed on XY plane gives parabola. When \(a>0\), the parabola opens upward and minimum value of \(ax^2+bx+c\) is y-coordinate of vertex, when \(a<0\), the parabola opens downward and maximum value of \(ax^2+bx+c\) is y-coordinate of vertex.


Examples:
Expression \(5x^2-10x+20\) reaches its minimum when \(x=-\frac{b}{2a}=-\frac{-10}{2*5}=1\), so minimum value is \(5x^2-10x+20=5*1^2-10*1+20=15\).

Expression \(-5x^2-10x+20\) reaches its maximum when \(x=-\frac{b}{2a}=-\frac{-10}{2*(-5)}=-1\), so maximum value is \(-5x^2-10x+20=-5*(-1)^2-10*(-1)+20=25\).

Back to the original question:
\(y= (x-1)(x+2)=x^2+x-2\) --> y reaches its minimum (as \(a=1>0\)) when \(x=-\frac{b}{2a}=-\frac{1}{2}\).

Therefore \(y_{min}=(-\frac{1}{2}-1)(-\frac{1}{2}+2)=-\frac{9}{4}\)


Answer: B.

Or:

Use derivative: \(y'=2x+1\) --> equate to 0: \(2x+1=0\) --> \(x=-\frac{1}{2}\) --> \(y_{min}=(-\frac{1}{2}-1)(-\frac{1}{2}+2)=-\frac{9}{4}\)

Answer: B.

Hope it's clear.