x^2 + x - 12
What is the minimum value of y?
This is a parabola.
(x+4)(x-3) = 0
x = -4 or x = 3 This is where you would be crossing the x axis.
You want the turning point which would be a minimum since this parabola opens upward. I know it opens up because the coeff. of x^2 is positive.
-b/2a = x -1/2 = -1/2 then find y
(-1/2)^2 -1/2 -12 = y