Question 156920
Graph the quadratic equation, label the ordered pairs for the vertex and the y-intercept 
y= x^2 + x-2
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*[invoke solve_quadratic_equation 1,1,-2]
The vertex is at the minimum.  To solve for the minimum, set the 1st derivative to 0.
2x+1 = 0
x = -1/2
Sub -1/2 for x into the eqn.
y(min) = (-1/2)^2 +(-1/2) -2
y(min) = 1/4 -1/2 - 2
= -2 1/4 = -9/4
So the vertex is (-1/2,-9/4)
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The y-intercept is easier, it's where x = 0.  Sub 0 for x in the original eqn:
y = 0 + 0 -2 = -2
So the point is (0,-2)