Question 156636
y= x^2 +2x
*[invoke solve_quadratic_equation 1,2,0]
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The y-intercept is at x=0.
At x=0, y=0 so the y-intercept is (0,0)
The vertex is at the minimum (in this example).
The 1st derivative is 2x+2
2x+2=0
x = -1
at x = -1, y = -1, so the vertex is (-1,-1)