Question 332261
the equation for a parabola:  y={{{a*x^2+b*x+c}}}

The vertex is at  x={{{-b/(2a)}}}, y={{{(4ac-b^2)/(4a)}}}

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this can be found from the quadratic solutions are
x={{{(-b-Sqrt(b^2-4ac))/(2a)}}} and 
x={{{(-b+Sqrt(b^2-4ac))/(2a)}}}
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The center is at value x={{{-b/(2a)}}} and then substituting this x value in the parabola equation to find y
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Try it: for parabola  Y={{{2*x^2+8*x+11}}}

x={{{-b/(2a)}}} = -8/4=-2
y={{{(4ac-b^2)/(4a)}}} = (4*2*11-8^2)/(4*2) = 24/8=3
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if you complete the square for Y={{{2*x^2+8*x+11}}}
y={{{2*(x+2)^2 +3}}} for which it is easy to see that the vertex is (-2,3)