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The average of the roots of the quadratic formula = -b/2a, assuming the discriminant>0, explain why h=-b/2a where V (h,k) is the vertex of this parabola.
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\[y=ax^2+bx+c\] \[y=a(x^2+\frac{b}{a}x)+c\] \[y=a(x+\frac{b}{2a})^2+k\] so vertex is at \[(-\frac{b}{2a},k)\]
|dw:1336594081547:dw| parabola is symmetric about its axis....hence just midway between both the roots lies the vertex.... so, midway is (x1 + x2)/2 and x1 + x2 is -b/a [sum of roots] so vertex is at -b/2a
Just curious what is the discriminant is = 0
if the discriminant is zero it means 1) you have a prefect square, like \(y=(x+3)^2\) and 2) there is only one zero, namely at the vertex
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