Derive the formula for the axis of symmetry to a quadratic equation.

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Derive the formula for the axis of symmetry to a quadratic equation.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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you just helped someone find the vertex of a parabola using the formula and now you want somemone to derive it for you?
Anytime I write a question it usually "for fun".. :)
U surely don't HAVE to.. :)

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okay thanks for the info
Find the derivative and equate it to zero solve for x to get ur ans
\[y=ax^2+bx+c\] \[y=a(x+\frac{b}{a})+c\] \[y=a(x+\frac{b}{2a})^2+\color{red}{\text{who cares}}\] since first term is a perfect square it will be greater than or equal to zero if a is positive, less that or equal to zero if a is negative and it will be zero (so that is the max or min) if \[x=-\frac{b}{2a}\]
derivative my foot
If you don't care BOUNCE
i don't care because if i need to find it i can alway replace x by \[-\frac{b}{2a}\] i just don't care to try to keep track of what i added and subtracted. waste of time since i can find it easily
I see.

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