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Mathematics 17 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

you just helped someone find the vertex of a parabola using the formula and now you want somemone to derive it for you?

OpenStudy (anonymous):

Anytime I write a question it usually "for fun".. :)

OpenStudy (anonymous):

U surely don't HAVE to.. :)

OpenStudy (anonymous):

okay thanks for the info

OpenStudy (anonymous):

Find the derivative and equate it to zero solve for x to get ur ans

OpenStudy (anonymous):

\[y=ax^2+bx+c\] \[y=a(x+\frac{b}{a})+c\] \[y=a(x+\frac{b}{2a})^2+\color{red}{\text{who cares}}\] 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 \[x=-\frac{b}{2a}\]

OpenStudy (anonymous):

derivative my foot

OpenStudy (anonymous):

If you don't care BOUNCE

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I see.

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