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