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LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0you just helped someone find the vertex of a parabola using the formula and now you want somemone to derive it for you?

101Ryan101
 3 years ago
Best ResponseYou've already chosen the best response.0Anytime I write a question it usually "for fun".. :)

101Ryan101
 3 years ago
Best ResponseYou've already chosen the best response.0U surely don't HAVE to.. :)

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0okay thanks for the info

ZakaullahUET
 3 years ago
Best ResponseYou've already chosen the best response.0Find the derivative and equate it to zero solve for x to get ur ans

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1\[y=ax^2+bx+c\] \[y=a(x+\frac{b}{a})+c\] \[y=a(x+\frac{b}{2a})^2+\color{red}{\text{who cares}}\] 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 \[x=\frac{b}{2a}\]

101Ryan101
 3 years ago
Best ResponseYou've already chosen the best response.0If you don't care BOUNCE

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1i 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
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