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anonymous
 5 years ago
find the zeros and the minimum or maximum for the relation y=x^(2)+2x24
anonymous
 5 years ago
find the zeros and the minimum or maximum for the relation y=x^(2)+2x24

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01) To find the zeros, set your function equal to zero: \[x^2+2x24=0 \implies (x+6)(x4)=0 \implies x=6,x=4\] So, y has to zeros x=6 and x=4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is this a calculus course?!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If so, then to find the extrema (either maximum or minimum), Do the following: 1) Find the derivative. 2) Set the derivative equal to zero, and solve for x. That value is a critical point (where extreme values occur). 3)Substitute this value in the original function, this will be your extreme value.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To see if it's a maximum or a minimum, just compare it with any other value of y. If it's greater, then it's a maximum. If it's smaller, then it's a minimum.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think he left for a break? ._. lol
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