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Mathematics 7 Online
OpenStudy (anonymous):

find the coordinates, of the top of this function: y = x^2 + 2x

OpenStudy (lgbasallote):

so you're looking for maximum huh...do you know calculus?

OpenStudy (anonymous):

just a little bit

OpenStudy (lgbasallote):

okay..good enough...do you know how to take the derivative of this thing?

OpenStudy (anonymous):

yes

OpenStudy (lgbasallote):

so what's the derivative?

OpenStudy (anonymous):

y' = 2x + 2

OpenStudy (lgbasallote):

right. now take the derivative of this thing again...if the result is positive that means it's a minimum....if it's negative then it's a maximum...

OpenStudy (anonymous):

Y'' = 2, so its a maximum?

OpenStudy (lgbasallote):

nope...it's positive..therefore it's a minimum

OpenStudy (lgbasallote):

that means this graph doesnt have a maximum

OpenStudy (lgbasallote):

here's a picture of the graph :D http://www.wolframalpha.com/input/?i=y+%3D+x%5E2+%2B+2x it doesnt have a "top"

OpenStudy (anonymous):

ow oke

OpenStudy (anonymous):

then i have to find the coordinates of the bottom

OpenStudy (lgbasallote):

oh goodie...then it's just the minimum..

OpenStudy (lgbasallote):

you have y' = 2x + 2 right?

OpenStudy (lgbasallote):

now equate 2x + 2 to 0 and solve for x what do you get?

OpenStudy (anonymous):

x = 1

OpenStudy (anonymous):

-1*

OpenStudy (anonymous):

ok, then just fill in -1, in the original function or in the derivative function?

OpenStudy (lgbasallote):

now substitute -1 for x in y = x^2 + 2x and solve for y

OpenStudy (lgbasallote):

original function

OpenStudy (anonymous):

ok thnx now I get it

OpenStudy (lgbasallote):

welcome

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