formula to find the vertex of y=2(x+1)square-1

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formula to find the vertex of y=2(x+1)square-1

Algebra
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Vertex for being y=a(x-h)^2 +k, you just need to plug in the numbers and do the math
I don't really get you question, are you trying to find the vertex?
normaly i m using -b over 2a to find the vertex, where should i use at here?

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it depends on what you are trying to find.
the vertex. I know y=(a)square+bx+c using -b over 2a to find the vertex, but how could i find the vertex at here?
well, the equation is already in vertex form, so just solve.
i don't get. I know the vertex is (-1,-1). But I need to show how i find the vertex
hang on, lemme put it in geo gebra
kk
that's it, the vertex is at( -1 -1) you had it. You just need to show how you get it?
yes, that is what i m asking
Well, how i did it is i put the equation into my graphing software and the lowest point of the graph is the vertex|dw:1441240202409:dw|
That help or no?
a little, i still need to ask my teacher tomorrow, but thank you:)
So all you have to do is graph the function and find the central point of the parabola, and that point is the vertex i believe.
you're welcome :)

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