burhan101
Increasing and decreasing intervals of 4x^2+4x-1
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etemplin
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do you know derivatives?
burhan101
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we have to complete the square
burhan101
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would it be \[\huge 4x^2+4x-4+4-1 \] for the first step of completing the square ?
etemplin
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yes
burhan101
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then \[\huge 4 (x-1)^2 -1\] ? :S
etemplin
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when you multiply that out it doesnt give you 4x^2+4x-4+4-1
etemplin
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try this one 4(x+(1/2))^2 -2
burhan101
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i dont get where the 1/2 comes from :S
etemplin
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i found the error: the leading coefficient must be 1 to complete the square
try setting it equal to 0, moving the -1 over, and dividing by 4
then complete the square
burhan101
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why does it have to be 1 ?
etemplin
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because thats what the formula says
burhan101
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oh so it applies at all equations ?
etemplin
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it applies to quadratics with a leading coefficient of 1
burhan101
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|dw:1351571054966:dw|
etemplin
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that works
move the -1/4 over
divide the middle term by 2 and square it
add that answer to both sides
factor
etemplin
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*by middle term, i meant the coefficient with the x^1
burhan101
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what do you mean by move the -1/4 over ? like why
etemplin
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because thats what the formula says to do
you'll end up with x^2+x=(1/4)
the term with x is 1
(1/2)^2 = 1/4
add (1/4) to both sides
end up with x^2+x+(1/4) = (1/2)
factor into (x+(1/2))^2 = (1/2)
solve for x
burhan101
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|dw:1351571511476:dw|
etemplin
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yes
now factor the left side