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f(x)=-(x+1)^2+4 find the zeroes
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\[-(x+1)^2+4=0\\ (x+1)^2=4\] would be a good first step
We then do the square root?
yes, don't forget the \(-\) as well
Wait, but isn't 0-4=-4
@satellite73
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I know one of the zeroes are going to be one, but what about the other?
hold on lets back up
Okay
\[-(x+1)^2+4=0\] add \((x+1)^2\) to both sides, that is what gives \[4=(x+1)^2\]
i just wrote it as \[(x+1)^2=4\] if you subtract \(4\) from both sides you get \[-(x+1)^2=-4\] like you said, but you then change the signs and get \[(x+1)^2=4\]
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Okay
that means \[x+1=2\] which gives \(x=1\) or \[x+1=-2\] which gives \(x=-3\)
I see
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