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ajdirig
Can anyone help find the inverse of the fxn: y=x^(2)+4x-6
it is not a one to one function, so it is unlikely that it has an inverse
there are 2 inverses, depending on the chosen domain
I have the answer, I just don't understand how to get there
ok then we can solve via completing the square \[y+6=x^2+4x\]\[y+6+4=(x+2)^2\] \[y+10=(x+2)^2\] then \[x+2=\pm\sqrt{y+10}\] and so \[x=-2\pm\sqrt{y+10}\]
where did the +4 come from?
because \(x^2+4x\neq (x+2)^2\) however \((x+2)^2=x^2+4x+4\) so when we replace \(x^2+4x\) by \((x+2)^2\) we were adding 4 therefore you have to add 4 to the other side as well
aka "completing the square"
ahh, thank you, I never would have thought to complete the square
satellite73 do you know the domains?
sure it is given by the plus minus part
the question is worded strangely, it asks which domains lead to the two inverses
vertex of the parabola \(y=x^2+4x-6\) is \((-2,-10)\) so if \(x<-2\) the "function" is decreasing, while if \(x>-2\) it is increasing
therefore if \(x<-2\) the inverse is the one with the negative radical, if \(x>-2\) use the one with the positive radical yw