anonymous
  • anonymous
Can anyone help find the inverse of the fxn: y=x^(2)+4x-6
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
it is not a one to one function, so it is unlikely that it has an inverse
anonymous
  • anonymous
there are 2 inverses, depending on the chosen domain
anonymous
  • anonymous
I have the answer, I just don't understand how to get there

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anonymous
  • anonymous
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 \[x=-2\pm\sqrt{y+10}\]
anonymous
  • anonymous
where did the +4 come from?
anonymous
  • anonymous
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
anonymous
  • anonymous
aka "completing the square"
anonymous
  • anonymous
ahh, thank you, I never would have thought to complete the square
anonymous
  • anonymous
yw
anonymous
  • anonymous
satellite73 do you know the domains?
anonymous
  • anonymous
sure it is given by the plus minus part
anonymous
  • anonymous
the question is worded strangely, it asks which domains lead to the two inverses
anonymous
  • anonymous
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
anonymous
  • anonymous
Ok, thank you
anonymous
  • anonymous
therefore if \(x<-2\) the inverse is the one with the negative radical, if \(x>-2\) use the one with the positive radical yw

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