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satellite73
 one year ago
Best ResponseYou've already chosen the best response.1it is not a one to one function, so it is unlikely that it has an inverse

ajdirig
 one year ago
Best ResponseYou've already chosen the best response.0there are 2 inverses, depending on the chosen domain

ajdirig
 one year ago
Best ResponseYou've already chosen the best response.0I have the answer, I just don't understand how to get there

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1ok 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 \[x=2\pm\sqrt{y+10}\]

ajdirig
 one year ago
Best ResponseYou've already chosen the best response.0where did the +4 come from?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1because \(x^2+4x\neq (x+2)^2\) 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1aka "completing the square"

ajdirig
 one year ago
Best ResponseYou've already chosen the best response.0ahh, thank you, I never would have thought to complete the square

ajdirig
 one year ago
Best ResponseYou've already chosen the best response.0satellite73 do you know the domains?

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1sure it is given by the plus minus part

ajdirig
 one year ago
Best ResponseYou've already chosen the best response.0the question is worded strangely, it asks which domains lead to the two inverses

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1vertex of the parabola \(y=x^2+4x6\) is \((2,10)\) so if \(x<2\) the "function" is decreasing, while if \(x>2\) it is increasing

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1therefore 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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