## ajdirig 2 years ago Can anyone help find the inverse of the fxn: y=x^(2)+4x-6

1. satellite73

it is not a one to one function, so it is unlikely that it has an inverse

2. ajdirig

there are 2 inverses, depending on the chosen domain

3. ajdirig

I have the answer, I just don't understand how to get there

4. satellite73

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 $x=-2\pm\sqrt{y+10}$

5. ajdirig

where did the +4 come from?

6. satellite73

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

7. satellite73

aka "completing the square"

8. ajdirig

ahh, thank you, I never would have thought to complete the square

9. satellite73

yw

10. ajdirig

satellite73 do you know the domains?

11. satellite73

sure it is given by the plus minus part

12. ajdirig

the question is worded strangely, it asks which domains lead to the two inverses

13. satellite73

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

14. ajdirig

Ok, thank you

15. satellite73

therefore if $$x<-2$$ the inverse is the one with the negative radical, if $$x>-2$$ use the one with the positive radical yw