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ajdirig

  • 3 years ago

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

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  1. anonymous
    • 3 years ago
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    it is not a one to one function, so it is unlikely that it has an inverse

  2. ajdirig
    • 3 years ago
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    there are 2 inverses, depending on the chosen domain

  3. ajdirig
    • 3 years ago
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    I have the answer, I just don't understand how to get there

  4. anonymous
    • 3 years ago
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    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
    • 3 years ago
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    where did the +4 come from?

  6. anonymous
    • 3 years ago
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    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. anonymous
    • 3 years ago
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    aka "completing the square"

  8. ajdirig
    • 3 years ago
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    ahh, thank you, I never would have thought to complete the square

  9. anonymous
    • 3 years ago
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    yw

  10. ajdirig
    • 3 years ago
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    satellite73 do you know the domains?

  11. anonymous
    • 3 years ago
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    sure it is given by the plus minus part

  12. ajdirig
    • 3 years ago
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    the question is worded strangely, it asks which domains lead to the two inverses

  13. anonymous
    • 3 years ago
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    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
    • 3 years ago
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    Ok, thank you

  15. anonymous
    • 3 years ago
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    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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