Question #b073a

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Answer 1 · 2016-07-30T22:56:43

#y = -3+-sqrt(10+x)#

When we find the inverse we effectively reflect the function across the line #y=x#. So, we swap y and x in the function.

#y=x^2+6x-1#

#therefore x = y^2 + 6y - 1#

#y^2 + 6y - (1+x) = 0#

#y = (-6+-sqrt(6^2+4(1)(1+x)))/2 = (-6+-sqrt(40+4x))/2 = (-6+-2sqrt(10+x))/2#

So #y = -3+-sqrt(10+x)#