Answer: Choice D) [tex]y = \pm \sqrt{x+5}[/tex]
The steps to finding the inverse will have us swap x and y. Afterward, we solve for y
[tex]y = x^2 - 5 \\\\x = y^2 - 5 \\\\x+5 = y^2 \\\\y^2 = x+5 \\\\y = \pm \sqrt{x+5} \\\\[/tex]
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Extra info:
The inverse relation is not a function because of the plus minus. For instance, plugging x = 4 into [tex]y = \pm \sqrt{x+5}[/tex] leads to y = -3 and y = 3 simultaneously. You would have to apply a domain restriction on [tex]y = x^2-5[/tex] to make it a one-to-one function, to make the inverse a function. One possible domain restriction is [tex]x > 0[/tex] which would lead to the inverse function [tex]y = \sqrt{x+5}[/tex]
