Question

Help please I’ll give you 25 points

Answer 1

Answer: Choice D) $y = \pm \sqrt{x+5}$

The steps to finding the inverse will have us swap x and y. Afterward, we solve for y

$y = x^2 - 5 \\\\x = y^2 - 5 \\\\x+5 = y^2 \\\\y^2 = x+5 \\\\y = \pm \sqrt{x+5}$

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Extra info:

The inverse relation is not a function because of the plus minus. For instance, plugging x = 4 into $y = \pm \sqrt{x+5}$ leads to y = -3 and y = 3 simultaneously. You would have to apply a domain restriction on $y = x^2-5$ to make it a one-to-one function, to make the inverse a function. One possible domain restriction is $x > 0$ which would lead to the inverse function $y = \sqrt{x+5}$