Question

Help please I’ll give you 25 points

Answer 1

Answer: Choice D)
`y = \pm √(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 √(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 √(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 = √(x+5)`