How to solve this function/relation?

Question

x = y^2 + 2y + 5

Domain=

Range=

Answer 1

Domain = all the valid values of x.

change the equation a bit to more easily see this.

Let's see what we can simplify the right side....

We know that

(y+1)^2 = y^2 + 2y + 1

so...

x = y^2 + 2y + 1 -1 + 5

x = (y + 1)^2 + 4

x - 4 = (y+1)^2

Since any number squared is positive, this limits the possible values of x.

x - 4 ≥ 0

x ≥ 4     ..... which is the domain

The range is all the valid values of y... which in this case, is all real numbers.