Answer
$ \sqrt[3] x + 2$ has a y-intercept at (0,2), and an x-intercept at (-8,0). It's domain is all real numbers, and it's range is all real numbers.
Work Step by Step
To get the y-intercept, we set $x=0$, and solve:
$\sqrt[3] 0 + 2 = 2$
To get the x-intercept, we set $y=0$, and solve:
$0 = \sqrt[3] x + 2$
$-2 = \sqrt[3] x$
$(-2)^{3} = x$
$x=-8$
The equation is solvable for all real numbers, so has a domain of all real numbers, and it's range increases to infinity as x gets larger, and decreases to negative infinity as x gets smaller, so has a range of all real numbers.
