What is the domain and range of $x = {(y + 2)}^{2}$ $x =(y+2)^2$ ?

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Answer 1 · 2017-02-15T04:24:10

Domain: $[0, \infty)$ $[0, oo)$ , Range: $[- 2, \infty)$ $[-2, oo)$

To graph, you need to solve for $y$ $y$ :

Square root both sides: $\sqrt{x} = y + 2$ $sqrt(x) = y+2$
Isolate the variable $y$ $y$ : $y = \sqrt{x} - 2$ $y =sqrt(x) -2$

Analytically finding the domain:
$\sqrt{x} \geq 0$ $sqrt(x) >=0$ which means $x \geq 0$ $x>=0$

If $x \geq 0$ $x>=0$ then $y \geq - 2$ $y >= -2$

From the graph: graph{sqrt(x) - 2 [-10, 10, -5, 5]}

What is the domain and range of x =(y+2)^2x=(y+2)2x =(y+2)^2?