# How do you find the domain and range of y = 3 sqrt (x-2) ?

Jul 17, 2015

For $x - 2$ to have a real square root, we require $x - 2 \ge 0$, hence $x \ge 2$.

Given $x \ge 2$ we find $y$ can have any positive value.

So the domain is $\left[2 , \infty\right)$ and range is $\left[0 , \infty\right)$

#### Explanation:

For $\sqrt{x - 2}$ to have a value in $\mathbb{R}$, we require $x - 2 \ge 0$

Add $2$ to both sides of this inequality to get:

$x \ge 2$.

If $x \ge 2$, then $\sqrt{x - 2} \ge 0$ is well defined and hence $y = 3 \sqrt{x - 2}$ is well defined.

So the domain is $\left[2 , \infty\right)$

$\sqrt{x - 2} \ge 0$ so $y = 3 \sqrt{x - 2} \ge 0$

In fact, for any $y \ge 0$, let $x = {\left(\frac{y}{3}\right)}^{2} + 2$.

Then $3 \sqrt{x - 2} = 3 \sqrt{{\left(\frac{y}{3}\right)}^{2}} = 3 \left(\frac{y}{3}\right) = y$

So the range is the whole of $\left[0 , \infty\right)$