# How do you graph y=1/4sqrt(x-1)+2, compare to the parent graph, and state the domain and range?

Jan 4, 2018

${D}_{f} = \left[1 , + \infty\right)$ , ${R}_{f} = \left[2 , + \infty\right)$

#### Explanation:

graph{sqrt(x-1)/4+2 [-10, 10, -5, 5]}

The graph is $\sqrt{x}$ shifted $1$ to the right and what comes out of that shifted $2$ times upwards.

${D}_{f}$$=${AAx$\in$$\mathbb{R}$: $x - 1 \ge 0$} $=$ $\left[1 , + \infty\right)$

I will find the range using Monotony and continuity.

$f \left(x\right) = \frac{1}{4} \sqrt{x - 1} + 2$ ,
$x \ge 1$

$f ' \left(x\right) = \frac{1}{4} \cdot \frac{\left(x - 1\right) '}{2 \sqrt{x - 1}}$ $=$

$\frac{1}{8 \sqrt{x - 1}}$ $> 0$ , $x$$\in$$\left(1 , + \infty\right)$

Therefore $f$ is strictly increasing $\uparrow$ in $\left[1 , + \infty\right)$

• ${R}_{f} = f \left({D}_{f}\right) = f \left(\left[1 , + \infty\right)\right) = \left[f \left(1\right) , {\lim}_{x \rightarrow + \infty} f \left(x\right)\right)$ $=$

$=$ $\left[2 , + \infty\right)$

because ${\lim}_{x \rightarrow + \infty} f \left(x\right) = {\lim}_{x \rightarrow + \infty} \left(\frac{\sqrt{x - 1}}{4} + 2\right)$ $=$ $+ \infty$

NOTE: ${\lim}_{x \rightarrow + \infty} \left(x - 1\right) = + \infty$ therefore, ${\lim}_{x \rightarrow + \infty} \sqrt{x - 1} = + \infty$