# Thread: Graphing square root functions

1. ## Graphing square root functions

Hi!

I can tell the domain from the graph of these functions but how do I tell the range from their graphs?
$h(x) = -\sqrt{x + 2}$ and $f(x) = \sqrt{2 - x} - 1$

I can find the domain easily: $h(x) = x + 2 \ge 0, x \ge -2$

Also, why does the graph of $h(x) = -\sqrt{x + 2}$ flip over the x-axis and the graph of $f(x) = \sqrt{2 - x} - 1$ flip over the y-axis?

2. ## Re: Graphing square root functions

Think about the transformations to a function f(x). -f(x) is a reflection over the x-axis while f(-x) is a reflection over the y-axis.

As for working out the range, for a standard function \displaystyle \begin{align*} f(x) = \sqrt{x} \end{align*}, the range is \displaystyle \begin{align*} f \in [0, \infty) \end{align*}. Why? Now what transformations have been applied and how does this affect the range?

3. ## Re: Graphing square root functions

Originally Posted by Prove It
As for working out the range, for a standard function \displaystyle \begin{align*} f(x) = \sqrt{x} \end{align*}, the range is \displaystyle \begin{align*} f \in [0, \infty) \end{align*}. Why? Now what transformations have been applied and how does this affect the range?
Since the range is dependent on the domain...
The range of the square root function lies in $f \in [0, \infty)$ because we are finding the principal root of the non-negative value $x$.

Transformation applied: square root function.
Effect: Take only principal root.

4. ## Re: Graphing square root functions

No, I meant you start with the square root function and then apply some transformations to get the functions you have been given. What are these transformations?

5. ## Re: Graphing square root functions

Originally Posted by Prove It
No, I meant you start with the square root function and then apply some transformations to get the functions you have been given. What are these transformations?
For $h(x) = -\sqrt{x + 2}$, means h(x) = -h(x).

Begin: $\sqrt{x}$
+ 2: Shifts left, (-2,0)
- : Flips over x-axis, point (-2,0) remains the same.

For $f(x) = \sqrt{2 - x} - 1$. $\Rightarrow \sqrt{-(x - 2)} - 1$
Begin: $\sqrt{x}$
$- 2$: Shifts right, (2,0)
$-$ inside radical : Flips over y-axis, (2,0)
$- 1$ : Pushes graph downward, (2,-1)

6. ## Re: Graphing square root functions

Correct, so now what are the ranges of these functions?

7. ## Re: Graphing square root functions

Range refers to y-values.

$h(x) = -\sqrt{x + 2}. \Rightarrow \text{Point}\ (2,0)$

Range: $\[0,\infty)$

$f(x) = \sqrt{2 - x} - 1. \Rightarrow \text{Point}\ (2, -1)$

Range: $\[-1, \infty)$

8. ## Re: Graphing square root functions

I agree with your range for f(x) but not your range for h(x).

9. ## Re: Graphing square root functions

Originally Posted by Prove It
I agree with your range for f(x) but not your range for h(x).
h(x) flips over the x-axis, so the y-values stop at 0.
Range: $\(-\infty,0]$

The range of h(x) seems less intuitive, maybe due to the graph flipping...your thoughts?

10. ## Re: Graphing square root functions

This is correct. And I don't think it's counter-intuitive at all actually, it should be obvious that when there's a reflection in one of the axes then either the domain or range will be inverted.