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?

$\displaystyle h(x) = -\sqrt{x + 2}$ and $\displaystyle f(x) = \sqrt{2 - x} - 1$

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

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

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

Re: Graphing square root functions

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Originally Posted by

**Prove It** As for working out the range, for a standard function $\displaystyle \displaystyle \begin{align*} f(x) = \sqrt{x} \end{align*}$, the range is $\displaystyle \displaystyle \begin{align*} f \in [0, \infty) \end{align*}$. 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 $\displaystyle f \in [0, \infty)$ because we are finding the *principal* root of the non-negative value $\displaystyle x$.

Transformation applied: square root function.

Effect: Take only *principal* root.

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?

Re: Graphing square root functions

Quote:

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 $\displaystyle h(x) = -\sqrt{x + 2}$, means h(x) = -h(x).

Begin: $\displaystyle \sqrt{x}$

+ 2: Shifts left, (-2,0)

- : Flips over x-axis, point (-2,0) remains the same.

For $\displaystyle f(x) = \sqrt{2 - x} - 1$. $\displaystyle \Rightarrow \sqrt{-(x - 2)} - 1$

Begin: $\displaystyle \sqrt{x}$

$\displaystyle - 2$: Shifts right, (2,0)

$\displaystyle -$ inside radical : Flips over y-axis, (2,0)

$\displaystyle - 1$ : Pushes graph downward, (2,-1)

Re: Graphing square root functions

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

Re: Graphing square root functions

Range refers to y-values.

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

Range: $\displaystyle \[0,\infty)$

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

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

Re: Graphing square root functions

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

Re: Graphing square root functions

Quote:

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: $\displaystyle \(-\infty,0]$

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

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.