Describing Transformations of Radical Functions

The radicals in the function rules are either square roots or cube roots. Thus, the parent function of each is either the square root of $x,$ or the cube root of $x.$ Let's start with Graph I.

From the image, we can see that the function corresponding to Graph I isn't defined for all real numbers. Thus, it must be one of the square root functions. Comparing it to the graph of $k(x)=\sqrt{x},$ can give us more information about its function rule.

It looks as though Graph I is a translation of $k(x)=\sqrt{x},$ both to the left and downward. Thus, its function rule is similar to $k(x),$ but with some number added to the input and some number subtracted from the output. Among the available choices, there is a match, $h(x).$ $$\text{I:}h(x)=\sqrt{x+1}-0.3$$

Graph II is very similar to Graph I, so let's compare them.

Here, II looks to be a reflection of I in the $x$-axis. If this is the case, one of the options must be equal to $\text{-}h(x).$

We can now identify that the function $$f(x)=\text{-}\sqrt{x+1}+0.3$$ is a reflection of $h$ in the $x$-axis. Thus, it corresponds to Graph II.

Graph III has to be the graph of one of the square root functions, as it's not defined for all real numbers. There is only one square root function remaining, $g,$ which must be its match. $$\text{III:}g(x)=\sqrt{\text{-}x+1}-0.3.$$ Looking at the graph, we can confirm that this is the case. Comparing $g(x)$ with $h(x),$ the sign of the input has been reversed. Thus $g$ is a reflection of $h$ in the $y$-axis. This is exactly what can be seen in the graphs when comparing I and III.

With only one rule remaining, Graph IV must correspond to $t.$ To confirm this, $t$ can be viewed as a translation of $l(x)=\sqrt[3]{x},$ $1$ unit to the left and $0.3$ units downward. This can also be seen in its graph.

Thus, we have matched all graphs and function rules. $$\begin{array}{rlrl}\text{I: }& h(x)& \text{II: }& f(x)\\ \text{III: }& g(x)& \text{IV: }& t(x)\end{array}$$

To write $g$ as a function of $x,$ we can first find an expression for $f(x+1),$ then one for $\text{-}f(x+1).$ This is done by replacing $x$ with $x+1$ in the function rule of $f.$ $$f(x+1)=\sqrt{2(x+1)}-1$$ Now that we have an expression for $f(x+1),$ we can find $\text{-}f(x+1),$ by multiplying the expression above by $\text{-}1.$ This gives the rule for $g.$ The function $g$ is now written in terms of $x$ without containing the function $f.$ We'll graph $f$ and $g.$

To graph $f,$ let's calculate some function values in a table. Since $f$ is a square root function, it's not defined when the expression under the radical sign is negative. Thus, a good starting value is $x=0.$

$x$	$\sqrt{2x}-1$	$f(x)$
$0$	$\sqrt{2(0)}-1$	$\text{-}1$
$1$	$\sqrt{2(1)}-1$	$\sim 0.41$
$2$	$\sqrt{2(2)}-1$	$1$
$3$	$\sqrt{2(3)}-1$	$\sim 1.45$
$4$	$\sqrt{2(4)}-1$	$\sim 1.83$

The $x$-values and the function values can now be plotted as points $(x,f(x))$ in a coordinate plane. We'll connect the points with a smooth curve. Note that the function is not defined for $x<0.$

We'll use the same procedure as we did for graphing $f.$ The function $g$ is also a square root function and is not defined for $x$-values less than $\text{-}1.$ Therefore, we will start at $x=\text{-}1$ for our table values.

$x$	$\text{-}\sqrt{2x+2}+1$	$f(x)$
$\text{-}1$	$\text{-}\sqrt{2(\text{-}1)+2}+1$	$1$
$0$	$\text{-}\sqrt{2(0)+2}+1$	$\sim \text{-}0.41$
$1$	$\text{-}\sqrt{2(1)+2}+1$	$\text{-}1$
$2$	$\text{-}\sqrt{2(2)+2}+1$	$\sim \text{-}1.45$
$3$	$\text{-}\sqrt{2(3)+2}+1$	$\sim \text{-}1.83$

The points can now be plotted in the same coordinate plane as $f,$ again connecting the points with a smooth curve.

To determine the transformations $f$ underwent to become $g,$ we'll study the function rule of $g\text{:}$ $$g(x)=\text{-}f(x+1).$$ This function can be seen as adding $1$ to the input of $f,$ and multiplying the function value by $\text{-}1.$ Therefore, the first transformation is a translation $1$ unit to the left.

Furthermore, the second transformation is a reflection in the $x$-axis.

Therefore, $f$ has undergone both a horizontal translation and a reflection in the $x$-axis to become $g.$