The index of a radical function can be any real number, but the most common are square roots with an index of 2, like y=\sqrt{x}, which are called a square root function, and cube roots with an index of 3, like y=\sqrt[3]{x}, called a cube root function.

Exploration

Consider the functions:

f\left(x\right)=\sqrt{x}
g\left(x\right)=\sqrt{x+4}
h\left(x\right)=\sqrt{x}-2
k\left(x\right)=\sqrt[3]{x}

Which functions have a real output for an input of x=-3? x=0? x=5?
Which functions have a domain of all real values of x?
Which functions go through the point \left(0,0\right)?
Which functions can have negative outputs?
Match each function to one of these graphs:

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Graph 1

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Graph 2

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Graph 3

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Graph 4

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Consider the graph of the square root function y=\sqrt{x} and its key features:

Domain: \left[0, \infty \right)
Range: \left[0, \infty \right)
x-intercept: \left(0,0\right)
y-intercept: \left(0,0\right)
Increasing over its domain
As x \to \infty, y \to \infty
Endpoint: \left(0,0\right)
Absolute minimum: \left(0,0\right)
Absolute maximum: none

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Consider the graph of the cube root function {y=\sqrt[3]{x}} and its key features:

Domain: \left(-\infty, \infty \right)
Range: \left(-\infty, \infty \right)
x-intercept: \left(0,0\right)
y-intercept: \left(0,0\right)
Point of inflection: \left(0,0\right)
Increasing over its domain
As x \to \infty, y \to \infty
As x \to -\infty, y \to -\infty

For cube root functions, the function increases (or decreases) at a fast rate, then the rate of change slows around a point called an inflection point. In other words, the function continues increasing (or decreasing), but the rate is slower around the point of inflection.

Radical functions can be transformed similarly to any transformation of the parent function, y= af \left[b\left(x-h \right)\right] +k.

	Square root	Cube root
\text{Parent function:}	y= \sqrt{x}	y= \sqrt[3]{x}
\text{Reflection across the }x\text{-axis:}	y=-\sqrt{x}	y=-\sqrt[3]{x}
\text{Reflection across the }y\text{-axis:}	y=\sqrt{-x}	y=\sqrt[3]{-x}
\text{Vertical stretch when } \left\|a\right\|>1 \\ \text{Vertical compression when } 0<\left\|a\right\|<1 \text{:}	y=a\sqrt{x}	y=a\sqrt[3]{x}
\text{Horizontal compression when } \left\|b\right\|>1 \\ \text{Horizontal stretch when } 0<\left\|b\right\|<1 \text{:}	y=\sqrt{bx}	y=\sqrt[3]{bx}
\text{Horizontal translation by } h \\ \text{Vertical translation by } k \text{:}	y=\sqrt{x-h} + k	y=\sqrt[3]{x-h} + k
(h, k) \text{:}	\text{Endpoint}	\text{Point of inflection}

The domain and range of the square root function will change with a reflection, or as h or k changes, while the domain and range of the cube root function will continue to be all real numbers.

Similarly, the absolute extremum of the square root function will change location when translated. If there are no reflections, the endpoint of the domain is an absolute minimum. If a vertical reflection occurs, it becomes an absolute maximum.

Square root functions do not have a relative extremum, and cube root functions have neither absolute nor relative extrema.

Examples

Example 1

For f\left(x\right) = -\sqrt{x + 2},

Describe the transformation that occurred to y=\sqrt{x} to give f\left(x\right).

Worked Solution

Create a strategy

The function is of the form f\left(x\right)=a\sqrt{x-h} with a=-1 and h=-2.

Apply the idea

The function has been translated to the left by 2 units and reflected across the x-axis.

Draw a graph of the function.

Worked Solution

Apply the idea

Using the key points of the parent function, y=\sqrt{x}, with the transformations identified in part (a), we can identify the key points of the given function.

The key points of the parent function are shown in the table.

x	0	1	4	9	16
y=\sqrt{x}	0	1	2	3	4

Reflecting across the x-axis results in:

x	0	1	4	9	16
y=-\sqrt{x}	0	-1	-2	-3	-4

Translating these points left 2 units gives us:

x	-2	-1	2	7	14
f\left(x\right)=-\sqrt{x+2}	0	-1	-2	-3	-4

Now, we can graph the function using these key points.

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Reflect and check

We can use technology to graph the given function and to confirm the transformations we identified.

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Here is the graph of the square root parent function y=\sqrt{x}.

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Here is the graph of the given function, f\left(x\right)=-\sqrt{x+2}. Notice that the graph does in fact show a reflection across the x-axis and a translation 2 units left.

This does not match the shape of our graph because we chose a different scale for the axes. This graph shows -5<x<5 and -5<y<5.

To match the graph that we drew, we can adjust the viewing window on the graphing calculator to -4<x<16 and -5<y<2.

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After adjusting the axes, we can see that this graph has the same shape as the graph we drew by hand. This confirms that the graph was not dilated and that the graph we drew is accurate.

Write the domain and range of f\left(x\right).

Worked Solution

Create a strategy

As the graph has been translated 2 units to the left, the domain will change. The range will also change since the function was reflected across the x-axis.

Apply the idea

Domain: \left[-2, \infty\right)

Range: \left(- \infty,0\right]

Reflect and check

In set notation, the domain can be written as \{x|x\geq -2\}, and the range can be written as \{y|y\leq 0\}.

Determine the intervals where f\left(x\right) is increasing, decreasing, or constant.

Worked Solution

Create a strategy

The function is increasing if the y-values increase as x increases.

The function is decreasing if the y-values decrease as x increases.

The function is constant if the y-values remain the same as x increases.

Apply the idea

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Observing the graph of f\left(x\right) = -\sqrt{x + 2}, we note that the y-values consistently decrease as the x-values get larger.

This function is strictly decreasing over its entire domain, \left[-2,\infty\right)

Example 2

Consider the graph of f \left( x \right).

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Write the equation that represents f\left(x\right).

Worked Solution

Create a strategy

First, we can look at the shape of the graph and determine the function family it belongs to. This graph is a cube root function, so the parent function is y=\sqrt[3]{x}.

Next, we want to identify the transformations applied to the function and write the equation in the form f\left(x\right)=a\sqrt[3]{x-h}+k.

Apply the idea

Comparing the given function to the parent cube root function, we can see that the function has been reflected about the x-axis and translated up by 3 units.

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Using the transformation form of the equation, f\left(x\right)=a\sqrt[3]{x-h}+k:

The reflection across the x-axis corresponds to a=-1
There was no horizontal shift, so h=0.
The vertical translation up 3 units corresponds to k=3.

This gives us the equation f\left(x\right)=-\sqrt[3]{x}+3.

Reflect and check

We can verify the equation by graphing it with technology. We can graph the equation {f\left(x\right)=-\sqrt[3]{x}+3} and adjust the viewing window to ensure it matches the given graph.

Write the domain and range of f\left(x\right).

Worked Solution

Create a strategy

The domain and range of the parent cube root function is all real x and y. Any type of transformation applied to the parent has no effect on the domain and range of the transformed function.

Apply the idea

Domain: \left(-\infty, \infty\right)

Range: \left(-\infty, \infty\right)

Reflect and check

The domain and range can also be represented with set notation:

Domain: \{x|-\infty \lt x\lt \infty\}
Range: \{y|-\infty \lt y\lt \infty\}

The symbol \Reals represents the set of real numbers. The symbol \in can be read as "belongs to". So another way of stating that the domain and range contain all real numbers in set notation is:

Domain: \{x| x\in \Reals \}
Range: \{y| y\in \Reals \}

Describe the end behavior of f\left(x\right).

Worked Solution

Create a strategy

To describe the end behavior, we want to determine if the y-values of the function approach a single value or if they continue increasing or decreasing indefinitely as the x-values approach positive and negative infinity.

Apply the idea

To see what happens as the x-values increase, we need to look at the right side of the graph. As the graph continues to the right indefinitely, the function decreases indefinitely. To describe this in words, we say "as the x-values approach positive infinity, the y-values approach negative infinity." In symbolic form, it is written as shown:

As x \to \infty,\, f\left(x\right) \to - \infty

To see what happens as the x-values decrease, we need to look at the left side of the graph. As the graph continues to the left indefinitely, our function increases indefinitely. To describe this in words, we say "as the x-values approach negative infinity, the y-values approach positive infinity." In symbolic form, it is written as shown:

As x \to - \infty,\, f\left(x\right) \to \infty

Find where f\left(x\right)=1.

Worked Solution

Create a strategy

This question is asking us to find the value(s) of x for which the y-values equal 1.

Apply the idea

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The graph has a y-value of 1 at x=8, therefore f\left(x\right) =1 at x=8.

Example 3

Consider the piecewise function shown in the graph:

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Identify the function families in the piecewise function.

Worked Solution

Create a strategy

We can start by using the general shape of the functions to identify the parent functions. We have worked with the following parent functions: constant, linear, quadratic, cubic, polynomial, square root, cube root, and exponential.

Apply the idea

Looking at the graphed function to the left of the y-axis we can see it is a straight line with a constant rate of change.

Looking at the graphed function to the right of the y-axis, we can see it is some form of a cube root function.

The piecewise function was created using a linear function and a cube root function.

Reflect and check

To take this one step further, we can use transformations to identify the equations that make up the piecewise function.

The linear equation as a y-intercept of 2 and a slope of \dfrac{1}{4}, and it is defined when x<0.

The cube root function has been translated 3 units right and is defined when x\geq 0.

Therefore, the equation of the piecewise function is:f\left(x\right) = \begin{cases} \sqrt[3]{x-3}, & \enspace x \geq 0 \\ \dfrac{x}{4} + 2 , & \enspace x < 0 \end{cases}

Find all zeros and intercepts of the piecewise function.

Worked Solution

Create a strategy

To find all zeros and intercepts of the piecewise function, we will analyze each piece of the function separately. Zeros of the function occur where f\left(x\right)=0, while intercepts involve finding points where the function crosses the x-axis (zeros) and the y-axis (y-intercept).

Apply the idea

Observing the graph of the piecewise function:

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Zeros:

x=-8, and x=3

y-intercept:

y=\sqrt[3]{-3}

Reflect and check

From the graph, notice that the linear portion does not include the point \left(0,2\right) since the circle is unfilled. This is due to the domain constraint x<0, which is why it is not included as a y-intercept.

Example 4

Compare the domain, range, and intercepts for each pair of functions.

f\left(x\right)=\sqrt{x+2}

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Worked Solution

Create a strategy

We can use our knowledge of the square root function family to determine the domain, range, and intercepts of f\left(x\right).

We can use the given graph of g\left(x\right) to identify its domain, range, and intercepts.

Apply the idea

We can use the fact that the radicand must be non-negative to find the domain of f\left(x\right):\begin{aligned}&x+2\geq 0\\&x\geq -2\end{aligned}

f\left(x\right) has a domain of \left[-2,\infty\right)
g\left(x\right) has a domain of \left(-\infty,0\right]

The domains are very different from each other. The only similarity is that both domains contain x\in\left[-2,0\right].

The square root of a real number is always non-negative, so the range of f\left(x\right) is f\left(x\right)\geq 0.

f\left(x\right) has a range of \left[0,\infty\right)
g\left(x\right) has a range of \left(-\infty,0\right]

The ranges are almost opposite of one another, but they both contain the value y=0.

To find the x-intercept of f\left(x\right), we have to substitute f\left(x\right)=0 into the equation and solve for x.\begin{aligned}0&=\sqrt{x+2}\\0^2&=x+2\\-2&=x\end{aligned}

f\left(x\right) has an x-intercept at (-2,0)
g\left(x\right) has an x-intercept at (0,0)

The x-intercept of f\left(x\right) is 2 units to the left of the x-intercept of g\left(x\right).

To find the y-intercept of f\left(x\right), we have to substitute x=0 into the equation and solve for y.f\left(0\right)=\sqrt{0+2}=\sqrt{2}

f\left(x\right) has a y-intercept at \left(0,\sqrt{2}\right)
g\left(x\right) has a y-intercept at \left(0,0\right)

The y-intercept of f\left(x\right) is \sqrt{2} units higher than the y-intercept of g\left(x\right).

f\left(x\right)=-\sqrt{x-1}-2

g\left(x\right)=\sqrt[3]{x}+2

Worked Solution

Create a strategy

We can compare the domain, range, and intercepts for each function algebraically or graphically. Let's graph the functions to help us visualize the functions, then use the graphs to compare the key features.

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Apply the idea

Looking at our graphs, we see

f\left(x\right) has a domain of \left[1,\infty\right)
g\left(x\right) has a domain of \left(-\infty,\infty\right)

The function g\left(x\right) has a domain of all real numbers while f\left(x\right) has a restricted domain. The only similarity is that both domains contain x\in\left[1,\infty\right).

Now, let's find the range using our graphs:

f\left(x\right) has a range of \left(-\infty,-2\right]
g\left(x\right) has a range of \left(-\infty,\infty\right)

The function g\left(x\right) has a range of all real numbers while f\left(x\right) has a restricted range. The only similarity is that both ranges contain y\in\left(-\infty,-2\right].

Now, let's compare the y-intercepts.

f\left(x\right) does not have a y-intercept
g\left(x\right) has a y-intercept at (0,2)

The function f\left(x\right) does not have a y-intercept, but g\left(x\right) does have a y-intercept.

Finally, we will compare the x-intercepts.

f\left(x\right) does not have an x-intercept
g\left(x\right) has an x-intercept at (-8,0)

The functions are clearly different in terms of x-intercepts. f\left(x\right) does not have an x-intercept while g\left(x\right) does.

Reflect and check

We could have also found the domain, range, and intercepts algebraically. Recall our functions:

f\left(x\right)=-\sqrt{x-1}-2

g\left(x\right)=\sqrt[3]{x}+2

Domain

We can use the fact that the radicand of a square root function must be non-negative to find the domain of f\left(x\right):\begin{aligned}&x-1\geq 0\\&x\geq 1\end{aligned}So, f\left(x\right) has a domain of \left[1,\infty\right).

Since g\left(x\right) is a cube root function, there are no domain restrictions. In other words, g\left(x\right) has a domain of \left(-\infty,\infty\right).

Range

To find the range of f\left(x\right), we can compare the transformations of the graph from the parent function, y=\sqrt{x}, which has a domain of \left[0,\infty\right). Our function, f\left(x\right)=-\sqrt{x-1}-2, has been reflected over the x-axis and translated right 1 unit and down 2 units. The range will be affected by the reflection and vertical translation, but not the horizontal translation.

The reflection will cause the graph to flip, which changes the range to \left(-\infty,0\right]. The translation down 2 units will shift the range, so it becomes \left(-\infty,-2\right].

Since g\left(x\right) is a cube root function, there is no restriction on its range. In other words, the range of g\left(x\right) is all real numbers.

Intercepts

To find the x-intercept of f\left(x\right), we have to substitute f\left(x\right)=0 into the equation and solve for x.\begin{aligned}0&=-\sqrt{x-1}-2\\2&=-\sqrt{x-1}\\-2&=\sqrt{x-1}\end{aligned}

There is no value that, when you take the square root, will result in a negative number. This equation has no solution, so f\left(x\right) has no x-intercept

To find the x-intercept of g\left(x\right), we have to substitute g\left(x\right)=0 into the equation and solve for x.\begin{aligned}0&=\sqrt[3]{x}+2\\-2&=\sqrt[3]{x}\\-8&=x\end{aligned}We have found out that g\left(x\right) has an x-intercept at (-8,0)

To find the y-intercept of f\left(x\right), we have to substitute x=0 into the equation and solve for y.f(0)=-\sqrt{0-1}-2\\-\sqrt{-1}-2We cannot take the square root of a negative number, so f\left(x\right) has no y-intercept.

To find the y-intercept of g\left(x\right), we have to substitute x=0 into the equation and solve for y.g(0)=\sqrt[3]{0}+2=2The y-intercept for g\left(x\right) is (0,2).

Idea summary

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Graph of y=\sqrt{x}

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Graph of y=\sqrt[3]{x}

The graphs of the square root and cube root parent functions are similar for x>0, but the domain of the square root function does not include negative values and the domain of the cube root function does.

Radical functions can be transformed in the following ways:

	Square root	Cube root
\text{Parent function:}	y= \sqrt{x}	y= \sqrt[3]{x}
\text{Reflection across the }x\text{-axis:}	y=-\sqrt{x}	y=-\sqrt[3]{x}
\text{Reflection across the }y\text{-axis:}	y=\sqrt{-x}	y=\sqrt[3]{-x}
\text{Vertical stretch when } \left\|a\right\|>1 \\ \text{Vertical compression when } 0<\left\|a\right\|<1 \text{:}	y=a\sqrt{x}	y=a\sqrt[3]{x}
\text{Horziontal compression when } \left\|b\right\|>1 \\ \text{Horizontal stretch when } 0<\left\|b\right\|<1 \text{:}	y=\sqrt{bx}	y=\sqrt[3]{bx}
\text{Horizontal translation by } h \\ \text{Vertical translation by } k \text{:}	y=\sqrt{x-h} + k	y=\sqrt[3]{x-h} + k
(h, k) \text{:}	\text{Endpoint}	\text{Point of inflection}

The domain and range of the square root function will change with a reflection, or as h or k changes, while the domain and range of the cube root function will continue to be all real numbers.

The absolute extremum of square root function will change location when translated. If there are no reflections, the endpoint of the domain is an absolute minimum. If a vertical reflection occurs, it becomes an absolute maximum.

Square root functions do not have a relative extremum, and cube root functions have neither absolute nor relative extrema.

Outcomes

A2.F.1

The student will investigate, analyze, and compare square root, cube root, rational, exponential, and logarithmic function families, algebraically and graphically, using transformations.

A2.F.1a

Distinguish between the graphs of parent functions for square root, cube root, rational, exponential, and logarithmic function families.

A2.F.1b

Write the equation of a square root, cube root, rational, exponential, and logarithmic function, given a graph, using transformations of the parent function, including f(x) + k; f(kx); f(x + k); and kf(x), where k is limited to rational values. Transformations of exponential and logarithmic functions, given a graph, should be limited to a single transformation.

A2.F.1c

Graph a square root, cube root, rational, exponential, and logarithmic function, given the equation, using transformations of the parent function including f(x) + k; f(kx); f(x + k); and kf(x), where k is limited to rational values. Use technology to verify transformations of the functions.

A2.F.1e

Compare and contrast the graphs, tables, and equations of square root, cube root, rational, exponential, and logarithmic functions.

A2.F.2

The student will investigate and analyze characteristics of square root, cube root, rational, polynomial, exponential, logarithmic, and piecewise-defined functions algebraically and graphically.

A2.F.2a

Determine and identify the domain, range, zeros, and intercepts of a function presented algebraically or graphically, including graphs with discontinuities.

A2.F.2b

Compare and contrast the characteristics of square root, cube root, rational, polynomial, exponential, logarithmic, and piecewise-defined functions.

A2.F.2c

Determine the intervals on which the graph of a function is increasing, decreasing, or constant.

A2.F.2d

Determine the location and value of absolute (global) maxima and absolute (global) minima of a function.

A2.F.2f

For any value, x, in the domain of f, determine f(x) using a graph or equation. Explain the meaning of x and f(x) in context, where applicable.

A2.F.2g

Describe the end behavior of a function.

4.05 Radical functions

Ideas

Radical functions

Exploration

Examples

Example 1

Example 2

Example 3

Example 4

Outcomes

A2.F.1

A2.F.1a

A2.F.1b

A2.F.1c

A2.F.1e

A2.F.2

A2.F.2a

A2.F.2b

A2.F.2c

A2.F.2d

A2.F.2f

A2.F.2g

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