# 11.1: Graphs of Square Root Functions

**At Grade**Created by: CK-12

## Learning Objectives

- Graph and compare square root functions.
- Shift graphs of square root functions.
- Graph square root functions using a graphing calculator.
- Solve real-world problems using square root functions.

## Introduction

In this chapter, you will be learning about a different kind of function called the **square root function**. You have seen that taking the square root is very useful in solving quadratic equations. For example, to solve the equation \begin{align*}x^2=25\end{align*} we take the square root of both sides \begin{align*} \sqrt{x^2}=\pm \sqrt{25}\end{align*} and obtain \begin{align*}x= \pm 5\end{align*}. A square root function has the form \begin{align*} y=\sqrt{f(x)}\end{align*}. In this type of function, the expression in terms of \begin{align*}x\end{align*} is found inside the square root sign (also called the “radical” sign).

## Graph and Compare Square Root Functions

The square root function is the first time where you will have to consider the domain of the function before graphing. The domain is very important because the function is undefined if the expression inside the square root sign is negative, and as a result there will be no graph in that region.

In order to is cover how the graphs of square root function behave, we should make a table of values and plot the points.

**Example 1**

*Graph the function \begin{align*} y=\sqrt{x}\end{align*}.*

**Solution**

Before we make a table of values, we need to find the domain of this square root function. The domain is found by realizing that the function is only defined when the expression inside the square root is greater than or equal to zero. We find that the domain is all values of \begin{align*}x\end{align*} such that \begin{align*}x \ge 0\end{align*}.

This means that when we make our table of values, we should pick values of \begin{align*}x\end{align*} that are greater than or equal to zero. It is very useful to include the value of zero as the first value in the table and include many values greater than zero. This will help us in determining what the shape of the curve will be.

\begin{align*}x\end{align*} | \begin{align*}y=\sqrt{x}\end{align*} |
---|---|

0 | \begin{align*}y=\sqrt{0}=0\end{align*} |

1 | \begin{align*}y=\sqrt{1}=1\end{align*} |

2 | \begin{align*}y=\sqrt{2}=1.4\end{align*} |

3 | \begin{align*}y=\sqrt{3}=1.7\end{align*} |

4 | \begin{align*}y=\sqrt{4}=2\end{align*} |

5 | \begin{align*}y=\sqrt{5}=2.2\end{align*} |

6 | \begin{align*}y=\sqrt{6}=22.4\end{align*} |

7 | \begin{align*}y=\sqrt{7}=2.6\end{align*} |

8 | \begin{align*}y=\sqrt{8}=2.8\end{align*} |

9 | \begin{align*}y=\sqrt{9}=3\end{align*} |

Here is what the graph of this table looks like.

The graphs of square root functions are always curved. The curve above looks like half of a parabola lying on its side. In fact the square root function we graphed above comes from the expression \begin{align*}y^2=x.\end{align*}

This is in the form of a parabola but with the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} switched. We see that when we solve this expression for \begin{align*}y\end{align*} we obtain two solutions \begin{align*} y=\sqrt{x}\end{align*} and \begin{align*}y=- \sqrt{x}\end{align*}. The graph above shows the positive square root of this answer.

**Example 2**

*Graph the function* \begin{align*} y=- \sqrt{x}\end{align*}.

**Solution**

Once again, we must look at the domain of the function first. We see that the function is defined only for \begin{align*}x \ge 0\end{align*}. Let’s make a table of values and calculate a few values of the function.

\begin{align*}x\end{align*} | \begin{align*}y=-\sqrt{x}\end{align*} |
---|---|

0 | \begin{align*}y=-\sqrt{0}=-0\end{align*} |

1 | \begin{align*}y=-\sqrt{1}=-1\end{align*} |

2 | \begin{align*}y=-\sqrt{2}=-1.4\end{align*} |

3 | \begin{align*}y=-\sqrt{3}=-1.7\end{align*} |

4 | \begin{align*}y=-\sqrt{4}=-2\end{align*} |

5 | \begin{align*}y=-\sqrt{5}=-2.2\end{align*} |

6 | \begin{align*}y=-\sqrt{6}=-22.4\end{align*} |

7 | \begin{align*}y=-\sqrt{7}=-2.6\end{align*} |

8 | \begin{align*}y=-\sqrt{8}=2.8\end{align*} |

9 | \begin{align*}y=-\sqrt{9}=-3\end{align*} |

Here is the graph from this table.

Notice that if we graph the two separate functions on the same coordinate axes, the combined graph is a parabola lying on its side.

Now let's compare square root functions that are multiples of each other.

**Example 3**

*Graph functions \begin{align*} y=\sqrt{x}, y=2 \sqrt{x}, y=3 \sqrt{x}, y=4 \sqrt{x}\end{align*} on the same graph.*

**Solution**

*Here we will show just the graph without the table of values.*

If we multiply the function by a constant bigger than one, the function increases faster the greater the constant is.

**Example 4**

*Graph functions \begin{align*} y=\sqrt{x}, y=\sqrt{2x}, y=\sqrt{3x}, y=\sqrt{4x}\end{align*} on the same graph.*

**Solution**

Notice that multiplying the expression inside the square root by a constant has the same effect as in the previous example but the function increases at a slower rate because the entire function is effectively multiplied by the square root of the constant. Also note that the graph of \begin{align*}\sqrt{4x}\end{align*} is the same as the graph of \begin{align*} 2 \sqrt{2x}\end{align*}. This makes sense algebraically since \begin{align*}\sqrt{4}=2\end{align*}.

**Example 5**

*Graph functions* \begin{align*} y=\sqrt{x}, y=\frac{1}{2}\sqrt{x}, y=\frac{1}{3}\sqrt{x}, y=\frac{1}{4}\sqrt{x}\end{align*} *on the same graph.*

**Solution**

If we multiply the function by a constant between 0 and 1, the function increases at a slower rate for smaller constants.

**Example 6**

*Graph functions \begin{align*} y =2 \sqrt{x}, y=-2 \sqrt{x}\end{align*} on the same graph.*

**Solution**

If we multiply the function by a negative function, the square root function is reflected about the \begin{align*}x-\end{align*}axis.

**Example 7**

*Graph functions \begin{align*} y=\sqrt{x}, y=\sqrt{-x}\end{align*} on the same graph.*

**Solution**

Notice that for function \begin{align*} y=\sqrt{x}\end{align*} the domain is values of \begin{align*}x \ge 0\end{align*}, and for function \begin{align*} y=\sqrt{-x}\end{align*} the domain is values of \begin{align*}x \le 0\end{align*}.

When we multiply the argument of the function by a negative constant the function is reflected about the \begin{align*}y-\end{align*}axis.

## Shift Graphs of Square Root Functions

Now, let’s see what happens to the square root function as we add positive and negative constants to the function.

**Example 8**

*Graph the functions* \begin{align*} y=\sqrt{x}, y=\sqrt{x}+ 2, y=\sqrt{x}- 2\end{align*}.

**Solution**

We see that the graph keeps the same shape, but moves up for positive constants and moves down for negative constants.

**Example 9**

*Graph the functions \begin{align*} y=\sqrt{x}, y=\sqrt{x - 2}, y=\sqrt{x + 2}\end{align*}.*

**Solution**

When we add constants to the argument of the function, the function shifts to the left for a positive constant and to the right for a negative constant because the domain shifts. There can't be a negative number inside the square root.

Now let’s graph a few more examples of square root functions.

**Example 10**

*Graph the function \begin{align*} y=2 \sqrt{3x - 1}+ 2\end{align*}.*

**Solution**

We first determine the domain of the function. The function is only defined if the expression inside the square root is positive \begin{align*}3x - 1 \ge 0\end{align*} or \begin{align*}x \ge \frac{1}{3}\end{align*}.

Make a table for values of \begin{align*}x\end{align*} greater than or equal to \begin{align*}\frac{1}{3}\end{align*}.

\begin{align*}x\end{align*} | \begin{align*}y =2 \sqrt{3x - 1}+ 2\end{align*} |
---|---|

\begin{align*}\frac{1}{3}\end{align*} | \begin{align*}y =2 \sqrt{3 \cdot \frac{1}{3}- 1}+ 2=2\end{align*} |

1 | \begin{align*}y =2 \sqrt{3(1) - 1}+ 2=4.8\end{align*} |

2 | \begin{align*}y =2 \sqrt{3(2) - 1}+ 2=6.5\end{align*} |

3 | \begin{align*}y =2 \sqrt{3(3) - 1}+ 2=7.7\end{align*} |

4 | \begin{align*}y =2 \sqrt{3(4) - 1}+ 2=8.6\end{align*} |

5 | \begin{align*}y =2 \sqrt{3(5) - 1}+ 2= 9.5\end{align*} |

Here is the graph.

You can also think of this function as a combination of shifts and stretches of the basic square root function \begin{align*} y=\sqrt{x}\end{align*}. We know that the graph of this function looks like the one below.

If we multiply the argument by 3 to obtain \begin{align*} y=\sqrt{3x}\end{align*}, this stretches the curve vertically because the value of \begin{align*}y\end{align*} increases faster by a factor of \begin{align*} y=\sqrt{3}\end{align*}.

Next, when we subtract the value of 1 from the argument to obtain \begin{align*} y= \sqrt{3x - 1}\end{align*} this shifts the entire graph to the left by one unit.

Multiplying the function by a factor of 2 to obtain \begin{align*} y=2 \sqrt{3x - 1}\end{align*} stretches the curve vertically again, and \begin{align*}y\end{align*} increases faster by a factor of 2.

Finally, we add the value of 2 to the function to obtain \begin{align*} y= \sqrt{3x - 1}+ 2\end{align*}. This shifts the entire function vertically by 2 units.

This last method of graphing showed a way to graph functions without making a table of values. If we know what the basic function looks like, we can use shifts and stretches to **transform** the function and get to the desired result.

## Graph Square Root Functions Using a Graphing Calculator

Next, we will demonstrate how to use the graphing calculator to plot square root functions.

**Example 11**

*Graph the following functions using a graphing calculator.*

a) \begin{align*} y=\sqrt{x + 5}\end{align*}

b) \begin{align*} y=\sqrt{9 - x^2}\end{align*}

**Solution:**

In all the cases we start by pressing the **[Y=button]** and entering the function on the function screen of the calculator:

We then press **[GRAPH]** to display the results. Make sure your window is set appropriately in order to view the function well. This is done by pressing the **[WINDOW]** button and choosing appropriate values for the Xmin, Xmax, Ymin and Ymax.

a)

The window of this graph is \begin{align*}-6 \le x \le 5;\ -5 \le y \le 5.\end{align*}

The domain of the function is \begin{align*}x \ge -5\end{align*}

b)

The window of this graph is \begin{align*}-5 \le x \le 5; -5 \le y \le 5.\end{align*}

The domain of the function is \begin{align*}-3 \le x \le 3\end{align*}

## Solve Real-World Problems Using Square Root Functions

**Pendulum**

Mathematicians and physicists have studied the motion of a pendulum in great detail because this motion explains many other behaviors that occur in nature. This type of motion is called **simple harmonic motion** and it is very important because it describes anything that repeats periodically. Galileo was the first person to study the motion of a pendulum around the year 1600. He found that the time it takes a pendulum to complete a swing from a starting point back to the beginning does not depend on its mass or on its angle of swing (as long as the angle of the swing is small). Rather, it depends only on the length of the pendulum.

The time it takes a pendulum to swing from a starting point and back to the beginning is called the **period** of the pendulum.

Galileo found that the period of a pendulum is proportional to the square root of its length \begin{align*} T=a \sqrt{L}\end{align*}. The proportionality constant depends on the acceleration of gravity \begin{align*} a=\frac{2\pi}{\sqrt{g}}\end{align*}. At sea level on Earth the acceleration of gravity is \begin{align*}g=9.81 \ m/s^2\end{align*} (meters per second squared). Using this value of gravity, we find \begin{align*}a=2.0\end{align*} with units of \begin{align*}\frac{s}{\sqrt{m}}\end{align*} (seconds divided by the square root of meters). Up until the mid 20th century, all clocks used pendulums as their central time keeping component.

**Example 12**

*Graph the period of a pendulum of a clock swinging in a house on Earth at sea level as we change the length of the pendulum. What does the length of the pendulum need to be for its period to be one second?*

**Solution**

The function for the period of a pendulum at sea level is: \begin{align*} T=2 \sqrt{L}\end{align*}.

We make a graph with the horizontal axis representing the length of the pendulum and with the vertical axis representing the period of the pendulum.

We start by making a table of values.

\begin{align*}L\end{align*} | \begin{align*}T =2 \sqrt{L}\end{align*} |
---|---|

0 | \begin{align*}T =2 \sqrt{0}=0\end{align*} |

1 | \begin{align*}T =2 \sqrt{1}=2\end{align*} |

2 | \begin{align*}T =2 \sqrt{2}=2.8\end{align*} |

3 | \begin{align*}T =2 \sqrt{3}=3.5\end{align*} |

4 | \begin{align*}T =2 \sqrt{4}=4\end{align*} |

5 | \begin{align*}T =2 \sqrt{5}=4.5\end{align*} |

Now let's graph the function.

We can see from the graph that a length of approximately \begin{align*}\frac{1}{4}\end{align*} meters gives a period of one second. We can confirm this answer by using our function for the period and plugging in \begin{align*}T=1 \ second\end{align*}.

\begin{align*} T=2 \sqrt{L}\Rightarrow 1=2 \sqrt{L}\end{align*}

\begin{align*}\text{Square both sides of the equation:}&&1 &=4L\\ \text{Solve for} \ L: && L &=\frac{1}{4} \ meters\end{align*}

**Example 13**

*“Square” TV screens have an aspect ratio of 4:3. This means that for every four inches of length on the horizontal, there are three inches of length on the vertical. TV sizes represent the length of the diagonal of the television screen. Graph the length of the diagonal of a screen as a function of the area of the screen. What is the diagonal of a screen with an area of \begin{align*}180 \ in^2\end{align*}?*

**Solution**

Let \begin{align*}d =\end{align*} length of the diagonal, \begin{align*}x =\end{align*} horizontal length

\begin{align*}4 \cdot \text{vertical length}&=3 \cdot \text{horizontal length}\\ & \text{Or},\\ \text{vertical length}&=\frac{3}{4}x.\end{align*}

The area of the screen is: \begin{align*}A=\text{length}\cdot \text{width}\end{align*} or \begin{align*}A=\frac{3}{4}x^{2}\end{align*}

Find how the diagonal length and the horizontal length are related by using the Pythagorean theorem, \begin{align*}a^{2}+ b^{2}=c^{2}\end{align*}.

\begin{align*}x^2 + \left( \frac{3}{4}x \right )^2 &=d^2 \\ x^2 + \frac{9}{16}x^2 &=d^2 \\ \frac{25}{16}x^2 &=d^2 \Rightarrow x^2 =\frac{16}{25}d^2 \Rightarrow x^2=\frac{4}{5}d\\ A &= \frac{3}{4}\left( \frac{4}{5}d \right)^2=\frac{3}{4}\cdot \frac{16}{25}d^2=\frac{12}{25}d^2\end{align*}

We can also find the diagonal length as a function of the area \begin{align*} d^2=\frac{25}{12}A\end{align*} or \begin{align*} d=\frac{5}{2 \sqrt{3}}\sqrt{A}\end{align*}.

Make a graph where the horizontal axis represents the area of the television screen and the vertical axis is the length of the diagonal. Let’s make a table of values.

\begin{align*}A\end{align*} | \begin{align*}d=\frac{5}{2 \sqrt{3}}\sqrt{A}\end{align*} |
---|---|

0 | 0 |

25 | 7.2 |

50 | 10.2 |

75 | 12.5 |

100 | 14.4 |

125 | 16.1 |

150 | 17.6 |

175 | 19 |

200 | 20.4 |

From the graph we can estimate that when the area of a TV screen is \begin{align*}180 \ in^2\end{align*} the length of the diagonal is approximately 19.5 inches inches. We can confirm this by substituting \begin{align*}a=180\end{align*} into the formula that relates the diagonal to the area.

\begin{align*} d=\frac{5}{2 \sqrt{3}}\sqrt{A}=\frac{5}{2 \sqrt{3}}\sqrt{180}=19.4\ inches\end{align*}

## Review Questions

Graph the following functions on the same coordinate axes.

- \begin{align*} y=\sqrt{x}, y=2.5 \sqrt{x}\end{align*} and \begin{align*} y=-2.5 \sqrt{x}\end{align*}
- \begin{align*} y=\sqrt{x}, y=0.3 \sqrt{x}\end{align*} and \begin{align*} y=0.6 \sqrt{x}\end{align*}
- \begin{align*} y=\sqrt{x}, y=\sqrt{x - 5}\end{align*} and \begin{align*} y=\sqrt{x + 5}\end{align*}
- \begin{align*} y=\sqrt{x}, y=\sqrt{x}+ 8\end{align*} and \begin{align*} y=\sqrt{x}- 8\end{align*}

Graph the following functions.

- \begin{align*} y=\sqrt{2x - 1}\end{align*}
- \begin{align*} y=\sqrt{4x + 4}\end{align*}
- \begin{align*} y=\sqrt{5 - x}\end{align*}
- \begin{align*} y=2 \sqrt{x}+ 5\end{align*}
- \begin{align*} y=3 - \sqrt{x}\end{align*}
- \begin{align*} y=4 + 2 \sqrt{x}\end{align*}
- \begin{align*} y=2 \sqrt{2x + 3}+ 1\end{align*}
- \begin{align*} y=4 + 2 \sqrt{2 - x}\end{align*}
- \begin{align*} y=\sqrt{x + 1}- \sqrt{4x - 5}\end{align*}
- The acceleration of gravity can also given in feet per second squared. It is \begin{align*}g=32 \ ft/s^2\end{align*} at sea level. Graph the period of a pendulum with respect to its length in feet. For what length in feet will the period of a pendulum be two seconds?
- The acceleration of gravity on the Moon is \begin{align*}1.6 \ m/s^2\end{align*}. Graph the period of a pendulum on the Moon with respect to its length in meters. For what length, in meters, will the period of a pendulum be 10 seconds?
- The acceleration of gravity on Mars is \begin{align*}3.69 \ m/s^2\end{align*}. Graph the period of a pendulum on the Mars with respect to its length in meters. For what length, in meters, will the period of a pendulum be three seconds?
- The acceleration of gravity on the Earth depends on the latitude and altitude of a place. The value of \begin{align*}g\end{align*} is slightly smaller for places closer to the Equator than places closer to the Poles, and the value of \begin{align*}g\end{align*} is slightly smaller for places at higher altitudes that it is for places at lower altitudes. In Helsinki, the value of \begin{align*}g=9.819 \ m/s^2\end{align*}, in Los Angeles the value of \begin{align*}g=9.796 \ m/s^2\end{align*} and in Mexico City the value of \begin{align*}g=9.779 \ m/s^2\end{align*}. Graph the period of a pendulum with respect to its length for all three cities on the same graph. Use the formula to find the length (in meters) of a pendulum with a period of 8 seconds for each of these cities.
- The aspect ratio of a wide-screen TV is 2.39:1. Graph the length of the diagonal of a screen as a function of the area of the screen. What is the diagonal of a screen with area \begin{align*}150 \ in^2\end{align*}?

Graph the following functions using a graphing calculator.

- \begin{align*} y=\sqrt{3x - 2}\end{align*}
- \begin{align*} y=4 + \sqrt{2 - x}\end{align*}
- \begin{align*} y=\sqrt{x^2 - 9}\end{align*}
- \begin{align*} y=\sqrt{x}- \sqrt{x + 2}\end{align*}

## Review Answers

- \begin{align*}L=3.25 \ feet\end{align*}
- \begin{align*}L=4.05 \ meters\end{align*}
- \begin{align*}L=0.84 \ meters\end{align*}

Note: The differences are so small that all of the lines appear to coincide on this graph. If you zoom (way) in you can see slight differences. The period of an 8 meter pedulum in Helsinki is 1.8099 seconds, in Los Angeles it is 1.8142 seconds, and in Mexico City it is 1.8173 seconds.

\begin{align*}D=20.5 \ inches\end{align*}

15.92 m Helsinki

15.88 m Los Angeles

15.85 m Mexico City

- Window \begin{align*}-1 \le x \le 5; -5 \le y \le 5\end{align*}
- Window \begin{align*}-5 \le x \le 5; 0 \le y \le 10\end{align*}
- Window \begin{align*}-6 \le x \le 6; -1 \le y \le 10\end{align*}
- Window \begin{align*}0 \le x \le 5; -3 \le y \le 1\end{align*}

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