# 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’ll learn about a different kind of function called the square root function. You’ve 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*}

A square root function is any function with the form: \begin{align*}y = a \sqrt{f(x)} + c\end{align*}

## Graph and Compare Square Root Functions

When working with square root functions, you’ll have to consider the domain of the function before graphing. The domain is very important because the function is undefined when the expression inside the square root sign is negative, and as a result there will be no graph in whatever region of \begin{align*}x-\end{align*}

To discover how the graphs of square root functions behave, let’s 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. Since the expression inside the square root is just \begin{align*}x\end{align*}

This means that when we make our table of values, we should pick values of \begin{align*}x\end{align*}

\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} = 2.4\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, and in fact it is. It’s half of the parabola that you would get if you graphed the expression \begin{align*}y^2 = x\end{align*}

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 2**

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

**Solution**

Here is 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 3**

*Graph the functions* \begin{align*}y=\sqrt{x}, y= \sqrt{2x}, y=\sqrt{3x},\end{align*}*and* \begin{align*}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 multiplying by a constant *outside* the square root; the function just 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*}

**Example 4**

*Graph the functions* \begin{align*}y=\sqrt{x}, y = \frac{1}{2} \sqrt{x}, y = \frac{1}{3} \sqrt{x},\end{align*}*and* \begin{align*}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 more slowly the smaller the constant is.

**Example 5**

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

**Solution**

If we multiply the whole function by -1, the graph is reflected about the \begin{align*}x-\end{align*}

**Example 6**

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

**Solution**

On the other hand, when just the \begin{align*}x\end{align*}

**Example 7**

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

**Solution**

When we add a constant to the right-hand side of the equation, the graph keeps the same shape, but shifts up for a positive constant or down for a negative one.

**Example 8**

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

**Solution**

When we add a constant to the **argument** of the function (the part under the radical sign), the function shifts to the left for a positive constant and to the right for a negative constant.

Now let’s see how to combine all of the above types of transformations.

**Example 9**

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

**Solution**

We can 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 that function looks like this:

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*}\sqrt{3}\end{align*}.

Next, when we subtract 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, because \begin{align*}y\end{align*} increases faster by a factor of 2.

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

Each step of this process is shown in the graph below. The purple line shows the final result.

Now we know how to graph square root 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.

## Solve Real-World Problems Using Square Root Functions

Mathematicians and physicists have studied the motion of pendulums 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 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 doesn’t 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 complete one whole back-and-forth swing 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, \begin{align*}a\end{align*}, depends on the acceleration of gravity: \begin{align*}a = \frac{2 \pi}{\sqrt{g}}\end{align*}. At sea level on Earth, 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 \begin{align*}20^{th}\end{align*} century, all clocks used pendulums as their central time keeping component.

**Example 10**

*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 start by making a table of values for this function:

\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*}y = 2 \sqrt{2} = 2.8\end{align*} |

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

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

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

Now let's graph the function. It makes sense to let the horizontal axis represent the length of the pendulum and the vertical axis represent the period of the pendulum.

We can see from the graph that a length of approximately \begin{align*}\frac{1}{4}\end{align*} meters gives a period of 1 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*}

For more equations that describe pendulum motion, check out http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html, where you can also find a tool for calculating the period of a pendulum in different gravities than Earth’s.

**Example 11**

*“Square” TV screens have an aspect ratio of 4:3; in other words, the width of the screen is* \begin{align*}\frac{4}{3}\end{align*} *the height. TV “sizes” are traditionally represented as 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*} width

Then 4 \begin{align*}\times\end{align*} height = 3 \begin{align*}\times\end{align*} width

Or, height = \begin{align*}\frac{3}{4}x\end{align*}.

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

Find how the diagonal length relates to the width by using the Pythagorean theorem:

\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 = \frac{4}{5}d\end{align*}

Therefore, the diagonal length relates to the area as follows: \begin{align*} 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 flip that around to 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*}.

Now we can make a graph where the horizontal axis represents the area of the television screen and the vertical axis is the length of the diagonal. First 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 180 \begin{align*}in^2\end{align*} the length of the diagonal is approximately 19.5 inches. We can confirm this by plugging in \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 + \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 2 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 3 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 for what length, in meters, will the period of a pendulum be 8 seconds in 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*}