# 11.1: Graphs of Square Root Functions

**Basic**Created by: CK-12

**Practice**Graphs of Square Root Functions

Suppose a clockmaker is trying to decide how large she should make the circular face of a clock. She knows that the radius of a circle as a function of its area is \begin{align*}r(a)=\frac{\sqrt{a}}{1.77}\end{align*}. How would she graph this function? In what quadrant(s) would the graph be located? Could she use a graphing calculator to graph the function? In this Concept, you'll learn how to graph square root functions like the one in this problem.

### Guidance

You have used squared roots many times in this text: to simplify, to evaluate, and to solve. This Concept will focus on the graph of the square root function.

The square root function is defined by \begin{align*}f(x)=\sqrt{x-h}+k\end{align*}, where \begin{align*}x-h\ge0\end{align*} and \begin{align*}(h,k)\end{align*} represents the origin of the curve.

#### Example A

The graph of the **parent function** \begin{align*}f(x)=\sqrt{x}\end{align*} is shown below. The function is not defined for negative values of \begin{align*}x\end{align*}; you cannot take the square root of a negative number and get a real value.

By shifting the square root function around the coordinate plane, you will change the origin of the curve.

#### Example B

Graph \begin{align*}f(x)=\sqrt{x}+4\end{align*} and compare it to the parent function.

**Solution:**

This graph has been shifted vertically upward four units from the parent function, \begin{align*}f(x)=\sqrt{x}\end{align*}. The graph is shown below.

**Graphing Square Root Functions Using a Calculator**

Graphing square root functions is similar to graphing linear, quadratic, or exponential functions.

#### Example C

Graph the parent function \begin{align*}f(x)=\sqrt{x}\end{align*} using the following steps.

### Video Review

### Guided Practice

*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 an area of \begin{align*}150 \ in^2\end{align*}?*

**Solution:**

Since the diagonal of the screen is the hypotenuse of a right triangle whose legs are the width and height, we can find the ratio of the diagonal using the Pythagorean Theorem:

\begin{align*}\text{The Pythagorean Theorem:} && a^2+b^2 &=c^2\\ \text{Substitute in the values of the width and height:} && 2.39^2 + 1^2 &=c^2\\ \text{Simplify:} &&6.71&\approx c^2 \\ \text{Take the square root of each side:}&& \sqrt{6.71}&\approx \sqrt{c^2}\Rightarrow 2.59\approx c\end{align*}

The ratio of the width to the height of the screen is 2.39 to 1. This means that any multiple of those numbers could be the actual lengths. We can write an expression for the width, \begin{align*}2.39x\end{align*}, and an expression for the height, \begin{align*}1x=x\end{align*}. From these expressions, the area is:

\begin{align*}A=w\cdot h=2.39x\cdot x=2.39x^2\end{align*}

We need a function where the area \begin{align*}A\end{align*} is the input, and the diagonal length is the output. Since the ratio of the diagonal is 2.59, then it's length is \begin{align*}2.59x\end{align*}.

If we create the function \begin{align*}f(x)=2.59x\end{align*}, this represents the diagonal of the screen as a function of \begin{align*}x\end{align*}, which is the height of the screen. We simply need to use the equation for area, but solve for \begin{align*}x\end{align*}:

\begin{align*}\text{Start with the equation for the area.} && A&=2.39x^2 \\ \text{Isolate x by dividing both sides by 2.39.} && \frac{A}{2.39}&=x^2\\ \text{Take the square root of each side.} && \sqrt{\frac{A}{2.39}}&=x\\ \text{Simplify by applying the square root to the top and bottom of the fraction.} && \frac{\sqrt{A}}{1.55}&=x\end{align*}

Now, substitute in \begin{align*}\frac{\sqrt{A}}{1.55}=x\end{align*} into \begin{align*}f(x)=2.59x\end{align*}:

\begin{align*} f(A)=2.59\left(\frac{\sqrt{A}}{1.55}\right)=1.67\sqrt{A}.\end{align*}

This is a function that takes the input \begin{align*}A\end{align*}, the area in square inches, and outputs \begin{align*}f(A)\end{align*}, the length of the diagonal in inches.

Graph the function and use it to find the length of the diagonal when the area is \begin{align*}150 \ in^2\end{align*}.

Find the value of the function when \begin{align*}A=150\end{align*}. It looks like the value of the function is between 20 and 21.

This is hard to see exactly from the graph. You can graph the function on your calculator and use the **TRACE** button to find the exact value. First, you must enter the function by pressing the **Y=** button, and then press **WINDOW** to set the range of \begin{align*}X\end{align*} and the range of \begin{align*}Y\end{align*} so that they include the point where \begin{align*}X=150\end{align*}.

Looking at the TRACE from the graphing calculator, you can see that:

\begin{align*}f(150)\approx 20.45\end{align*}.

### Practice

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Graphs of Square Root Functions (15:01)

- In the definition of a square root function, why must \begin{align*}(x-h) \ge 0\end{align*}?
- What is the domain and range of the parent function \begin{align*}f(x)=\sqrt{x}\end{align*}?

Identify the ordered pair of the origin of each square root function.

- \begin{align*}f(x)=\sqrt{x-2}\end{align*}
- \begin{align*}g(x)=\sqrt{x+4}+6\end{align*}
- \begin{align*}h(x)=\sqrt{x-1}-1\end{align*}
- \begin{align*}y=\sqrt{x}+3\end{align*}
- \begin{align*}f(x)=\sqrt{2x}+4\end{align*}

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*}

In 12-20, graph the function.

- \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 length between any two consecutive bases of a baseball diamond is 90 feet. How much shorter is it for the catcher to walk along the diagonal from home plate to second base than the runner running from second to home?
- The units of acceleration of gravity are 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 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 for 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 is \begin{align*}g=9.819 \ m/s^2\end{align*}, in Los Angeles the value is \begin{align*}g=9.796 \ m/s^2\end{align*}, and in Mexico City the value is \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.

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*}

**Mixed Review**

- Solve \begin{align*}16=2x^2-3x+4\end{align*}.
- Write an equation for a line with a slope of 0.2 containing the point (1, 10).
- Are these lines parallel, perpendicular, or neither: \begin{align*}x+5y=16\end{align*} and \begin{align*}y=5x-3\end{align*}?
- Which of the following vertices minimizes the expression \begin{align*}20x+32y\end{align*}?
- (50, 0)
- (0, 60)
- (15, 30)

- Is the following graph a function? Explain your reasoning.
- Between which two consecutive integers is \begin{align*}\sqrt{205}\end{align*}?

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### Image Attributions

Here you'll learn about square root functions and their graphs.