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# 7.4: Graphing Square Root Functions

Difficulty Level: At Grade Created by: CK-12
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Practice Graphs of Square Root Functions

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Mrs. Garcia has assigned her student the function y=x+23\begin{align*}y = -\sqrt{x + 2} - 3\end{align*} to graph for homework. The next day, she asks her students which quadrant(s) their graph is in.

Alendro says that because it is a square root function, it can only have positive values and therefore his graph is only in the first quadrant.

Dako says that because of the two negative sign, all y values will be positive and therefore his graph is in the first and second quadrants.

Marisha says they are both wrong. Because it is a negative square root function, her graph is in the third and fourth quadrants.

Which one of them is correct?

### Graphing Square Root Functions

A square root function has the form y=axh+k\begin{align*}y=a \sqrt{x-h}+k\end{align*}, where y=x\begin{align*}y=\sqrt{x}\end{align*} is the parent graph. Graphing the parent graph, we have:

x y
16 4
9 3
4 2
1 1
0 0
-1 und

Notice that this shape is half of a parabola, lying on its side. For y=x\begin{align*}y= \sqrt{x}\end{align*}, the output is the same as the input of y=x2\begin{align*}y=x^2\end{align*}. The domain and range of y=x\begin{align*}y= \sqrt{x}\end{align*} are all positive real numbers, including zero. x\begin{align*}x\end{align*} cannot be negative because you cannot take the square root of a negative number.

Let's graph y=x2+5\begin{align*}y= \sqrt{x-2}+5\end{align*} without a calculator.

To graph this function, draw a table. x=2\begin{align*}x=2\end{align*} is a critical value because it makes the radical zero.

x y
2 5
3 6
6 7
11 8

After plotting the points, we see that the shape is exactly the same as the parent graph. It is just shifted up 5 and to the right 2. Therefore, we can conclude that h\begin{align*}h\end{align*} is the horizontal shift and k\begin{align*}k\end{align*} is the vertical shift.

The domain is all real numbers such that x2\begin{align*}x \ge 2\end{align*} and the range is all real numbers such that y5\begin{align*}y \ge 5\end{align*}.

Now's let's graph y=3x+1\begin{align*}y=3 \sqrt{x+1}\end{align*} and find the domain and range.

From the previous problem, we already know that there is going to be a horizontal shift to the left one unit. The 3 in front of the radical changes the width of the function. Let’s make a table.

x y
1\begin{align*}-1\end{align*} 0
0 3
3 6
8 9
15 12

Notice that this graph grows much faster than the parent graph. Extracting (h,k)\begin{align*}(h, k)\end{align*} from the equation, the starting point is (1,0)\begin{align*}(-1, 0)\end{align*} and then rather than increase at a “slope” of 1, it is three times larger than that.

Finally, let's graph f(x)=x2+3\begin{align*}f(x)=- \sqrt{x-2}+3\end{align*}.

Extracting (h,k)\begin{align*}(h, k)\end{align*} from the equation, we find that the starting point is (2,3)\begin{align*}(2, 3)\end{align*}. The negative sign in front of the radical indicates a reflection. Let’s make a table. Because the starting point is (2,3)\begin{align*}(2, 3)\end{align*}, we should only pick x\begin{align*}x\end{align*}-values after x=2\begin{align*}x=2\end{align*}.

x y
2 3
3 2
6 1
11 0
18 -1

The negative sign in front of the radical, we now see, results in a reflection over x\begin{align*}x\end{align*}-axis.

Using the graphing calculator: If you wanted to graph this function using the TI-83 or 84, press Y=\begin{align*}Y=\end{align*} and clear out any functions. Then, press the negative sign, (-) and 2nd x2\begin{align*}x^2\end{align*}, which is \begin{align*}\sqrt{\;\;}\end{align*}. Then, type in the rest of the function, so that Y=(X2)+3\begin{align*}Y=- \sqrt{\;\;}(X-2)+3\end{align*}. Press GRAPH and adjust the window.

### Examples

#### Example 1

Earlier, you were asked to determine which student was correct.

If you graph the function y=x+23\begin{align*}y = -\sqrt{x + 2} - 3\end{align*}, you will see that its domain is x2\begin{align*}x \ge -2\end{align*}, which makes all of the quadrants possibilities. But its range is y3\begin{align*}y \le -3\end{align*}, limiting the graph to the third and fourth quadrants. Therefore, Marisha is correct.

#### Example 2

Evaluate y=2x5+8\begin{align*}y=-2 \sqrt{x-5}+8\end{align*} when x=9\begin{align*}x=9\end{align*}.

Plug in x=9\begin{align*}x=9\end{align*} into the equation and solve for y\begin{align*}y\end{align*}.

y=295+8=24+8=2(2)+8=4+8=4\begin{align*}y=-2 \sqrt{9-5}+8=-2 \sqrt{4}+8=-2(2)+8=-4+8=-4\end{align*}

Graph the following square root functions. Describe the relationship to the parent graph and find the domain and range. Use a graphing calculator for Example 5.

#### Example 3

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

Here, the negative is under the radical. This graph is a reflection of the parent graph over the y\begin{align*}y\end{align*}-axis.

The domain is all real numbers less than or equal to zero. The range is all real numbers greater than or equal to zero.

#### Example 4

Graph f(x)=12x+3\begin{align*}f(x)= \frac{1}{2} \sqrt{x+3}\end{align*}.

The starting point of this function is (3,0)\begin{align*}(-3, 0)\end{align*} and it is going to “grow” half as fast as the parent graph.

The domain is all real numbers greater than or equal to -3. The range is all real numbers greater than or equal to zero.

#### Example 5

Graph f(x)=4x5+1\begin{align*}f(x)=-4 \sqrt{x-5}+1\end{align*}.

Using the graphing calculator, the function should be typed in as: Y=4(X5)+1\begin{align*}Y=-4 \sqrt{\;\;}(X-5) + 1\end{align*}. It will be a reflection over the x\begin{align*}x\end{align*}-axis, have a starting point of (5,1)\begin{align*}(5, 1)\end{align*} and grow four times as fast as the parent graph.

### Review

Evaluate the function, f(x)=x4+3\begin{align*}f(x)=-\sqrt{x-4}+3\end{align*} for the following values of x.

1. f(3)\begin{align*}f(3)\end{align*}
2. f(6)\begin{align*}f(6)\end{align*}
3. f(13)\begin{align*}f(13)\end{align*}
4. What is the domain of this function?

Graph the following square root functions and find the domain and range. Use your calculator to check your answers.

1. f(x)=x+2\begin{align*}f(x)=\sqrt{x+2}\end{align*}
2. y=x52\begin{align*}y=\sqrt{x-5}-2\end{align*}
3. \begin{align*}y=-2 \sqrt{x+1}\end{align*}
4. \begin{align*}f(x)=1+ \sqrt{x-3}\end{align*}
5. \begin{align*}f(x)=\frac{1}{2} \sqrt{x+8}\end{align*}
6. \begin{align*}f(x)=3 \sqrt{x+6}\end{align*}
7. \begin{align*}y=2 \sqrt{1-x}\end{align*}
8. \begin{align*}y=\sqrt{x+3}-5\end{align*}
9. \begin{align*}f(x)=4 \sqrt{x+9}-8\end{align*}
10. \begin{align*}y=- \frac{3}{2} \sqrt{x-3}+6\end{align*}
11. \begin{align*}y=-3 \sqrt{5-x}+7\end{align*}
12. \begin{align*}f(x)=2 \sqrt{3-x}-9\end{align*}

To see the Review answers, open this PDF file and look for section 7.4.

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### Vocabulary Language: English

TermDefinition
General Equation for a Square Root Function The general equation for a square root function is $f(x)=a \sqrt{x-h}+k$ where $h$ is the horizontal shift and $k$ is the vertical shift.
square root function A square root function is a function with the parent function $y=\sqrt{x}$.
starting point The starting point is the initial point of a square root function, $(h, k)$.

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