Mrs. Garcia has assigned her student the function \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?
Guidance
A square root function has the form @$\begin{align*}y=a \sqrt{x-h}+k\end{align*}@$ , where @$\begin{align*}y=\sqrt{x}\end{align*}@$ 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 @$\begin{align*}y= \sqrt{x}\end{align*}@$ , the output is the same as the input of @$\begin{align*}y=x^2\end{align*}@$ . The domain and range of @$\begin{align*}y= \sqrt{x}\end{align*}@$ are all positive real numbers, including zero. @$\begin{align*}x\end{align*}@$ cannot be negative because you cannot take the square root of a negative number.
Example A
Graph @$\begin{align*}y= \sqrt{x-2}+5\end{align*}@$ without a calculator.
Solution: To graph this function, draw a table. @$\begin{align*}x=2\end{align*}@$ 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 @$\begin{align*}h\end{align*}@$ is the horizontal shift and @$\begin{align*}k\end{align*}@$ is the vertical shift .
The domain is all real numbers such that @$\begin{align*}x \ge 2\end{align*}@$ and the range is all real numbers such that @$\begin{align*}y \ge 5\end{align*}@$ .
Example B
Graph @$\begin{align*}y=3 \sqrt{x+1}\end{align*}@$ . Find the domain and range.
Solution: From the previous example, 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 |
---|---|
@$\begin{align*}-1\end{align*}@$ | 0 |
0 | 3 |
3 | 6 |
8 | 9 |
15 | 12 |
Notice that this graph grows much faster than the parent graph. Extracting @$\begin{align*}(h, k)\end{align*}@$ from the equation, the starting point is @$\begin{align*}(-1, 0)\end{align*}@$ and then rather than increase at a “slope” of 1, it is three times larger than that.
Example C
Graph @$\begin{align*}f(x)=- \sqrt{x-2}+3\end{align*}@$ .
Solution: Extracting @$\begin{align*}(h, k)\end{align*}@$ from the equation, we find that the starting point is @$\begin{align*}(2, 3)\end{align*}@$ . The negative sign in front of the radical indicates a reflection. Let’s make a table. Because the starting point is @$\begin{align*}(2, 3)\end{align*}@$ , we should only pick @$\begin{align*}x\end{align*}@$ -values after @$\begin{align*}x=2\end{align*}@$ .
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 @$\begin{align*}x\end{align*}@$ -axis.
Using the graphing calculator: If you wanted to graph this function using the TI-83 or 84, press @$\begin{align*}Y=\end{align*}@$ and clear out any functions. Then, press the negative sign, (-) and 2nd @$\begin{align*}x^2\end{align*}@$ , which is @$\begin{align*}\sqrt{\;\;}\end{align*}@$ . Then, type in the rest of the function, so that @$\begin{align*}Y=- \sqrt{\;\;}(X-2)+3\end{align*}@$ . Press GRAPH and adjust the window.
Intro Problem Revisit If you graph the function @$\begin{align*}y = -\sqrt{x + 2} - 3\end{align*}@$ , you will see that its domain is @$\begin{align*}x \ge -2\end{align*}@$ , which makes all of the quadrants possibilities. But its range is @$\begin{align*}y \le -3\end{align*}@$ , limiting the graph to the third and fourth quadrants. Therefore, Marisha is correct.
Guided Practice
1. Evaluate @$\begin{align*}y=-2 \sqrt{x-5}+8\end{align*}@$ when @$\begin{align*}x=9\end{align*}@$ .
Graph the following square root functions. Describe the relationship to the parent graph and find the domain and range. Use a graphing calculator for #3.
2. @$\begin{align*}y=\sqrt{-x}\end{align*}@$
3. @$\begin{align*}f(x)= \frac{1}{2} \sqrt{x+3}\end{align*}@$
4. @$\begin{align*}f(x)=-4 \sqrt{x-5}+1\end{align*}@$
Answers
1. Plug in @$\begin{align*}x=9\end{align*}@$ into the equation and solve for @$\begin{align*}y\end{align*}@$ .
@$\begin{align*}y=-2 \sqrt{9-5}+8=-2 \sqrt{4}+8=-2(2)+8=-4+8=-4\end{align*}@$
2. Here, the negative is under the radical. This graph is a reflection of the parent graph over the @$\begin{align*}y\end{align*}@$ -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.
3. The starting point of this function is @$\begin{align*}(-3, 0)\end{align*}@$ 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.
4. Using the graphing calculator, the function should be typed in as: @$\begin{align*}Y=-4 \sqrt{\;\;}(X-5) + 1\end{align*}@$ . It will be a reflection over the @$\begin{align*}x\end{align*}@$ -axis, have a starting point of @$\begin{align*}(5, 1)\end{align*}@$ and grow four times as fast as the parent graph.
Explore More
Evaluate the function, @$\begin{align*}f(x)=-\sqrt{x-4}+3\end{align*}@$ for the following values of x .
- @$\begin{align*}f(3)\end{align*}@$
- @$\begin{align*}f(6)\end{align*}@$
- @$\begin{align*}f(13)\end{align*}@$
- 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.
- @$\begin{align*}f(x)=\sqrt{x+2}\end{align*}@$
- @$\begin{align*}y=\sqrt{x-5}-2\end{align*}@$
- @$\begin{align*}y=-2 \sqrt{x+1}\end{align*}@$
- @$\begin{align*}f(x)=1+ \sqrt{x-3}\end{align*}@$
- @$\begin{align*}f(x)=\frac{1}{2} \sqrt{x+8}\end{align*}@$
- @$\begin{align*}f(x)=3 \sqrt{x+6}\end{align*}@$
- @$\begin{align*}y=2 \sqrt{1-x}\end{align*}@$
- @$\begin{align*}y=\sqrt{x+3}-5\end{align*}@$
- @$\begin{align*}f(x)=4 \sqrt{x+9}-8\end{align*}@$
- @$\begin{align*}y=- \frac{3}{2} \sqrt{x-3}+6\end{align*}@$
- @$\begin{align*}y=-3 \sqrt{5-x}+7\end{align*}@$
- @$\begin{align*}f(x)=2 \sqrt{3-x}-9\end{align*}@$