such familiar shapes. Until we learn to recognize the kinds of graphs that other equations have, we graph other equations by calculating enough ordered pairs to determine the shape of the graph. When you graph equations, try to anticipate what the graph will look like, and after the graph is drawn, pause to reflect on the shape of the graph and the type of equation that produced it. You might wish to look ahead to the Function Gallery on pages 140 and 141, which shows the basic functions that we will be studying.
Of course, a graphing calculator can speed up this process. Remember that a graphing calculator shows only finitely many points and a graph usually consists of infinitely many points. After looking at the display of a graphing calculator, you must still decide what the entire graph looks like.
Figure 2.10 y
–1 x –1
3 1
Domain (–∞, ∞)
]Range
[0,
∞)
–2 2 2 3
– 3 (0, 0)
(1, 1) (2, 4)
(–1, 1)
(– 2, 4) 4
5
y = x2
Figure 2.11
⫺3 3
⫺1 5
x is any real number
y ⱖ 0
x 0 1 -1 2 -2
y= x2 0 1 1 4 4
EXAMPLE 1 The square function
Graph the equation y =x2 and state the domain and range. Determine whether the relation is a function.
Solution
Make a table of ordered pairs that satisfy y= x2:
These ordered pairs indicate a graph in the shape shown in Fig. 2.10. The do- main is 1-⬁, ⬁2 because any real number can be used for x in y =x2. Since all y-coordinates are nonnegative, the range is 30, ⬁2. Because no vertical line crosses this curve more than once, y = x2 is a function.
The calculator graph shown in Fig. 2.11 supports these conclusions.
䉴TRY THIS. Determine whether y = 12 x2 is a function, graph it, and state the do-
main and range. ■
The graph of y = x2 is called a parabola. We study parabolas in detail in Chapter 3. The graph of the square-root function y = 1x is half of a parabola.
EXAMPLE 2 The square-root function
Graph y = 1x and state the domain and range of the relation. Determine whether the relation is a function.
Solution
Make a table listing ordered pairs that satisfy y = 1x:
x 0 1 4 9
y= 1x 0 1 2 3
y ⱖ 0 x ⱖ 0
Plotting these ordered pairs suggests the graph shown in Fig. 2.12. The domain of the relation is 30, ⬁2 and the range is 30, ⬁2. Because no vertical line can cross this graph more than once, y = 1x is a function.
The calculator graph in Fig. 2.13 supports these conclusions.
䉴TRY THIS. Determine whether y = 11 -x is a function, graph it, and state the
domain and range. ■
In the next example we graph x = y2 and see that its graph is also a parabola.
Figure 2.12
]
y
– 1 x –1
3 1
Domain [0, )
Range
[0,
∞)
2 2 3
(0, 0) (1, 1)
(4, 2)
4 5
1
]
y = √x
∞
Figure 2.13
⫺2 9
⫺1 3
Figure 2.15
⫺1 5
⫺3 3
Figure 2.14 x = y2 y
–1 x – 1
3 1
Domain [ 0, ∞)
Range
(–∞, ∞)
2 2 3
(0, 0) (1, 1)
(4, 2)
– 2 – 3
4 5
1
]
(1, –1) (4, – 2)
EXAMPLE 3 A parabola opening to the right
Graph x = y2 and state the domain and range of the relation. Determine whether the relation is a function.
Solution
Make a table listing ordered pairs that satisfy x= y2. In this case choose y and cal- culate x:
x =y2 0 1 1 4 4
y 0 1 -1 2 -2
y is any real number x ⱖ 0
Note that these ordered pairs are the same ones that satisfy y =x2 except that the coordinates are reversed. For this reason the graph of x = y2 in Fig. 2.14 has the same shape as the parabola in Fig. 2.10 and it is also a parabola. The domain of x = y2 is 30, ⬁2 and the range is 1-⬁, ⬁2. Because we can draw a vertical line that crosses this parabola twice, x= y2 does not define y as a function of x. Because x = y2 is equivalent to y = {1x, the top half of the graph of x =y2 is y = 1x and the bottom half is y = -1x.
To support these conclusions with a graphing calculator, graph y1 = 1x and y2 = -1x as shown in Fig. 2.15.
䉴TRY THIS. Determine whether x= -y2 is a function, graph it, and state the do-
main and range. ■
In the next example we graph the cube function y= x3 and the cube-root function y = 13x.
These ordered pairs indicate a graph in the shape shown in Fig. 2.16. The domain is 1-⬁, ⬁2 and the range is 1-⬁, ⬁2. By the vertical line test, this graph is the graph of a function because no vertical line crosses the curve more than once.
b. Make a table listing ordered pairs that satisfy y = 13x. Note that this table is simply the table for y =x3 with the coordinates reversed.
y is any real number x is any real number
x -2 -1 0 1 2
y =x3 -8 -1 0 1 8
x -8 -1 0 1 8
y= 23x -2 -1 0 1 2
x is any real number
y is any real number
EXAMPLE 4 Cube and cube-root functions
Graph each equation. State the domain and range.
a. y = x3 b. y = 23x Solution
a. Make a table listing ordered pairs that satisfy y =x3:
These ordered pairs indicate a graph in the shape shown in Fig. 2.17. The domain is 1-⬁, ⬁2 and the range is 1-⬁, ⬁2. Because no vertical line crosses this graph more than once, y = 13x is a function.
The calculator graph in Fig. 2.18 supports these conclusions.
Figure 2.16
6 8
4 2
–4 –6 –8 –2
2
–2 1
y
x
Domain (–∞, ∞)
Range (–∞, ∞)
y = x3
–1
Figure 2.18
⫺8 8
⫺2 2
Figure 2.17 y
–2 6 x
1
8
–2 4 –4 2
2
– 6 – 8
3
Range
(–)∞, ∞
Domain (–∞, ∞) y = √x
䉴TRY THIS. Determine whether y = -13x is a function, graph it, and state the
domain and range. ■
Semicircles
The graph of x2 +y2 =r21r 7 02 is a circle centered at the origin of radius r. A circle does not pass the vertical line test, and is not the graph of a function. We can find an equivalent equation by solving for y:
x2 +y2 =r2 y2 =r2- x2
y = {2r2- x2
The equation y = 2r2 -x2 does define y as a function of x. Because y is nonneg- ative in this equation, the graph is the top semicircle in Fig. 2.19. The top semicircle passes the vertical line test. Likewise, the equation y = -2r2 - x2 defines y as a function of x, and its graph is the bottom semicircle in Fig. 2.19.
Figure 2.19 y
x r
r –r
–r x2
x2 y = √r2–
y = –√r2–
Figure 2.20 y
– 1 x – 1
– 3
1 3
1
] ]
Domain [– 2, 2]
] ]
Range [–2, 0]
– y = √4–x2
Figure 2.21 y
– 1 x – 1
1 3
1
] ]
Domain [– 3, 3]
] ]Range
[0, 3
]
– 2 2 2 4
–3
y = 9√ –x2
EXAMPLE 5 Graphing a semicircle
Sketch the graph of each function and state the domain and range of the function.
a. y = -24 -x2 b. y = 29 - x2 Solution
a. Rewrite the equation in the standard form for a circle:
y = -24- x2
y2= 4 -x2 Square each side.
x2+ y2= 4 Standard form for the equation of a circle
The graph of x2+ y2= 4 is a circle of radius 2 centered at 10, 02. Since y must be negative in y = -24- x2, the graph of y = -24 - x2 is the semicircle shown in Fig. 2.20. We can see from the graph that the domain is 3-2, 24 and the range is 3-2, 04.
b. Rewrite the equation in the standard form for a circle:
y = 29 - x2
y2 =9 - x2 Square each side.
x2 +y2 =9 Standard form for the equation of a circle
The graph of x2+ y2= 9 is a circle with center 10, 02 and radius 3. But this equa- tion is not equivalent to the original. The value of y in y= 29 - x2 is nonnegative.
So the graph of the original equation is the semicircle shown in Fig. 2.21. We can read the domain 3-3, 34 and the range 30, 34 from the graph.
䉴TRY THIS. Graph y = -29 -x2 and state the domain and range. ■
Piecewise Functions
For some functions, different formulas are used in different regions of the domain.
Since these functions are pieced together from two or more functions, they are called piecewise functions. The simplest example of such a function is the absolute value function ƒ 1x2 = 0x0, which can be written as
ƒ 1x2 = e x for x Ú0 -x for x 60.
For x Ú0 the equation ƒ 1x2= x is used to obtain the second coordinate, and for x 6 0 the equation ƒ 1x2 = -x is used. The graph of the absolute value function is shown in the next example. Note how the graph is pieced together from the graphs of y =x and y = -x.
EXAMPLE 6 The absolute value function
Graph the equation y= 0x0 and state the domain and range. Determine whether the relation is a function.
Solution
If xÚ 0 use y =x to determine ordered pairs that satisfy y = 0x0. If x 60 use y = -x :
Figure 2.25 y
–1 x – 1
3 1
2 3
– 2 – 3
4 5
1 y = x – 2
for x > 2
y = x2 – 4 for – 2 ≤ x ≤ 2 Figure 2.22
y = ⏐x⏐ y
–1 x –1
3 1
Domain (–∞, ∞)
]Range
[0,
∞)
–2 2 2 3
– 3 (0, 0)
(1, 1) (2, 2)
(3, 3)
(–1, 1) (– 2, 2)
(– 3, 3)
⫺3 3
⫺1 3
Figure 2.23
Plot these ordered pairs to get the V-shaped graph shown in Fig. 2.22. Because no verti- cal line crosses the graph more than once, y = 0x0 is a function. Note that x can be any real number, but y is nonnegative. So the domain is 1-⬁, ⬁2 and the range is 30, ⬁2.
To support these conclusions with a graphing calculator, graph y1 = abs1x2 as shown in Fig. 2.23.
䉴TRY THIS. Determine whether y = 0x0 +2 is a function, graph it, and state the
domain and range. ■
In the next example we graph two more piecewise functions.
For x in the interval 3-2, 24 the graph is a portion of a parabola as shown in Fig. 2.25. For x 7 2, the graph is a portion of a straight line through 12.1, 0.12, 13, 12, 14, 22, and 15, 32. The domain is 3-2, ⬁2, and the range is 3-4, ⬁2.
To graph a piecewise function on your calculator, use the inequality sym- bols from the TEST menu as shown in Fig. 2.26(a). On the calculator, an inequal- ity or a compound inequality has a value of 1 when it is satisfied and 0 when it is not satisfied. So y1 = 1x2 - 42>1x Ú -2 and x … 22 will be graphed as the parabola y = x2 -4 only when both inequalities are satisfied. The word “and”
is found in the TEST LOGIC menu. To graph y = x -2 only for x7 2 enter y2= 1x - 22>1x 722. The calculator graph is shown in Fig. 2.26(b).
EXAMPLE 7 Graphing a piecewise function
Sketch the graph of each function and state the domain and range.
a. ƒ 1x2 = e 1 for x6 2
-1 for xÚ 2 b. ƒ 1x2 = ex2 - 4 for -2… x … 2 x -2 for x 7 2 Solution
a. For x 62 the graph is the horizontal line y = 1. For xÚ 2 the graph is the horizontal line y = -1. Note that 12, -12 is on the graph shown in Fig. 2.24 but 12, 12 is not, because when x =2 we have y= -1. The domain is the interval 1-⬁, ⬁2 and the range consists of only two numbers, -1 and 1. The range is not an interval. It is written in set notation as 5-1, 16. Note that the graph consists of two separate sections that do not touch each other.
b. Make a table of ordered pairs using y = x2 -4 for x between -2 and 2 and y = x -2 for x7 2.
x 1x Ú02 0 1 2 3
y = x 0 1 2 3
x 1x6 02 -1 -2 -3
y= -x 1 2 3
x -2 -1 0 1 2
y =x2-4 0 -3 -4 -3 0
x 2.1 3 4 5
y =x - 2 0.1 1 2 3
Figure 2.24 y
– 1 x – 3
–1 –2
1 2 3
– 2 2
f(x) = –1 for x ≥ 2 f(x) = 1
for x < 2
Figure 2.26 (a)
4
–6
6 –3
(b)
䉴TRY THIS. Graph y = e x for x Ú 0
-2x for x 60 and state the domain and range. ■
Piecewise functions are often found in shipping charges. For example, if the weight in pounds of an order is in the interval 10, 14, the shipping and handling charge is $3. If the weight is in the interval 11, 24, the shipping and handling charge is $4, and so on. The next example is a function that is similar to a shipping and handling charge. This function is referred to as the greatest integer function and is written ƒ 1x2= Œxœ or ƒ 1x2= int1x2. The symbol Œxœ is defined to be the largest integer that is less than or equal to x. For example, Œ5.01œ = 5, because the great- est integer less than or equal to 5.01 is 5. Likewise, Œ3.2œ = 3, Œ-2.2œ = -3, and
Œ7œ =7.
Figure 2.27 y
–1 x –3 – 4
– 2 – 3 – 4
1 2 3
–2 2 1
4 3
f(x) = [[ x]]
Figure 2.28
⫺4 4
⫺4 4
EXAMPLE 8 Graphing the greatest integer function
Sketch the graph of ƒ 1x2 = Œxœ and state the domain and range.
Solution
For any x in the interval 30, 12 the greatest integer less than or equal to x is 0. For any x in 31, 22 the greatest integer less than or equal to x is 1. For any x in 3-1, 02 the greatest integer less than or equal to x is -1. The definition of Œxœ causes the function to be constant between the integers and to “jump” at each integer. The graph of ƒ 1x2 = Œxœ is shown in Fig. 2.27. The domain is 1-⬁, ⬁2, and the range is the set of integers.
The calculator graph of y1 = int1x2 looks best in dot mode as in Fig. 2.28, because in connected mode the calculator connects the disjoint pieces of the graph.
The calculator graph in Fig. 2.28 supports our conclusion that the graph of this func- tion looks like the one drawn in Fig. 2.27. Note that the calculator is incapable of showing whether the endpoints of the line segments are included.
䉴TRY THIS. Graph y = -Œxœ and state the domain and range. ■ In the next example we vary the form of the greatest integer function, but the graph is still similar to the graph in Fig. 2.27.
EXAMPLE 9 A variation of the greatest integer function
Sketch the graph of ƒ 1x2 = Œx - 2œ for 0 … x …5.
Solution
If x=0, ƒ 102= Œ-2œ = -2. If x=0.5, ƒ 10.52= Œ-1.5œ = -2. In fact, ƒ 1x2= -2 for any x in the interval 30, 12. Similarly, ƒ 1x2 = -1 for any x in the interval 31, 22.
This pattern continues with ƒ 1x2 = 2 for any x in the interval 34, 52, and ƒ 1x2 = 3 for x = 5. The graph of ƒ 1x2= Œx -2œ is shown in Fig. 2.29.
䉴TRY THIS. Graph y = Œx+ 2œ and state the domain and range. ■
Increasing, Decreasing, and Constant
Imagine a point moving from left to right along the graph of a function. If the y- coordinate of the point is getting larger, smaller, or staying the same, then the func- tion is said to be increasing, decreasing, or constant, respectively. More precisely we make the following definition.
Figure 2.29 for 0 ≤ x ≤ 5
f(x) = [[x – 2]]
y
– 1 x –1 –2 –3
1 2 3
2 1
4 5
3
6
Definition: Increasing, Decreasing, Constant
If a 6 b implies ƒ 1a2 6 ƒ 1b2 for any a and b in the domain of ƒ, then ƒ is an increasing function.
If a6 b implies ƒ 1a27 ƒ 1b2 for any a and b in the domain of ƒ, then ƒ is a decreasing function.
If a6 b implies ƒ 1a2 =ƒ 1b2 for any a and b in the domain of ƒ, then ƒ is a constant function.
Solution
If a point is moving left to right along the graph of ƒ 1x2= 2x, then its y-coordinate is increasing. So ƒ 1x2 = 2x is an increasing function. If a point is moving left to right along the graph of ƒ 1x2= -2x, then its y-coordinate is decreasing. So ƒ 1x2 = -2x is a decreasing function. The function ƒ 1x2 =2 has a constant y-coordinate and it is a constant function.
䉴TRY THIS. Determine whether ƒ 1x2 = -3x is increasing, decreasing, or con-
stant by examining its graph. ■
We can also discuss whether a function is increasing, decreasing, or constant on a subset of its domain. For example, the absolute value function ƒ 1x2 = 0x0 , which we graphed in Example 6, is decreasing on the interval 1-⬁, 04 and increasing on the interval 30, ⬁2.
y
x f(x) = 2x
1 2 –1 –2
2 1 3 4
– 2 – 3
y
x f(x) = – 2x
1 2 –1 –2
2 3 4
– 2 – 1 – 3
y
x f(x) = 2
1 2 –1 –2
1 3 4
– 2 – 1
– 3
EXAMPLE 10 Increasing, decreasing, or constant functions
Determine whether each function is increasing, decreasing, or constant by examin- ing its graph.
a. b. c.
䉴TRY THIS. Graph ƒ1x2= e x for x Ú 0
-2x for x 6 0 and identify any intervals on which the function is increasing, decreasing, or constant. ■
Figure 2.30 y
–1 x –3 – 4
–5 – 1
– 2 – 3
1 3
2 1 5
4 5 3
(2, 0) (– 2, 0)
(1, 3) (0, 4) (–1, 3) f(x) = 4 – x2
Figure 2.31 g(x) = 0
for x 0 y
– 1 x – 3
– 4 –1
–2 –3
1 3
– 2 2 1 4 5
4 5 3
6 g(x) = 2
for x 4
g(x) = for 0 < x < 4
≥
≤
√x
EXAMPLE 11 Increasing, decreasing, or constant on an interval
Sketch the graph of each function and identify any intervals on which the function is increasing, decreasing, or constant.
a. ƒ 1x2= 4 -x2 b. g 1x2 = c
0 for x … 0 1x for 0 6 x 64 2 for x Ú 4 Solution
a. The graph of ƒ 1x2 = 4 -x2 includes the points 1-2, 02, 1-1, 32, 10, 42, 11, 32, and 12, 02. The graph is shown in Fig. 2.30. The function is increasing on the in- terval 1-⬁, 04 and decreasing on 30, q2.
b. The graph of g is shown in Fig. 2.31. The function g is constant on the intervals 1-⬁, 04 and 34, ⬁2, and increasing on 30, 44.
1. The range of y = -x2 is 1-⬁, 04.
2. The function y = -1x is increasing on 30, ⬁2.
3. The function ƒ 1x2= 23x is increasing on 1-⬁, ⬁2.
4. If ƒ 1x2 = Œx +3œ, then ƒ 1-4.52= -2.
5. The range of the function ƒ 1x2 = 0x0
x is the interval 1-1, 12.
6. The range of ƒ 1x2 = Œx- 1œ is the set of integers.
7. The only ordered pair that satisfies 1x- 522+ 1 y +622= 0 is 15, -62.
8. The domain of the function y = 24- x2 is the interval 3-2, 24.
9. The range of the function y= 216 - x2 is the interval 30, ⬁2.
10. The function y = 24- x2 is increasing on 3-2, 04 and decreasing on 30, 24.
FOR thought … True or False? Explain
Fill in the blank.
1. The graph of y =x2 is a ________.
2. The absolute value function is an example of a ________
function.
Make a table listing ordered pairs that satisfy each equation. Then graph the equation. Determine the domain and range, and whether y is a function of x.
3. y =2x 4. x = 2y 5. x- y= 0
6. x- y= 2 7. y =5 8. x= 3
9. y =2x2 10. y =x2 -1 11. y =1 -x2 12. y = -1 -x2 13. y =1+ 1x 14. y =2 - 1x 15. x =y2 +1 16. x = 1-y2 17. x= 1y 18. x -1 = 1y 19. y = 23x+ 1 20. y = 23x -2 21. x = 23y 22. x = 23y -1 23. y2 =1- x2 24. x2+ y2=4 25. y = 21-x2 26. y = -225-x2 27. y =x3 28. y = -x3 29. y =2 0x0 30. y = 0x -10 31. y = - 0x0 32. y = - 0x+ 10 33. x = 0y0 34. x = 0y0 + 1
Make a table listing ordered pairs for each function. Then sketch the graph and state the domain and range.
35. ƒ 1x2 = e 2 for x6 -1 -2 for xÚ -1 36. ƒ 1x2 = e3 for x 62
1 for x Ú2 37. ƒ 1x2 = ex +1 for x 71
x -3 for x …1 38. ƒ 1x2 = e5- x for x …2 x +1 for x 72 39. ƒ 1x2 = e1x +2 for -2…x …2
4 -x for x7 2 40. ƒ 1x2 = e1x for xÚ 1
-x for x6 1 41. ƒ 1x2 = e1-x for x 60
1x for x Ú0 42. ƒ 1x2 = e 3 for x 60
3+ 1x for x Ú0
43. ƒ 1x2 = e x2 for x 6 -1 -x for xÚ -1 44. ƒ 1x2 = e4 -x2 for -2 …x …2
x -2 for x 72 45. ƒ 1x2 = Œx+ 1œ
46. ƒ 1x2 =2Œxœ
47. ƒ 1x2 = Œxœ + 2 for 0… x6 4 48. ƒ 1x2=Œx-3œ for 0 6x …5
From the graph of each function in Exercises 49–56, state the do- main, the range, and the intervals on which the function is increas- ing, decreasing, or constant.
49. a.