EXAMPLE 2 Graphing the Absolute Value Function DE GRAFICAS.pdffoundation for further graphing techniques. Constant Function f1x2 = b b is a real number x y f(x ) = b (0,b) Figure 31

236 CHAPTER 3 Functions and Their Graphs

Seeing the ConceptGraph on a square screen and compare what you see with Figure 30. Note that some graphingcalculators use for absolute value.abs1x2y = ƒx ƒ

Properties of

1. The domain is the set of all real numbers. The range of f is .2. The x-intercept of the graph of is 0. The y-intercept of the

graph of is also 0.3. The graph is symmetric with respect to the y-axis. The function is even.4. The function is decreasing on the interval It is increasing on the

interval 5. The function has an absolute minimum of 0 at x = 0.

10, q2. 1-q , 02.f1x2 = ƒx ƒ

f1x2 = ƒx ƒ5y|y Ú 06f (x) # ƒx ƒ

Below is a list of the key functions that we have discussed. In going through this list,pay special attention to the properties of each function, particularly to the shape ofeach graph. Knowing these graphs along with key points on each graph will lay thefoundation for further graphing techniques.

Constant Function

f1x2 = b b is a real number

x

y

f (x ) = b(0,b)

Figure 31 Constant Function

Graphing the Absolute Value Function

(a) Determine whether is even, odd, or neither. State whether the graph ofis symmetric with respect to the y-axis or symmetric with respect to the origin.

(b) Determine the intercepts, if any, of the graph of (c) Graph f1x2 = ƒx ƒ .

f1x2 = ƒx ƒ .f

f1x2 = ƒx ƒEXAMPLE 2

Solution (a) Because

the function is even. The graph of is symmetric with respect to the y-axis.(b) The y-intercept is The x-intercept is found by solving the

equation So the x-intercept is 0.(c) Use the function to form Table 5 and obtain some points on the graph. Because

of the symmetry with respect to the y-axis, we need to find only points forwhich Figure 30 shows the graph of f1x2 = ƒx ƒ .x Ú 0.

1x, y2f1x2 = 0 or ƒx ƒ = 0.f102 = ƒ0 ƒ = 0.

f

= ƒx ƒ = f1x2 f1-x2 = ƒ -x ƒ

From the results of Example 2 and Figure 30, we have the following propertiesof the absolute value function.

x

y3

2

1 (1, 1) ($1, 1)

(2, 2) ($2, 2)

(3, 3) ($3, 3)

(0, 0)$1$2$3

$131 2

Figure 30x

0 0 (0, 0)

1 1 (1, 1)

2 2 (2, 2)

3 3 (3, 3)

1x, y2y # f (x) # |x|Table 5

!

See Figure 31.

SECTION 3.4 Library of Functions; Piecewise-defined Functions 237

The domain of a constant function is the set of all real numbers; its range is theset consisting of a single number b. Its graph is a horizontal line whose y-intercept is b.The constant function is an even function.

Identity Function

f1x2 = x

See Figure 32.The domain and the range of the identity function are the set of all real

numbers. Its graph is a line whose slope is 1 and whose y-intercept is 0. The lineconsists of all points for which the x-coordinate equals the y-coordinate. Theidentity function is an odd function that is increasing over its domain. Note that thegraph bisects quadrants I and III.

Square Function

f1x2 = x2

See Figure 33.The domain of the square function is the set of all real numbers; its range is

the set of nonnegative real numbers. The graph of this function is a parabola whoseintercept is at The square function is an even function that is decreasing onthe interval and increasing on the interval 10, q2.1-q , 0210, 02. f

x

y

3

3

–3

(1, 1)

(–1, –1)

f (x ) = x

(0, 0)

Figure 32 Identity Function

x

y

4

4

–4

(2, 4)

(0, 0)

(–2, 4)

f (x ) = x2

(1, 1)(–1, 1)

Figure 33 Square Function

Cube Function

f1x2 = x3

See Figure 34.The domain and the range of the cube function are the set of all real numbers.

The intercept of the graph is at The cube function is odd and is increasing onthe interval 1-q , q2. 10, 02.

Square Root Function

f1x2 = 1xSee Figure 35.The domain and the range of the square root function are the set of nonnega-

tive real numbers. The intercept of the graph is at The square root function isneither even nor odd and is increasing on the interval 10, q2.10, 02.

Cube Root Function

f1x2 = 13 x

x

y

4

4

!4

f (x ) = x 3

(1, 1)

(0, 0)(!1, !1)

!4

Figure 34 Cube Function

x

y

5

2

!1

f (x ) = x

(1, 1)

(0, 0)

(4, 2)

Figure 35 Square Root Function

See Figure 36.The domain and the range of the cube root function are the set of all real

numbers. The intercept of the graph is at The cube root function is an oddfunction that is increasing on the interval 1-q , q2.10, 02.

( , )x

y3

(1, 1)

(!1, !1)

(2, 2 )

(0, 0)!3

!3

3

3

(!2,! 2 )3

1–8

1–2

(! ,! )1–8

1–2

Figure 36 Cube Root Function

238 CHAPTER 3 Functions and Their Graphs

DEFINITION Greatest Integer Function

f1x2 = int1x2* = greatest integer less than or equal to x

We obtain the graph of by plotting several points. See Table 6. Forvalues of x, the value of is for values of x,

the value of is 0. See Figure 39 for the graph.f0 … x 6 1,-1;f1x2 = int1x2-1 … x 6 0,

f1x2 = int1x2

x

y4

2

!2 2 4

!3

Figure 39 Greatest Integer Function

The domain of the greatest integer function is the set of all real numbers; itsrange is the set of integers. The y-intercept of the graph is 0. The x-intercepts lie inthe interval The greatest integer function is neither even nor odd. It isconstant on every interval of the form for k an integer. In Figure 39, we usea solid dot to indicate, for example, that at the value of is we usean open circle to illustrate that the function does not assume the value of 0 at x = 1.

f112 = 1;fx = 13k, k + 12,30, 12.

Table 6

y # f (x)# int(x) (x, y)x

0 0

0

0

0 a34

, 0b34

a12

, 0b12

a14

, 0b14

(0, 0)

a - 14

, -1b-1- 14

a - 12

, -1b-1- 12

(-1, -1)-1-1

* Some books use the notation instead of int1x2.f1x2 = 3x4

Reciprocal Function

f1x2 = 1x

Absolute Value Function

f1x2 = ƒx ƒ

Refer to Example 12, page 164, for a discussion of the equation SeeFigure 37.

The domain and the range of the reciprocal function are the set of all nonzeroreal numbers. The graph has no intercepts. The reciprocal function is decreasing onthe intervals and and is an odd function.10, q21-q , 02

y = 1x

.

See Figure 38.The domain of the absolute value function is the set of all real numbers; its range is

the set of nonnegative real numbers. The intercept of the graph is at Ifthen and the graph of is part of the line if thenand the graph of is part of the line The absolute value function is an evenfunction; it is decreasing on the interval and increasing on the interval

The notation stands for the largest integer less than or equal to x. Forexample,

This type of correspondence occurs frequently enough in mathematics that we giveit a name.

int112 = 1, int12.52 = 2, inta12b = 0, inta -

34b = -1, int1p2 = 3

int1x2 10, q2.1-q , 02y = -x.ff1x2 = -x,x 6 0,y = x;ff1x2 = x,

x Ú 0,10, 02.

, 2( )1–2

( )1–2

x

y

2

2f (x ) =

(1, 1)

(!1, !1)

!2

1––x

!2

!2, !

Figure 37 Reciprocal Function

x

y

3

3

!3

f (x ) = x

(1, 1)

(0, 0)

(!1, 1)

(2, 2)(!2, 2)

Figure 38 Absolute Value Function

SECTION 6.3 Exponential Functions 425

is . Therefore, the exponential function that models the data is

.

(c) See Table 2(c). For this function, the average rate of change from to 0 is 2,and the average rate of change from 0 to 1 is 3. Because the average rate ofchange is not constant, the function is not a linear function. The ratio ofconsecutive outputs from to 0 is 2, and the ratio of consecutive outputs from 0

to 1 is . Because the ratio of consecutive outputs is not a constant, the function

is not an exponential function.

Now Work P R O B L E M 2 5

74

-1

-1

g1x2 = Cax = 16 # a12bxC = 16

#

x

y

3

(2, 4)

(1, 2)

(0, 1)

( )–2,

6

3

1–4 ( )–1, 1–

2( )–3, 1–8

y " 0

Figure 18

Graph Exponential FunctionsIf we know how to graph an exponential function of the form , then wecould use transformations (shifting, stretching, and so on) to obtain the graph of anyexponential function.

First, we graph the exponential function f1x2 = 2x.

f1x2 = ax2

The domain of is the set of all real numbers. We begin by locating somepoints on the graph of as listed in Table 3.

Since for all x, the range of is From this, we conclude that thegraph has no x-intercepts, and, in fact, the graph will lie above the x-axis for all x.As Table 3 indicates, the y-intercept is 1. Table 3 also indicates that as thevalues of get closer and closer to 0.We conclude that the x-axis isa horizontal asymptote to the graph as This gives us the end behavior forx large and negative.

To determine the end behavior for x large and positive, look again at Table 3.Asgrows very quickly, causing the graph of to rise very

rapidly. It is apparent that is an increasing function and hence is one-to-one.Using all this information, we plot some of the points from Table 3 and connect

them with a smooth, continuous curve, as shown in Figure 18.

ff1x2 = 2xx: q , f1x2 = 2x

x: -q .1y = 02f1x2 = 2x

x: -q

10, q2.f2x 7 0f1x2 = 2x,

f1x2 = 2xSolution

Table 3 x

!10

!3

!2

!1

0

1

2

3

10 210 = 1024

23 = 8

22 = 4

21 = 2

20 = 1

2-1 = 12

2-2 = 14

2-3 = 18

2-10 L 0.00098

f(x) ! 2x

As we shall see, graphs that look like the one in Figure 18 occur very frequentlyin a variety of situations. For example, the graph in Figure 19 illustrates the number

Graphing an Exponential Function

Graph the exponential function: f1x2 = 2xEXAMPLE 3

#

SECTION 6.3 Exponential Functions 429

One way of arriving at this important number e is given next.

DEFINITION The number e is defined as the number that the expression

(2)

approaches as In calculus, this is expressed using limit notation as

e = limn:q

a1 + 1nbn

n: q .

a1 + 1nbn

Table 5 illustrates what happens to the defining expression (2) as n takes onincreasingly large values. The last number in the right column in the table is correctto nine decimal places and is the same as the entry given for e on your calculator (if expressed correctly to nine decimal places).

The exponential function whose base is the number e, occurs withsuch frequency in applications that it is usually referred to as the exponential function. Indeed, most calculators have the key or , which may beused to evaluate the exponential function for a given value of x.*

! exp1x2 !! ex !

f1x2 = ex,

Seeing the ConceptGraph and compare what you see to Figure 27. Use eVALUEate or TABLE to verify the points onthe graph shown in Figure 27. Now graph and on the same screen as Noticethat the graph of lies between these two graphs.Y1 = ex

Y1 = ex.Y3 = 3xY2 = 2xY1 = ex

Table 5n

1 1 2 2

2 0.5 1.5 2.25

5 0.2 1.2 2.48832

10 0.1 1.1 2.59374246

100 0.01 1.01 2.704813829

1,000 0.001 1.001 2.716923932

10,000 0.0001 1.0001 2.718145927

100,000 0.00001 1.00001 2.718268237

1,000,000 0.000001 1.000001 2.718280469

1,000,000,000 2.7182818271 + 10-910-9

a1 "1nbn

1 "1n

1n

Now use your calculator to approximate for and as we have done to create Table 6.The graph of the exponential function

is given in Figure 27. Since the graph of lies betweenthe graphs of and Do you see why? (Refer to Figures 18 and 20.)y = 3x.y = 2x

y = ex2 6 e 6 3,f1x2 = exx = 2,

x = 1,x = -2, x = -1, x = 0,ex

Table 6

x

2

1

0

1

2 e2 L 7.39

e1 L 2.72

e0 L 1

e-1 L 0.37-

e-2 L 0.14-

ex

Begin with the graph of Figure 28 shows the stages.y = ex.Solution

* If your calculator does not have one of these keys, refer to your Owner’s Manual.

(1, e)

0 x

y

3

6

3

(2, e2)

(0, 1)

y " 0 ( )–2, 1–e2

( )–1, 1–e

Figure 27y = ex

Graphing Exponential Functions Using Transformations

Graph and determine the domain, range, and horizontal asymptoteof f.

f1x2 = -ex-3

EXAMPLE 6

SECTION 6.4 Logarithmic Functions 441

If the base of a logarithmic function is the number e, then we have the naturallogarithm function. This function occurs so frequently in applications that it is givena special symbol, ln (from the Latin, logarithmus naturalis). That is,

(1)

Since and the exponential function are inverse functions, wecan obtain the graph of by reflecting the graph of about the line

See Figure 33.Using a calculator with an key, we can obtain other points on the graph of

See Table 7.f1x2 = ln x.! ln !

y = x.y = exy = ln x

y = exy = ln x

y = ln x if and only if x = ey

In Wordsis written y = ln xy = logex

( )(

!1,

x

y"xy"e x

y"In x

y " 0

x " 0

y

!1

5

!3 3

0, 1

)1, 0

(1, e)

(e, 1))

(

1–e

( ),!11–e

Figure 33 Table 7 x ln x

2 0.69

3 1.10

-0.6912

Seeing the ConceptGraph and on thesame square screen. Use eVALUEate toverify the points on the graph given inFigure 33. Do you see the symmetry ofthe two graphs with respect to the liney = x?

Y2 = ln xY1 = ex

(a) The domain of f consists of all x for which The domain of f is in interval notation.

(b) To obtain the graph of we begin with the graph of and use transformations. See Figure 34.

y = ln xy = - ln1x - 22,{x ƒx 7 2} or 12, q2 x - 2 7 0 or, equivalently, x 7 2.Solution

x x x

y

3 3(1, 0)

, !1)(

(e, 1)

(e, !1)

3

(a) y " In x

y

(1, 0)!1 !1

!1 !1!1

1 1 1

3

(b) y " !In x

1–e

1–e)(1–

e , 1( )

y

531

(3, 0)

(e#2, !1)

3

(c) y " !In (x ! 2)

# 2,1

x " 2

x " 0x " 0

Multiply by ! 1;reflectabout x -axis

Replace x byx ! 2; shiftright 2 units.

Figure 34

(c) The range of is the set of all real numbers. The verticalasymptote is [Do you see why? The original asymptote is shiftedto the right 2 units.]

1x = 02x = 2.f1x2 = - ln1x - 22

Graphing a Logarithmic Function and Its Inverse

(a) Find the domain of the logarithmic function (b) Graph f.(c) From the graph, determine the range and vertical asymptote of f.(d) Find , the inverse of f.(e) Find the domain and the range of .(f) Graph .f-1

f-1f-1

f1x2 = - ln1x - 22.EXAMPLE 6

SECTION 7.6 Graphs of the Sine and Cosine Functions 561

Now Work the ‘Are You Prepared?’ problems on page 571.

OBJECTIVES 1 Graph Functions of the Form Using Transformations (p. 562)

2 Graph Functions of the Form Using Transformations (p. 564)

3 Determine the Amplitude and Period of Sinusoidal Functions (p. 565)

4 Graph Sinusoidal Functions Using Key Points (p. 566)

5 Find an Equation for a Sinusoidal Graph (p. 570)

y = A cos1vx2y = A sin1vx2

7.6 Graphs of the Sine and Cosine Functions*

• Graphing Techniques: Transformations (Section 3.5, pp. 244–253)

PREPARING FOR THIS SECTION Before getting started, review the following:

Since we want to graph the trigonometric functions in the xy-plane, we shall use thetraditional symbols x for the independent variable (or argument) and y for thedependent variable (or value at x) for each function. So we write the six trigonometricfunctions as

Here the independent variable x represents an angle, measured in radians. In calculus,x will usually be treated as a real number. As we said earlier, these are equivalentways of viewing x.

The Graph of the Sine Function Since the sine function has period , we only need to graph on theinterval .The remainder of the graph will consist of repetitions of this portionof the graph.

We begin by constructing Table 7, which lists some points on the graph of, . As the table shows, the graph of ,

begins at the origin. As x increases from 0 to , the value of increases

from 0 to 1; as x increases from to to , the value of y decreases from 1 to 0

to as x increases from to , the value of y increases from to 0. If we

plot the points listed in Table 7 and connect them with a smooth curve, we obtain thegraph shown in Figure 77.

-12p3p2

-1;

3p2

pp

2

y = sin xp

2

y = sin x, 0 … x … 2p0 … x … 2py = sin x

30, 2p4 y = sin x2p

y ! sin x

y = f1x2 = csc x y = f1x2 = sec x y = f1x2 = cot x y = f1x2 = sin x y = f1x2 = cos x y = f1x2 = tan x

!––6

x

y

1

! 2!

"1

( , )1–2 ( , )1–

2

x2!

( , 1)

(0, 0)(!, 0) (2!, 0)

!––2

5!–––6

( , " )1–2

7!–––6

( , " )1–2

11!––––6

( , "1)3!–––2

!––2

3!–––2

The graph in Figure 77 is one period, or cycle, of the graph of .To obtaina more complete graph of , we continue the graph in each direction, asshown in Figure 78.

y = sin xy = sin x

* For those who wish to include phase shifts here, Section 7.8 can be covered immediately after Section 7.6without loss of continuity.

Table 7

Figure 77y = sin x, 0 … x … 2p

x

0 0

1

0

0 (2p, 0)2p

a11p6

, - 12b- 1

211p

6

a3p2

, -1b-13p2

a7p6

, - 12b- 1

27p6

(p, 0)p

a5p6

, 12b1

25p6

ap2

, 1bp

2

ap6

, 12b1

2p

6

10, 02(x, y)y ! sin x

562 CHAPTER 7 Trigonometric Functions

The graph of illustrates some of the facts that we already know aboutthe sine function.

y = sin x

x! !" 2

!

" "

( , !1)

( , 1)

( , !1)3"–––2

( , 1)5"–––2

y

1

!1

"––2

"––2

"––2

!"––2

3"–––2

5"–––2

Properties of the Sine Function y ! sin x

1. The domain is the set of all real numbers.2. The range consists of all real numbers from to 1, inclusive.3. The sine function is an odd function, as the symmetry of the graph with

respect to the origin indicates.4. The sine function is periodic, with period .5. The x-intercepts are ; the y-intercept is 0.

6. The maximum value is 1 and occurs at ; the

minimum value is and occurs at .x = Á , -p2

, 3p2

, 7p2

, 11p

2, Á-1

x = Á , - 3p2

, p

2,

5p2

, 9p2

, Á

Á , -2p, -p, 0, p, 2p, 3p, Á2p

-1

Now Work P R O B L E M 9

Figure 79 illustrates the steps.Solution

x2! " "

"––2( , 3)

( , 1)"––2 ( , 1)5"–––2

y

1

"––2

"––2

3"–––2

5"–––2

x

y

!" "

!3

3

1

"–2

3"–––2

"–––2

(a) y # sin x

Multiply by 3;vertical stretchby a factor of 3 (b) y # 3 sin x

!!"

(! , !1)"––2 ( , !1) 3"–––

2 ( , !3) 3"–––2!1

Graph Functions of the Form A sin( )Using Transformations

Vxy !1

Figure 78y = sin x, -q 6 x 6 q

!

Graphing Functions of the Form A sin( )Using Transformations

Graph using transformations.y = 3 sin x

Vxy !EXAMPLE 1

Figure 79

SECTION 7.6 Graphs of the Sine and Cosine Functions 563

Graphing Functions of the Form A sin(Using Transformations

Graph using transformations.y = -sin(2x)

Vx)y !EXAMPLE 2

Solution Figure 80 illustrates the steps.

Notice in Figure 80(c) that the period of the function is due to

the horizontal compression of the original period by a factor of .

Now Work P R O B L E M 3 7 U S I N G T R A N S F O R M A T I O N S

12

2p

py = -sin(2x)

Replace x by 2x ;Horizontal compressionby a factor of 1–2

!––4"!––

43!–––4

x

y

"1

1

" !

( , "1)

"!––2

!––2"

(c) y # ! sin (2x )

( , 1) 3 !–––4!––4(" , 1)

Period !

!––4

Multiply by "1;Reflect about thex"axis

!––2

!––2

3!–––2

5!–––2

x

y

"1

1

!

( , "1) !––2

"

(b) y # " sin x

!––2(" , 1) 3!–––

2( , 1)

"!

Period 2!

2!

(a) y # sin x

x" !

( , 1)!––2 ( , 1)

( , "1) 3 !–––2

y

1

!––2

!––2

3!–––2

5!–––2

5!–––2

"1!––2(" , "1)

2!"!

!––3

x! 2!

( , )1–2 ( , )1–

2

(0, 1) (2!, 1)

(!, "1)

5!–––3

( , " )1–2

2!–––3 ( , " )1–

24!–––

3

!––2

3!–––2

y

1

"1

x"!

!

2" ! !

(" , "1) !( , "1)

!(2 , 1)y1

"1!––2

!––2

3!–––2

5!–––2

A more complete graph of is obtained by continuing the graph ineach direction, as shown in Figure 82.

y = cos x

The graph of illustrates some of the facts that we already know aboutthe cosine function.

y = cos x

"

The Graph of the Cosine FunctionThe cosine function also has period . We proceed as we did with the sine functionby constructing Table 8, which lists some points on the graph of ,

. As the table shows, the graph of , begins at the

point . As x increases from 0 to to , the value of y decreases from 1 to 0 to

; as x increases from to to , the value of y increases from to 0 to 1. As

before, we plot the points in Table 8 to get one period or cycle of the graph. SeeFigure 81.

-12p3p2

p-1

pp

210, 12 y = cos x, 0 … x … 2p0 … x … 2p

y = cos x2p

Figure 80

Figure 81y = cos x, 0 … x … 2p

Figure 82y = cos x, -q 6 x 6 q

Table 8

x

0 1

0

0

1 (2p, 1)2p

a5p3

, 12b1

25p3

a3p2

, 0b3p2

a4p3

, - 12b- 1

24p3

(p, -1)-1p

a2p3

, - 12b- 1

22p3

ap2

, 0bp

2

ap3

, 12b1

2p

3

(0, 1)

(x, y)y ! cos x

SECTION 7.7 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions 577

Table 9

x

y

3

1

!1

3

!

! 3!! 3

( , 1)

(0, 0)

( , )

( , )

( , )

!( , !1)

!( , ! )

"––2! "––

3! "––6! "––

6"––6

"––6

"––2

"––4

"––4

"––3

"––3

"––3

3––3

3––3

3––3

3––3

x

0 0

1

ap3

, 23b23 L 1.73p

3

ap4

, 1bp

4

ap6

, 23

3b23

3L 0.58

p

6

(0, 0)

a- p6

, - 233b- 23

3L -0.58- p

6

a- p4

, -1b-1- p4

a- p3

, -23b-23 L -1.73- p3

(x, y)y ! tan x Figure 95y = tan x, - p

3… x … p

3

positive and large. In fact, the closer x gets to , the closer sin x gets to 1 and cos x

gets to 0, so tan x approaches . In other words, the vertical line

is a vertical asymptote to the graph of .y = tan xx = p2

q a limx:p2 -

tan x = qbp

2

Table 10 x sin x cos x

1.5 0.9975 0.0707 14.1

1.57 0.9999 1255.8

1.5707 0.9999 10,381

1 0 Undefinedp

2L 1.5708

9.6 * 10-5

7.96 * 10-4

23 L 1.7312

232

p

3L 1.05

y ! tan x

If x is close to , but remains greater than , then sin x will be close to

and will be positive and close to 0. The ratio approaches

. In other words, the vertical line is also a vertical

asymptote to the graph.With these observations, we can complete one period of the graph. We obtain

the complete graph of by repeating this period, as shown in Figure 96.y = tan x

x = -p2

a limx:-p2 +

tan x = -qb -qsin xcos x

cos x-1

- p2

- p2

x

y

1

!1

!2" 2"!" "

"––2

"––2!! 5"–––

23"–––2

5"–––2 !3"–––

2x # x # x # x # x # x #

(! , !1)"––4

( , 1)"––4

Figure 96not equal

to odd multiples of p

2, -q 6 y 6 q

y = tan x, -q 6 x 6 q , x

Check: Graph andcompare the result with Figure 96.Use TRACE to see what happensas x gets close to , but is less than .

p

2

p

2

Y1 = tan x

Properties of the Tangent Function

1. The domain is the set of all real numbers, except odd multiples of .2. The range is the set of all real numbers.3. The tangent function is an odd function, as the symmetry of the graph

with respect to the origin indicates.4. The tangent function is periodic, with period .5. The x-intercepts are the y-intercept is 0.

6. Vertical asymptotes occur at .x = Á , - 3p2

, - p2

, p

2,

3p2

, Á

. . . , -2p, -p, 0, p, 2p, 3p, . . . ;p

p

2


Now Work P R O B L E M S 7 A N D 1 5

Graph Functions of the Form and For tangent functions, there is no concept of amplitude since the range of thetangent function is .The role of A in is to provide themagnitude of the vertical stretch. The period of , so the period of

, caused by the horizontal compression of the graph by a

factor of . Finally, the presence of B indicates that a vertical shift is required.1v

y = A tan 1vx2 + B is p

v

y = tan x is py = A tan 1vx2 + B1-q, q2

y ! A cot(Vx) " By ! A tan(Vx) " B1

Graphing Functions of the Form

Graph: . Use the graph to determine the domain and the range of.y = 2 tan x - 1

y = 2 tan x - 1

y ! A tan (Vx) " BEXAMPLE 1

Figure 97 shows the steps using transformations.Solution

yy

x

1

'

'––2

'––2% 3'–––

2

( , 1)

(0, 0) (', 0) (0, 0) (', 0)

'––4

(% , %1)'––4

(a) y & tan x (b) y & 2 tan x

x

1

21

2

%1

2

%2

'

'––2

'––2% 3'–––

2

( , 2)'––4

(% , %2)'––4

Multiply by 2;Vertical stretchby a factor of 2

Subtract 1;vertically shiftdown one unit

x & x & x & x & x & x & y

(c) y & 2 tan x %1

x%1 (', %1)

'––2

'––2% 3'–––

2

( , 1)

(0, %1)

'––4

(% , %3)'––4

x & x & x &

The domain of is is an odd integer , and the

range is the set of all real numbers, or 1-q , q2. fex ` x Z kp2

, ky = 2 tan x - 1

"

Figure 97

Figure 98 shows the steps using transformations.Solution

Graphing Functions of the Form

Graph . Use the graph to determine the domain and the range of.y = 3 tan12x2y = 3 tan12x2 y ! A tan(Vx) " BEXAMPLE 2

The graph of in Figure 96 on page 577 illustrates the following properties.y = tan x

Check: Graph to verify the graph shown inFigure 97(c).

Y1 = 2 tan x - 1

SECTION 7.7 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions 579

y

x

1(0, 0) (!, 0)

! !

!––2

!––2" 3!–––

2

( , 1)!––4

(" , "1)!––4

Multiply by 3;Vertical stretch bya factor of 3

Replace x by 2x ;Horizontal compressionby a factor of

(a) y # tan x (c) y # 3 tan (2x )

x # x # x #y

(0, 0)

3

"3

3

y

(b) y # 3 tan x

x

(0, 0) (!, 0)

!––2

!––2" 3!–––

2( , 3)5!–––4

( , 3)5!–––8

( , "3)3!–––8

1––2

( , 3)!––4

( ,"3)3!––4(" , "3)

!––4

x #x # !––

4x #

x #

x

!––4" 3!–––

4

(" , "3)!––8

( , 3)!––8

( , 0)!––2

x # x #

( , 1)5!–––4

( , "1)3!–––4

Figure 98

Table 11

The domain of is , k is an odd integer , and the

range is the set of all real numbers or .

Check: Graph to verify the graph in Figure 98(c).

Notice in Figure 98(c) that the period of is due to the

compression of the original period by a factor of . Notice that the asymptotes

are , and so on, also due to the compression.


x = - p4

, x = p4

, x = 3p4

12

p

p

2y = 3 tan12x2

Y1 = 3 tan12x2 1-q , q2 fex ` x Z kp4

y = 3 tan 12x2

The Graph of the Cotangent FunctionWe obtain the graph of as we did the graph of . The period of

is . Because the cotangent function is not defined for integer multiplesof , we will concentrate on the interval . Table 11 lists some points on thegraph of , . As x approaches 0, but remains greater than 0, thevalue of cos x will be close to 1 and the value of sin x will be positive and close to 0.

Hence, the ratio will be positive and large; so as x approaches 0, with

, approaches . Similarly, as x approaches , but

remains less than , the value of cos x will be close to , and the value of sin x will

be positive and close to 0. So the ratio will be negative and will

approach as x approaches . Figure 99 shows the graph.1 limx:p-cot x = -q2p-q

cos xsin x

= cot x

-1p

pq1 limx:0+cot x = q2cot xx 7 0

cos xsin x

= cot x

0 6 x 6 py = cot x10, p2p

py = cot xy = tan xy = cot x

x

! 2!–2! –!y

–1

1

3!––2

– –!–2

!–2

3!––2

3!–––4

5!––2

x # x # x #x # 0

( , "1)

!––4

( , 1)

x #Figure 99, x not equal

to integer multiples of ,-q 6 y 6 q

py = cot x, -q 6 x 6 q

x

1

0

a5p6

, -23b-235p6

a3p4

, -1b-13p4

a2p3

, - 233b- 23

32p3

ap2

, 0bp

2

ap3

, 23

3b23

3p

3

ap4

, 1bp

4

ap6

, 23b23p

6

(x, y)y ! cot x

"


(pp. 576–578) Domain: , except odd integer multiples of

Range:

Periodic:

Odd function

Vertical asymptotes at odd integer multiples of

(pp. 579–580) Domain: , except integer multiples of (180°)

Range:

Periodic:

Odd function

Vertical asymptotes at integer multiples of

(p. 580) Domain: , except integer multiples of

Range:

Periodic:

Odd function

Vertical asymptotes at integer multiples of

(p. 580) Domain: , except odd integer multiples of

Range:

Periodic:

Even function

Vertical asymptotes at odd integer multiples of

Sinusoidal graphs

(pp. 566, 584)

(pp. 566, 584)

(p. 584)

y = A cos1vx - f2 + B = A cos cvax -f

vb d + B

Phase shift =f

vy = A sin1vx - f2 + B = A sin cvax -

f

vb d + B

Amplitude = ƒA ƒy = A cos1vx2 + B, v 7 0

Period = 2pv

y = A sin1vx2 + B, v 7 0

p

2

period = 2p 1360°2ƒy ƒ Ú 1

p

2 190°2-q 6 x 6 qy = sec x

p

period = 2p 1360°2ƒy ƒ Ú 1

p 1180°2-q 6 x 6 qy = csc x

p

period = p 1180°2-q 6 y 6 qp-q 6 x 6 qy = cot x

p

2

period = p 1180°2-q 6 y 6 q

p

2 190°2-q 6 x 6 qy = tan x

ObjectivesSection You should be able to: Example(s) Review Exercises

7.1 1 Convert between decimals and degrees, minutes, seconds measuresfor angles (p. 506) 2 86

2 Find the length of an arc of a circle (p. 508) 3 87, 883 Convert from degrees to radians and from radians to degrees (p. 508) 4–6 1–84 Find the area of a sector of a circle (p. 511) 7 875 Find the linear speed of an object traveling in circular motion (p. 512) 8 89–92

7.2 1 Find the values of trigonometric functions of acute angles (p. 517) 1 792 Use the Fundamental Identities (p. 519) 2, 3 21–243 Find the values of the remaining trigonometric functions, given the

value of one of them (p. 521) 4, 5 31–324 Use the Complementary Angle Theorem (p. 523) 6, 7 25, 26

!––2$ $ $ !–

2x3!––

25!––2

5!––2

3!––2

y

$1

1

$!–2

!–2

x3!–––2

5!–––2

$ 3!–––2

y

$1

1

!–2

x"3!–––2

y

"1

1y # sin x

y # csc x

"!––2

!–2

x3!––2

" 3!–––2

y

"1

1

y # sec x

y # cos x

Chapter Review 595

!–2

x!$! 2! 5!––2

y

$1

1

$!–2

!–2

x! 3!––2

$! 2! 5!––2

y

$1

1

Reference angle of (p. 545) The acute angle formed by the terminal side of and either the positive or negative x-axis

Periodic function (p. 557) , for all , where the smallest such p is the fundamental period

Formulas

(p. 505)(p. 509)

(p. 508) is measured in radians; s is the length of the arc subtended by the central angle ofthe circle of radius r.

(p. 511) A is the area of the sector of a circle of radius r formed by a central angle of radians.

(p. 512) is the linear speed along the circle of radius r ; is the angular speed (measured in radians per unit time).

vvv = rv

uA = 12

r2 u

uus = r u

= 2p radians revolution = 360° 1 counterclockwise

u, p 7 0f1u + p2 = f1u2uu

(Radians) (Degrees)

0 0° 0 1 0 Not defined 1 Not defined

30° 2

45° 1 1

60° 2

90° 1 0 Not defined 1 Not defined 0

180° 0 0 Not defined Not defined

270° 0 Not defined Not defined 0-1-13p2

-1-1p

p

2

233

2233

2312

232

p

3

222222

222

2p

4

23223

323

323

212

p

6

cot Usec Ucsc Utan Ucos Usin UUU

Fundamental identities (p. 520)

Properties of the trigonometric functions(p. 562) Domain:

Range:

Periodic:

Odd function

(p. 564) Domain:

Range:

Periodic:

Even function

period = 2p 1360°2-1 … y … 1

-q 6 x 6 qy = cos x

period = 2p 1360°2-1 … y … 1

-q 6 x 6 qy = sin x

sin2 u + cos2 u = 1 tan2 u + 1 = sec2 u cot2 u + 1 = csc2 u

csc u = 1sin u

sec u = 1cos u

cot u = 1tan u

tan u = sin ucos u

cot u = cos usin u

Table of Values

778 CHAPTER 11 Analytic Geometry

We stated earlier (Section 4.3) that the graph of a quadratic function is a parabola.In this section, we give a geometric definition of a parabola and use it to obtain anequation.

DEFINITION A parabola is the collection of all points P in the plane that are the samedistance d from a fixed point F as they are from a fixed line D. The point F iscalled the focus of the parabola, and the line D is its directrix. As a result, aparabola is the set of points P for which

(1)d1F, P2 = d1P, D2Figure 3 shows a parabola (in blue). The line through the focus F and

perpendicular to the directrix D is called the axis of symmetry of the parabola. Thepoint of intersection of the parabola with its axis of symmetry is called the vertex V.

Because the vertex V lies on the parabola, it must satisfy equation (1):The vertex is midway between the focus and the directrix. We

shall let a equal the distance from F to V. Now we are ready to derive anequation for a parabola. To do this, we use a rectangular system of coordinatespositioned so that the vertex V, focus F, and directrix D of the parabola are convenientlylocated.

d1F, V2d1F, V2 = d1V, D2.

Analyze Parabolas with Vertex at the OriginIf we choose to locate the vertex V at the origin we can conveniently positionthe focus F on either the x-axis or the y-axis. First, consider the case where thefocus F is on the positive x-axis, as shown in Figure 4. Because the distance from Fto V is a, the coordinates of F will be with Similarly, because the distancefrom V to the directrix D is also a and, because D must be perpendicular to the x-axis (since the x-axis is the axis of symmetry), the equation of the directrix D mustbe

Now, if is any point on the parabola, P must obey equation (1):

So we have

Use the Distance Formula.

Square both sides.

Remove parentheses.

Simplify. y2 = 4ax

x2 - 2ax + a2 + y2 = x2 + 2ax + a2

1x - a22 + y2 = 1x + a22 41x - a22 + 1y - 022 = ƒx + a ƒ

d1F, P2 = d1P, D2P = 1x, y2x = -a.

a 7 0.1a, 0210, 02,1

aa

2a

Axis of symmetry

F

P

d (F, P )d (P, D )

V

Directrix D

Figure 3

y

(0, 0)

P ! (x, y )

Vd (F, P )

d (P, D )

F ! (a, 0)

("a, y)

D : x ! "a

x

Figure 4


OBJECTIVES 1 Analyze Parabolas with Vertex at the Origin (p. 778)

2 Analyze Parabolas with Vertex at (h, k) (p. 781)

3 Solve Applied Problems Involving Parabolas (p. 783)

11.2 The Parabola

• Distance Formula (Section 2.1, p. 151)• Symmetry (Section 2.2, pp. 160–162)• Square Root Method (Section 1.2,

pp. 94–95)

• Completing the Square (Chapter R, Review,Section R.5, p. 56)

• Graphing Techniques: Transformations (Section 3.5,pp. 244–253)


SECTION 11.2 The Parabola 779

Recall that a is the distance from the vertex to the focus of a parabola. Whengraphing the parabola it is helpful to determine the “opening” by findingthe points that lie directly above or below the focus . This is done by letting

in so or The line segment joining thetwo points, and is called the latus rectum; its length is 4a.1a, -2a2,1a, 2a2 y = ;2a.y2 = 4a1a2 = 4a2,y2 = 4ax,x = a

1a, 02y2 = 4ax

THEOREM Equation of a Parabola: Vertex at (0, 0), Focus at (a, 0), a 0

The equation of a parabola with vertex at focus at and directrixis

(2)y2 = 4ax

x = -a, a 7 0,1a, 02,10, 02, >

F ! (3, 0)

D : x ! "3

x

y

V

(3, 6)

Latusrectum

(0, 0)

(3, "6)

6"6

6

"6

Figure 5

D: x ! "2

x

y

"5

"5

5

5

V F ! (2, 0)

(2, 4)

(2, "4)

Latus rectum

(0, 0)

Figure 6

Solution The equation is of the form where , so Consequently, the graph of the equation is a parabola with vertex at andfocus on the positive x-axis at The directrix is the vertical line

The two points that determine the latus rectum are obtained by lettingThen so The points and determine the latus

rectum. See Figure 6 for the graph.12, 4212, –42y = ;4.y2 = 16,x = 2.

x = -2.1a, 02 = 12, 02. 10, 02a = 2.4a = 8y2 = 4ax,y2 = 8x

The distance from the vertex to the focus is Based on equation (2),the equation of this parabola is

To graph this parabola, we find the two points that determine the latus rectum byletting Then

Solve for y.

The points and determine the latus rectum. These points help ingraphing the parabola because they determine the “opening.” See Figure 5.

13, -6213, 62 y = ;6 y2 = 12x = 12132 = 36

x = 3.

a = 3 y2 = 12x y2 = 4ax

a = 3.13, 0210, 02Solution

COMMENT To graph the paraboladiscussed in Example 1, we need

to graph the two functionsand Do this and comparewhat you see with Figure 5. !

Y2 = -112x .Y1 = 112x

y 2 = 12x


By reversing the steps used to obtain equation (2), it follows that the graph of anequation of the form of equation (2), is a parabola; its vertex is at its focus is at its directrix is the line and its axis of symmetry is the x-axis.

For the remainder of this section, the direction “Analyze the equation” willmean to find the vertex, focus, and directrix of the parabola and graph it.

x = -a,1a, 02, 10, 02,y2 = 4ax,

Finding the Equation of a Parabola and Graphing It

Find an equation of the parabola with vertex at and focus at Graph theequation.

13, 02.10, 02EXAMPLE 1

Analyzing the Equation of a Parabola

Analyze the equation: y2 = 8x

EXAMPLE 2

"

"


Recall that we obtained equation (2) after placing the focus on the positivex-axis. If the focus is placed on the negative x-axis, positive y-axis, or negative y-axis,a different form of the equation for the parabola results. The four forms of theequation of a parabola with vertex at and focus on a coordinate axis a distance afrom are given in Table 1, and their graphs are given in Figure 7. Notice thateach graph is symmetric with respect to its axis of symmetry.

10, 02 10, 02

x

(d) x 2 ! "4ay

Vx

y

(c) x 2 ! 4ay

V

y

(b) y 2 ! "4ax

Vx

y

V

(a) y 2 ! 4ax

x

y

F ! (0, "a)

D: y ! aF ! (0, a)

D: y ! "a

D: x ! a

F ! ("a, 0)

D: x ! "a

F ! (a, 0)

Figure 7

Table 1 Equations of a Parabola: Vertex at (0, 0); Focus on an Axis; a 0

Vertex Focus Directrix Equation Description

Axis of symmetry is the x-axis, opens right

Axis of symmetry is the x-axis, opens left

Axis of symmetry is the y-axis, opens up

Axis of symmetry is the y-axis, opens downx2 = -4ayy = a(0, -a)(0, 0)

x2 = 4ayy = -a(0, a)(0, 0)

y2 = -4axx = a(-a, 0)(0, 0)

y2 = 4axx = -a(a, 0)(0, 0)

>

The equation is of the form with Consequently, thegraph of the equation is a parabola with vertex at focus at , and directrixthe line The parabola opens down, and its axis of symmetry is the y-axis. Toobtain the points defining the latus rectum, let Then so The points determine the latus rectum. See Figure 8 for thegraph.


1–6, –32 and 16, –32 x = ;6.x2 = 36,y = -3.y = 3.

10, -3210, 02, a = 3.x2 = -4ay,x2 = -12y

Solution A parabola whose focus is at and whose directrix is thehorizontal line will have its vertex at (Do you see why? The vertexis midway between the focus and the directrix.) Since the focus is on the positive y-axis at the equation of this parabola is of the form with that is,

The points 8, 4 and –8, 4 determine the latus rectum. Figure 9 shows the graph ofx2 = 16y.

2121 a = 4 c

x2 = 4ay = 4142y = 16y

a = 4;x2 = 4ay,10, 42, 10, 02.y = -410, 42

D: y ! 3

F ! (0, "3)(6, "3)

("6, "3)

(0, 0) x

y

"6 6

6

V

Figure 8

x

y

V10(0, 0)

(!8, 4) (8, 4)

!10

10

!10

D : y " !4

F " (0,4)

Figure 9

Solution


Analyze the equation: x2 = -12y

EXAMPLE 3

Finding the Equation of a Parabola

Find the equation of the parabola with focus at and directrix the line Graph the equation.

y = -4.10, 42EXAMPLE 4

"

"

SECTION 11.2 The Parabola 781

The vertex is at the origin, the axis of symmetry is the x-axis, and the graph containsa point in the second quadrant, so the parabola opens to the left.We see from Table 1that the form of the equation is

Because the point is on the parabola, the coordinates must

satisfy . Substituting and into this equation, we find

The equation of the parabola is

The focus is at and the directrix is the line Let Then so The points and determine the latus rectum.

See Figure 10.


1-2, -421-2, 42y = ;4.y2 = 16,x = -2.x = 2.1-2, 02 y2 = -4122x = -8x

a = 2

y2 = -4ax; x = - 12

, y = 2 4 = -4aa- 12b y = 2x = - 1

2y2 = -4ax

x = - 12

, y = 2a- 12

, 2b y2 = -4ax

Solution

Analyze Parabolas with Vertex at (h, k)If a parabola with vertex at the origin and axis of symmetry along a coordinate axisis shifted horizontally h units and then vertically k units, the result is a parabola withvertex at and axis of symmetry parallel to a coordinate axis. The equations ofsuch parabolas have the same forms as those in Table 1, but with x replaced by

(the horizontal shift) and y replaced by (the vertical shift). Table 2gives the forms of the equations of such parabolas. Figures 11(a)–(d) on page 782,illustrate the graphs for h 7 0, k 7 0.

y - kx - h

1h, k22

NOTE It is not recommended that Table 2 be memorized. Rather use the ideas of transformations(shift horizontally h units, vertically k units) along with the fact that a represents the distancefrom the vertex to the focus to determine the various components of a parabola. It is also helpfulto remember that parabolas of the form “ ” will open up or down, while parabolas of the form“ ” will open left or right. !y 2 =

x 2 =

x

y

V

(!2, 4)

(!2, !4)

(0, 0) 5!5

5

!5

1–2(! , 2)

F " (!2, 0)

D : x " 2Figure 10

Table 2 Equations of a Parabola: Vertex at (h, k); Axis of Symmetry Parallel to a Coordinate Axis; a 0

Vertex Focus Directrix Equation Description

Axis of symmetry is parallel to the x-axis,opens right

Axis of symmetry is parallel to the x-axis,opens left

Axis of symmetry is parallel to the y-axis,opens up

Axis of symmetry is parallel to the y-axis,opens down

(x - h)2 = -4a(y - k)y = k + a(h, k - a)(h, k)

(x - h)2 = 4a(y - k)y = k - a(h, k + a)(h, k)

(y - k)2 = -4a(x - h)x = h + a(h - a, k)(h, k)

(y - k)2 = 4a(x - h)x = h - a(h + a, k)(h, k)

>

Finding the Equation of a Parabola

Find the equation of a parabola with vertex at if its axis of symmetry is the

x-axis and its graph contains the point Find its focus and directrix, andgraph the equation.

a- 12

, 2b .

10, 02EXAMPLE 5

"


y

V " (h, k)

Axis ofsymmetryx " h

x

y

V " (h, k)

Axis ofsymmetryx " h

y

V " (h, k)Axis ofsymmetryy " k

x

y

V " (h, k)Axis ofsymmetryy " k

F " (h, k # a)

D : y " k ! aF " (h, k ! a)

D : y " k # a

F " (h ! a, k)

D : x " h # a

F " (h # a, k)

D : x " h ! a

(c) (x ! h)2 " 4a(y ! k) (d) (x ! h)2 " !4a(y ! k)(b) (y ! k)2 " !4a(x ! h)(a) (y ! k)2 " 4a(x ! h)

xx

Figure 11

The vertex and focus both lie on the horizontal line (the axis ofsymmetry). The distance a from the vertex to the focus is Also, because the focus lies to the right of the vertex, the parabola opens to theright. Consequently, the form of the equation is

where and Therefore, the equation is

To find the points that define the latus rectum, let so that Then so or The points and determine thelatus rectum; the line is the directrix. See Figure 12.


Polynomial equations define parabolas whenever they involve two variablesthat are quadratic in one variable and linear in the other.

x = -410, 7210, -12y = 7.y = -1y - 3 = ;4,

1y - 322 = 16.x = 0,

1y - 322 = 81x + 22 1y - 322 = 4 # 23x - 1-224a = 2.1h, k2 = 1-2, 32 1y - k22 = 4a1x - h2a = 2.10, 321-2, 32 y = 310, 321-2, 32Solution

To analyze the equation complete the square involving thevariable x.

Isolate the terms involving x on the left side.

Complete the square on the left side.

Factor.

This equation is of the form with andThe graph is a parabola with vertex at that opens up. The

focus is at and the directrix is the line See Figure 13.


y = -2.1-2, 02, 1h, k2 = 1-2, -12a = 1.h = -2, k = -1,1x - h22 = 4a1y - k2, 1x + 222 = 41y + 12 x2 + 4x + 4 = 4y + 4

x2 + 4x = 4y x2 + 4x - 4y = 0

x2 + 4x - 4y = 0,Solution

x

y

V " (!2, 3)

(0, !1)

(0, 7)8

!4

6!6

Axis ofsymmetryy " 3

D : x " !4

F " (0, 3)

Figure 12

x!4 4

!3

Axis ofsymmetryx " !2

V " (!2, !1)

y

4

D : y " !2

F " (!2, 0)(!4, 0)(0, 0)

Figure 13

Finding the Equation of a Parabola, Vertex Not at the Origin

Find an equation of the parabola with vertex at and focus at Graphthe equation.

10, 32.1-2, 32EXAMPLE 6


Analyze the equation: x2 + 4x - 4y = 0

EXAMPLE 7

"

"


To obtain points on the ellipse off the x-axis, it must be that To see why,look again at Figure 18. Then

The sum of the lengths of two sides of a triangleis greater than the length of the third side.

Since we also have so Let Thenand equation (1) can be written as

Divide each side by

As you can verify, the graph of this equation has symmetry with respect to the x-axis, y-axis, and origin.

Because the major axis is the x-axis, we find the vertices of this ellipse by

letting The vertices satisfy the equation the solutions of which are

Consequently, the vertices of this ellipse are and The y-intercepts of the ellipse, found by letting have coordinates and

These four intercepts, and are used to graphthe ellipse.

10, -b2,10, b2,1a, 02, 1-a, 02,10, b2. 10, -b2x = 0,V2 = 1a, 02.V1 = 1-a, 02x = ;a.

x2

a2 = 1,y = 0.

a2 b2.

x2

a2 +y2

b2 = 1

b2 x2 + a2

y2 = a2 b2

a 7 bb2 = a2 - c2, b 7 0.a2 - c2 7 0.a2 7 c2,a 7 c 7 0,

a 7 cd(F1 , P) + d(F2 , P) = 2a, d(F1 , F2) = 2c 2a 7 2c

d1F1 , P2 + d1F2 , P2 7 d1F1 , F22a 7 c.

THEOREM Equation of an Ellipse: Center at (0, 0); Major Axis along the x-Axis

An equation of the ellipse with center at foci at and andvertices at and is

(2)

The major axis is the x-axis. See Figure 19.

x2

a2 +y2

b2 = 1, where a 7 b 7 0 and b2 = a2 - c2

1a, 021-a, 02 1c, 02,1-c, 0210, 02,x

y

(0, b)

(0, !b)

b acV1 " (!a, 0) V2 " (a, 0)

F1 " (!c, 0) F2 " (c, 0)

Figure 19

Sum of the distances from Pto the foci equals a constant,


Isolate one radical.

Square both sides.

Remove parentheses.

Simplify; isolate the radical.

Divide each side by 4.

Square both sides again.

Remove parentheses.

Rearrange the terms.

Multiply each side by factor on the right side. (1)a2

- 1; 1a2 - c22x2 + a2 y2 = a21a2 - c22 1c2 - a22x2 - a2 y2 = a2

c2 - a4

c2 x2 - 2a2 cx + a4 = a21x2 - 2cx + c2 + y22 1cx - a222 = a231x - c22 + y24 cx - a2 = -a41x - c22 + y2

4cx - 4a2 = -4a41x - c22 + y2

+ x2 - 2cx + c2 + y2

x2 + 2cx + c2 + y2 = 4a2 - 4a41x - c22 + y2

+ 1x - c22 + y2

1x + c22 + y2 = 4a2 - 4a41x - c22 + y2

41x + c22 + y2 = 2a - 41x - c22 + y2

41x + c22 + y2 + 41x - c22 + y2 = 2a2a.

d1F1 , P2 + d1F2 , P2 = 2a

convenient to let 2a denote the constant distance referred to in the definition. Then,if is any point on the ellipse, we haveP = 1x, y2

SECTION 11.3 The Ellipse 789

Notice in Figure 19 the right triangle formed by the points andBecause (or ), the distance from the focus at

to the point is a.This can be seen another way. Look at the two right triangles in Figure 19. They

are congruent. Do you see why? Because the sum of the distances from the foci to apoint on the ellipse is 2a, it follows that the distance from (c, 0) to (0, b) is a.

10, b2 1c, 02b2 + c2 = a2b2 = a2 - c210, b2. 10, 02, 1c, 02,

The ellipse has its center at the origin and, since the given focus and vertex lie on thex-axis, the major axis is the x-axis. The distance from the center, to one of thefoci, is The distance from the center, to one of the vertices,

is From equation (2), it follows that

so an equation of the ellipse is

Figure 20 shows the graph.

In Figure 20, the intercepts of the equation are used to graph the ellipse. Followingthis practice will make it easier for you to obtain an accurate graph of an ellipsewhen graphing.

COMMENT The intercepts of the ellipse also provide information about how to set the viewing rectangle for graphing an ellipse. To graph the ellipse

discussed in Example 1, set the viewing rectangle using a square screen that includes the intercepts,perhaps Then proceed to solve the equation for y:

Subtract from each side.

Multiply both sides by 7.

Take the square root of each side.

Now graph the two functions

and

Figure 21 shows the result. !


Y2 = -C7¢ 1 - x2

16!Y1 = C7¢ 1 - x2

16!

y = ;B7¢ 1 - x2

16!

y 2 = 7¢ 1 - x 2

16!

x2

16 y 2

7= 1 - x 2

16

x2

16+

y2

7= 1

–4.5 … x … 4.5, –3 … y … 3.

x2

16+

y2

7= 1

x2

16+y2

7= 1

b2 = a2 - c2 = 16 - 9 = 7

a = 4.1-4, 02, 10, 02,c = 3.13, 02, 10, 02,Solution

V1 " (!4, 0) V2 " (4, 0)

x

y

!5 5

5

!5

(0, )7

(0, ! )7

F2 " (3, 0)F1 " (!3, 0)

Figure 20

x 2

16+

y 2

7= 1

3x 216

!3

!4.5 4.5

Y1 " 7 1 !( )

x 216Y2 " ! 7 1 !( )

Figure 21

Finding an Equation of an Ellipse

Find an equation of the ellipse with center at the origin, one focus at and avertex at Graph the equation.1-4, 02. 13, 02,EXAMPLE 1

"


An equation of the form of equation (2), with is the equation of an ellipse with center at the origin, foci on the x-axis at and where

and major axis along the x-axis.For the remainder of this section, the direction “Analyze the equation” will

mean to find the center, major axis, foci, and vertices of the ellipse and graph it.

c2 = a2 - b2,1c, 02,1–c, 02a2 7 b2,

x

y

!6 6

6

V2 " (5, 0)F2 " (4, 0)F1 " (!4, 0)V1 " (!5, 0)

(0, 3)

(0, !3)

Figure 22

xb

ac

y

(b, 0)(!b, 0)

V2 " (0, a )

V1 " (0, !a)

F2 " (0, c)

F1 " (0, !c)

Figure 23

The given equation is of the form of equation (2), with and Theequation is that of an ellipse with center and major axis along the x-axis. Thevertices are at Because we find that

The foci are at Figure 22 shows the graph.1;c, 02 = 1;4, 02.c2 = a2 - b2 = 25 - 9 = 16

b2 = a2 - c2,1;a, 02 = 1;5, 02. 10, 02 b2 = 9.a2 = 25Solution


If the major axis of an ellipse with center at lies on the y-axis, the foci areat and Using the same steps as before, the definition of an ellipseleads to the following result:

10, c2.10, -c2 10, 02THEOREM Equation of an Ellipse: Center at (0, 0); Major Axis along the y-Axis

An equation of the ellipse with center at foci at and andvertices at and is

(3)

The major axis is the y-axis.

x2

b2 +y2

a2 = 1 where a 7 b 7 0 and b2 = a2 - c2

10, a210, -a2 10, c2,10, -c210, 02,

Figure 23 illustrates the graph of such an ellipse. Again, notice the right triangleformed by the points at and , so that

Look closely at equations (2) and (3). Although they may look alike, there is adifference! In equation (2), the larger number, is in the denominator of the

so the major axis of the ellipse is along the x-axis. In equation (3), the largernumber, is in the denominator of the so the major axis is along they-axis.

y2-term,a2,x2-term,

a2,

a2 = b2 + c2 (or b2 = a2 - c2).10, c210, 02, 1b, 02,

Analyzing the Equation of an Ellipse

Analyze the equation:x2

25+y2

9= 1

EXAMPLE 2

"

SECTION 11.3 The Ellipse 791

To put the equation in proper form, divide each side by 9.

The larger denominator, 9, is in the so, based on equation (3), this is theequation of an ellipse with center at the origin and major axis along the y-axis.Also, we conclude that and The verticesare at and the foci are at Figure 24 showsthe graph.


10, ;c2 = 10, ;2122.10, ;a2 = 10, ;32, 9 - 1 = 8.c2 = a2 - b2 =a2 = 9, b2 = 1,

y2-term

x2 +y2

9= 1

Solution

By plotting the given focus and vertices, we find that the major axis is the y-axis.Because the vertices are at and the center of this ellipse is at theirmidpoint, the origin.The distance from the center, to the given focus, is

The distance from the center, to one of the vertices, is So The form of the equation of this ellipse is given byequation (3).



The circle may be considered a special kind of ellipse. To see why, let inequation (2) or (3). Then

This is the equation of a circle with center at the origin and radius a. The value of c is

We conclude that the closer the two foci of an ellipse are to the center, the morethe ellipse will look like a circle.

a = bc

c2 = a2 - b2 = 0

x2 + y2 = a2

x2

a2 +y2

a2 = 1

a = b

x2

5+y2

9= 1

x2

b2 +y2

a2 = 1

b2 = a2 - c2 = 9 - 4 = 5.a = 3.10, 32,10, 02,c = 2.10, 22,10, 02,10, 32,10, -32Solution

Analyze Ellipses with Center at (h, k)If an ellipse with center at the origin and major axis coinciding with a coordinate axis isshifted horizontally h units and then vertically k units, the result is an ellipse with centerat and major axis parallel to a coordinate axis. The equations of such ellipseshave the same forms as those given in equations (2) and (3), except that x is replacedby (the horizontal shift) and y is replaced by (the vertical shift). Table 3gives the forms of the equations of such ellipses, and Figure 26 shows their graphs.

y - kx - h

1h, k22

x

y

3

3

!3

!3

V1 " (0, !3)

V2 " (0, 3)

(1, 0)(!1, 0)

F1 " (0, !2 2)

F2 " (0, 2 2)

Figure 24

x

y

(! , 0)

!3 3

3

!3

V2 " (0, 3)

V1 " (0, !3)

5 ( , 0)5

F2 " (0, 2)

F1 " (0, !2)

Figure 25

Analyzing the Equation of an Ellipse

Analyze the equation: 9x2 + y2 = 9

EXAMPLE 3

Finding an Equation of an Ellipse

Find an equation of the ellipse having one focus at and vertices at and Graph the equation.10, 32. 10, -3210, 22EXAMPLE 4

"

"


NOTE It is not recommended thatTable 3 be memorized. Rather, use the ideas of transformations (shifthorizontally h units, vertically k units)along with the fact that a representsthe distance from the center to thevertices, c represents the distancefrom the center to the foci, and

(or !a2 - b2).c2 =b2 = a2 - c2

x

(h # a, k)(h , k)(h ! a, k) x

(h, k # a)

(h , k)

(h, k ! a)

(a) (b)

Major axis

(h # c, k)(h ! c, k)

Major axis

(h, k # c)

(h, k ! c)

y

y

(x ! h)2––––––

a2

(y ! k )2––––––

b2# " 1

(x ! h)2––––––

b2

(y ! k )2––––––

a2# " 1

Figure 26

Table 3 Equations of an Ellipse: Center at (h, k); Major Axis Parallel to a Coordinate Axis

Center Major Axis Foci Vertices Equation

Parallel to the x-axis

and

Parallel to the y-axis

and b2 = a2 - c2a 7 b 7 0(h, k - a)(h, k - c)

(x - h)2

b2 +(y - k)2

a2 = 1,(h, k + a)(h, k + c)(h, k)

b2 = a2 - c2a 7 b 7 0(h - a, k)(h - c, k)

(x - h)2

a2 +(y - k)2

b2 = 1,(h + a, k)(h + c, k)(h, k)

The center is at so and If we plot the center, focus,and vertex, we notice that the points all lie on the line so the major axis isparallel to the x-axis. The distance from the center to a focus is

the distance from the center to a vertex is ThenThe form of the equation is

To graph the equation, use the center to locate the vertices.The major axis is parallel to the x-axis, so the vertices are units left and rightof the center Therefore, the vertices are

Since and the major axis is parallel to the x-axis, the foci are 1 unit left andright of the center. Therefore, the foci are

Finally, use the value of to find the two points above and below the center.



A2, -3 - 222 B and A2, -3 + 222 Bb = 212

F1 = 12 - 1, -32 = 11, -32 and F2 = 12 + 1, -32 = 13, -32c = 1

V1 = 12 - 3, -32 = 1-1, -32 and V2 = 12 + 3, -32 = 15, -3212, -32. a = 31h, k2 = 12, -32

1x - 2229

+1y + 322

8= 1

1x - h22a2 +

1y - k22b2 = 1 where h = 2, k = -3, a = 3, b = 222

b2 = a2 - c2 = 9 - 1 = 8.a = 3.15, -3212, -32c = 1;13, -3212, -32y = -3,

k = -3.h = 21h, k2 = 12, -32,Solution

x

y

6!2

2(2, !3 # 2 )

2(2, !3 ! 2 )

(2, !3)

V2 " (5, !3)V1 " (!1, !3)

2

F1 F2

Figure 27

Finding an Equation of an Ellipse, Center Not at the Origin

Find an equation for the ellipse with center at one focus at and onevertex at Graph the equation.15, -32. 13, -32,12, -32,EXAMPLE 5

"

SECTION 11.4 The Hyperbola 797

Explaining Concepts: Discussion and Writing

85. The eccentricity e of an ellipse is defined as the number where a is the distance of a vertex from the center and c is the distance

of a focus from the center. Because it follows that Write a brief paragraph about the general shape of each of thefollowing ellipses. Be sure to justify your conclusions.

(a) Eccentricity close to 0 (b) (c) Eccentricity close to 1Eccentricity = 0.5

e 6 1.a 7 c,

ca

,

‘Are You Prepared?’ Answers

1. 2. 3. 4. 5. left; 1; down: 4 6. 1x - 222 + 1y + 322 = 112, 521-2, 02, 12, 02, 10, -42, 10, 4294

213

84. Show that the graph of an equation of the form

where A and C are of the same sign,

(a) Is an ellipse if is the same sign as A.D2

4A+ E

2

4C- F

Ax2 + Cy2 + Dx + Ey + F = 0, A Z 0, C Z 0(b) Is a point if

(c) Contains no points if is of opposite signto A.

D2

4A+ E

2

4C- F

D2

4A+ E

2

4C- F = 0.

Figure 31 illustrates a hyperbola with foci and The line containing thefoci is called the transverse axis. The midpoint of the line segment joining the foci isthe center of the hyperbola. The line through the center and perpendicular to thetransverse axis is the conjugate axis. The hyperbola consists of two separate curves,called branches, that are symmetric with respect to the transverse axis, conjugateaxis, and center. The two points of intersection of the hyperbola and the transverseaxis are the vertices, and of the hyperbola.V2 ,V1

F2 .F1

F2

Transverseaxis

Conjugateaxis

F1

CenterV1

V2

Figure 31

DEFINITION A hyperbola is the collection of all points in the plane, the difference of whosedistances from two fixed points, called the foci, is a constant.

Analyze Hyperbolas with Center at the OriginWith these ideas in mind, we are now ready to find the equation of a hyperbolain the rectangular coordinate system. First, place the center at the origin. Next,

1


OBJECTIVES 1 Analyze Hyperbolas with Center at the Origin (p. 797)

2 Find the Asymptotes of a Hyperbola (p. 802)

3 Analyze Hyperbolas with Center at (h, k) (p. 804)

4 Solve Applied Problems Involving Hyperbolas (p. 805)

11.4 The Hyperbola

• Distance Formula (Section 2.1, p. 151)• Completing the Square (Chapter R, Section R.5,

p. 56)• Intercepts (Section 2.2, pp. 159–160)• Symmetry (Section 2.2, pp. 160–162)

• Asymptotes (Section 5.2, pp. 345–348)• Graphing Techniques: Transformations (Section 3.5,

pp. 244–253)• Square Root Method (Section 1.2, pp. 94–95)


x

y

P " (x, y )Transverseaxis d (F2, P )

d (F1, P )

F2 " (c, 0)F1

" (!c, 0)

Figure 32d1F1, P2 - d1F2, P2 = ;2a


position the hyperbola so that its transverse axis coincides with a coordinate axis.Suppose that the transverse axis coincides with the x-axis, as shown in Figure 32.

If c is the distance from the center to a focus, one focus will be at and the other at . Now we let the constant difference of the distancesfrom any point on the hyperbola to the foci and be denoted by (If P is on the right branch, the sign is used; if P is on the left branch,the sign is used.) The coordinates of P must satisfy the equation

Difference of the distances from P to the foci equals


Isolate one radical.

Square both sides.

Next we remove the parentheses.

Simplify; isolate the radical.

Divide each side by 4.

Square both sides.

Simplify.

Remove parentheses and simplify.

Rearrange terms.

Factor on the right side. (1)

To obtain points on the hyperbola off the x-axis, it must be that . To seewhy, look again at Figure 32.

Use triangle

P is on the right branch, so

Since we also have so Let Then equation (1) can be written as

Divide each side by

To find the vertices of the hyperbola defined by this equation, let

The vertices satisfy the equation the solutions of which are

Consequently, the vertices of the hyperbola are andNotice that the distance from the center to either vertex is a.10, 02 V2 = 1a, 02.V1 = 1-a, 02 x = ;a.x2

a2 = 1,

y = 0.

a2 b2.

x2

a2 -y2

b2 = 1

b2 x2 - a2

y2 = a2 b2

b2 = c2 - a2, b 7 0.c2 - a2 7 0.a2 6 c2,a 6 c,

a 6 cd1F1, F22 = 2c.d(F1 , P) - d(F2 , P) = 2a;

2a 6 2c d1F1 , P2 - d1F2 , P2 6 d1F1 , F22 F1 PF2 . d1F1 , P2 6 d1F2 , P2 + d1F1 , F22

a 6 c

a2 1c2 - a22x2 - a2 y2 = a21c2 - a22 1c2 - a22x2 - a2 y2 = a2

c2 - a4

c2 x2 + a4 = a2 x2 + a2

c2 + a2 y2

c2 x2 - 2ca2 x + a4 = a21x2 - 2cx + c2 + y22 1cx - a222 = a231x - c22 + y24 cx - a2 = ;a41x - c22 + y2

4cx - 4a2 = ;4a41x - c22 + y2

x2 + 2cx + c2 + y2 = 4a2 ; 4a41x - c22 + y2 + x2 - 2cx + c2 + y2

+ 1x - c22 + y2

1x + c22 + y2 = 4a2 ; 4a41x - c22 + y2

41x + c22 + y2 = ;2a + 41x - c22 + y2

41x + c22 + y2 - 41x - c22 + y2 = ;2a;2a.

d1F1 , P2 - d1F2 , P2 = ;2a

-+;2a.

F2F1P = 1x, y2F2 = 1c, 02 F1 = 1-c, 02


THEOREM Equation of a Hyperbola: Center at (0, 0); Transverse Axis along the x-Axis

An equation of the hyperbola with center at foci at and and vertices at and is

(2)

The transverse axis is the x-axis.

x2

a2 -y2

b2 = 1, where b2 = c2 - a2

1a, 021-a, 02 1c, 02,1-c, 0210, 02,

See Figure 33. As you can verify, the hyperbola defined by equation (2) issymmetric with respect to the x-axis, y-axis, and origin. To find the y-intercepts, if

any, let in equation (2). This results in the equation which has no

real solution, so the hyperbola defined by equation (2) has no y-intercepts. In fact,

since it follows that There are no points on the graph for

-a 6 x 6 a.

x2

a2 Ú 1.x2

a2 - 1 =y2

b2 Ú 0,

y2

b2 = -1,x = 0F2 " (c, 0)F1

" (!c, 0) x

y

V2 " (a, 0)

V1 " (!a, 0)

Transverseaxis

Figure 33

x2

a2 -y2

b2 = 1, b2 = c2 - a2

The hyperbola has its center at the origin. Plot the center, focus, and vertex. Sincethey all lie on the x-axis, the transverse axis coincides with the x-axis. One focus isat so One vertex is at so From equation (2), it follows that so an equation of thehyperbola is

To graph a hyperbola, it is helpful to locate and plot other points on the graph.For example, to find the points above and below the foci, we let Then

The points above and below the foci are and These points

determine the “opening” of the hyperbola. See Figure 34.

a ;3, - 52b .a ;3,

52b

y = ;52

y2 = 254

y2

5= 5

4

94

-y2

5= 1

x = ;3 1;322

4-y2

5= 1

x2

4-y2

5= 1

x = ;3.

x2

4-y2

5= 1

b2 = c2 - a2 = 9 - 4 = 5,a = 2.1-a, 02 = 1-2, 02,c = 3.1c, 02 = 13, 02,Solution

Finding and Graphing an Equation of a Hyperbola

Find an equation of the hyperbola with center at the origin, one focus at andone vertex at Graph the equation.1-2, 02. 13, 02,EXAMPLE 1

"

x

y

F 2 " (3, 0)F1

" (!3, 0)!5 5

5

!5

V2 " (2, 0)

( )3,

V1 " (!2, 0)

5–2

( )3,! 5–2( )!3,! 5–2

( ) !3,5–2

Figure 34


xV1 = (0, –2)

V2 = (0, 2)

y

–5 5

5

25( ), 32

5(– ), 3

25( ), –32

5(– ), –3

F1 = (0, –3)

F2 = (0, 3)

–5

Figure 38

Figure 36 shows the graph of a typical hyperbola defined by equation (3).

An equation of the form of equation (2), is the equation of a

hyperbola with center at the origin, foci on the x-axis at and whereand transverse axis along the x-axis.

An equation of the form of equation (3), is the equation of a

hyperbola with center at the origin, foci on the y-axis at and whereand transverse axis along the y-axis.

Notice the difference in the forms of equations (2) and (3). When the issubtracted from the the transverse axis is along the x-axis. When the

is subtracted from the the transverse axis is along the y-axis.y2-term,x2-termx2-term,

y2-termc2 = a2 + b2,

10, c2,10, -c2y2

a2 - x2

b2 = 1,

c2 = a2 + b2,1c, 02,1-c, 02x2

a2 -y2

b2 = 1,

xV1 " (0, !a )

V2 " (0, a)

y

F2 " (0, c )

F 1 " (0, !c )

Figure 36

y2

a2 - x2

b2 = 1, b2 = c2 - a2

To put the equation in proper form, divide each side by 4:

Since the is subtracted from the the equation is that of a hyperbolawith center at the origin and transverse axis along the y-axis. Also, comparing the above equation to equation (3), we find and Thevertices are at and the foci are at

To locate other points on the graph, let Then

Four other points on the graph are and See Figure 37.1;2, -2152.1;2, 2152 y = ;225

y2 = 20

y2 - 16 = 4

x = ;2 y2 - 41;222 = 4

y2 - 4x2 = 4

x = ;2.10, ;c2 = 10, ;152.10, ;a2 = 10, ;22, c2 = a2 + b2 = 5.a2 = 4, b2 = 1,

y2-term,x2-term

y2

4- x2 = 1

Solution

xV1 = (0, –2)

V2 = (0, 2)

F1 = (0, – )

y

–5 5

5

–5

5

F2 = (0, )5

(2, 2 )5(–2, 2 )5

(–2, –2 )5 (2, –2 )5

Figure 37

Since the foci are at and the center of the hyperbola, which is attheir midpoint, is the origin. Also, the transverse axis is along the y-axis. The giveninformation also reveals that and Theform of the equation of the hyperbola is given by equation (3):

Let to obtain points on the graph on either side of each focus. See Figure 38.


y = ;3

y2

4- x

2

5= 1

y2

a2 - x2

b2 = 1

b2 = c2 - a2 = 9 - 4 = 5.c = 3, a = 2,

10, 32,10, -32Solution

Analyzing the Equation of a Hyperbola

Analyze the equation: y2 - 4x2 = 4

EXAMPLE 3

Finding an Equation of a Hyperbola

Find an equation of the hyperbola having one vertex at and foci at and Graph the equation.10, 32. 10, -3210, 22EXAMPLE 4

"

"

Look at the equations of the hyperbolas in Examples 2 and 4. For the hyperbolain Example 2, and so for the hyperbola in Example 4,and so We conclude that, for hyperbolas, there are no requirementsinvolving the relative sizes of a and b. Contrast this situation to the case of an ellipse,in which the relative sizes of a and b dictate which axis is the major axis. Hyperbolashave another feature to distinguish them from ellipses and parabolas: Hyperbolashave asymptotes.

a 6 b.b2 = 5,a2 = 4a 7 b;b2 = 4,a2 = 16

Proof We begin by solving for y in the equation of the hyperbola.

Since we can rearrange the right side in the form

Now, as or as the term approaches 0, so the expression under the

radical approaches 1. So, as or as the value of y approaches that is, the graph of the hyperbola approaches the lines

These lines are oblique asymptotes of the hyperbola. !

The asymptotes of a hyperbola are not part of the hyperbola, but they do serveas a guide for graphing a hyperbola. For example, suppose that we want to graphthe equation

x2

a2 -y2

b2 = 1

y = - ba

x and y = ba

x

; bxa

;x: q ,x: -q

a2

x2x: q ,x: -q

y = ;bxa

A1 - a2

x2

y2 = b2

x2

a2 ¢1 - a2

x2 !x Z 0,

y2 = b2¢x2

a2 - 1! y2

b2 = x2

a2 - 1

x2

a2 -y2

b2 = 1

Find the Asymptotes of a HyperbolaRecall from Section 5.2 that a horizontal or oblique asymptote of a graph is a linewith the property that the distance from the line to points on the graph approaches 0as or as Asymptotes provide information about the end behaviorof the graph of a hyperbola.

x: q .x: -q

2

THEOREM Asymptotes of a Hyperbola

The hyperbola has the two oblique asymptotes

(4)y = ba

x and y = - ba

x

x2

a2 -y2

b2 = 1



x

y

(0, b)

(0, !b)V1 " (!a, 0)

V2 " (a, 0)

y " xb–a

y " ! xb–a

Figure 39

x2

a2 -y2

b2 = 1

Begin by plotting the vertices and Then plot the points andand use these four points to construct a rectangle, as shown in Figure 39.

The diagonals of this rectangle have slopes and and their extensions are the

asymptotes and of the hyperbola. If we graph the asymptotes, we

can use them to establish the “opening” of the hyperbola and avoid plotting other points.

y = - ba

xy = ba

x

- ba

,ba

10, b2 10, -b21a, 02.1-a, 02

You are asked to prove this result in Problem 84.For the remainder of this section, the direction “Analyze the equation” will mean

to find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola andgraph it.

THEOREM Asymptotes of a Hyperbola

The hyperbola has the two oblique asymptotes

(5)y = ab

x and y = - ab

x

y2

a2 - x2

b2 = 1

Solution Since the is subtracted from the the equation is ofthe form of equation (3) and is a hyperbola with center at the origin and transverseaxis along the y-axis. Also, comparing this equation to equation (3), we findthat and The vertices are atand the foci are at Using equation (5) with and ,

the asymptotes are the lines and Form the rectangle

containing the points and The extensions ofthe diagonals of this rectangle are the asymptotes. Now graph the rectangle, theasymptotes, and the hyperbola. See Figure 40.

1;b, 02 = 1;1, 02.10, ;a2 = 10, ;22 y = - ab

x = -2x.y = ab

x = 2x

b = 1a = 210, ;c2 = 10, ;152. 10, ;a2 = 10, ;22,c2 = a2 + b2 = 5.a2 = 4, b2 = 1,

y2-term,x2-term

Divide each side of the equation by 36 to put the equation in proper form.

The center of the hyperbola is the origin. Since the is first in the equation,the transverse axis is along the x-axis and the vertices and foci will lie on the x-axis.Using equation (2), we find and The vertices are

units left and right of the center at the foci are c = 1131;a, 02 = 1;2, 02,a = 2c2 = a2 + b2 = 13.a2 = 4, b2 = 9,

x2-term

x2

4-y2

9= 1

Solution

xV1 ! (0, " 2)

V2 ! (0, 2)

y ! "2x y ! 2xy

"5

"5

5

5

F2 ! (0, 5)

F1 ! (0, " 5)

Figure 40


Analyze the equation:y2

4- x2 = 1

EXAMPLE 5


Analyze the equation: 9x2 - 4y2 = 36

EXAMPLE 6

"

Analyze Hyperbolas with Center at (h, k)If a hyperbola with center at the origin and transverse axis coinciding with acoordinate axis is shifted horizontally h units and then vertically k units, the result isa hyperbola with center at and transverse axis parallel to a coordinate axis.The equations of such hyperbolas have the same forms as those given in equations(2) and (3), except that x is replaced by (the horizontal shift) and y is replacedby (the vertical shift). Table 4 gives the forms of the equations of suchhyperbolas. See Figure 42 for typical graphs.y - k

x - h

1h, k23

x

y

"5

"5

5

5

(0, 3)

V2 ! (2, 0)V1 ! ("2, 0)

(0, "3)

x3–2

y ! " y !x3–2

F2 ! ( 13, 0)F1 ! (" 13, 0)

Figure 41 units left and right of the center at and the asymptotes havethe equations

To graph the hyperbola, form the rectangle containing the points andthat is, and The extensions of the diagonals

of this rectangle are the asymptotes. See Figure 41 for the graph.


10, 32.1-2, 02, 12, 02, 10, -32,10, ;b2, 1;a, 02y = ba

x = 32

x and y = - ba

x = - 32

x

1;c, 02 = 1;113, 02,

Equations of a Hyperbola: Center at (h, k); Transverse Axis Parallel to a Coordinate Axis

CenterTransverse Axis Foci Vertices Equation Asymptotes

(h, k)Parallel to the x-axis

Parallel to they-axis y - k = ; a

b (x - h)

(y - k)2

a2 - (x - h)2

b2 = 1, b2 = c2 - a2(h, k ; a)(h, k ; c)(h, k)

y - k = ; ba

(x - h)(x - h)2

a2 -(y - k)2

b2 = 1, b2 = c2 - a2(h ; a, k)(h ; c, k)

Table 4

x

y

V2

V1

(h, k )

Transverseaxis

(h, k )

Transverseaxis

x

y

V2V1

(b) ! " 1(y ! k )2 –––––––

a2(x ! h)2 ––––––

b2(a) ! " 1

(x ! h )2 –––––––a2

(y ! k)2 ––––––b2

F2

F1

F2F1

Figure 42

NOTE It is not recommended thatTable 4 be memorized. Rather use theideas of transformations (shifthorizontally h units, vertically k units)along with the fact that a representsthe distance from the center to thevertices, c represents the distancefrom the center to the foci, andb2 = c2 - a2 (or c2 = a2 ! b2). !

The center is at so and Since the center, focus, andvertex all lie on the line the transverse axis is parallel to the x-axis. Thedistance from the center to the focus is the distance fromc = 3;14, -2211, -22y = -2,

k = -2.h = 11h, k2 = 11, -22,Solution


Finding an Equation of a Hyperbola, Center Not at the Origin

Find an equation for the hyperbola with center at one focus at andone vertex at Graph the equation.13, -22. 14, -22,11, -22,EXAMPLE 7

"

344 CHAPTER 5 Polynomial and Rational Functions

Table 9

x

x ! 0

y ! 0 y ! 03%3

y

5

2,

(1, 1)1–4

, 41–2

(%2,

(%1, 1)1–4

(% , 41–2 ) ( )

( ))

Figure 28

H(x) = 1

x2

x ! 0

x

y

%2 3

Replace x by x % 2;shift right2 units

1

Add 1;shift up1 unit

x ! 2x ! 2

y ! 1(1, 1)(%1, 1)

x

y

5

3

(3, 1)(1, 1)

x

y

5

3(3, 2)

(1, 2)

(b) y ! 1(x – 2)2(a) y ! 1

x 2 (c) y ! " 1 1(x % 2)2

y ! 0 y ! 0

Figure 29

Graphing

Analyze the graph of .H1x2 = 1x2

y !1x2

EXAMPLE 2

Using Transformations to Graph a Rational Function

Graph the rational function: R1x2 = 11x - 222 + 1

EXAMPLE 3

Solution The domain of R is the set of all real numbers except . To graph R, start with

the graph of . See Figure 29 for the steps.y = 1x2

x = 2

x

4

10,000

100,000,000

1 1

2

100

10,0001

100,000,000

110,000

14

110,000

1100

12

H(x) !1

x2

Solution The domain of is the set of all real numbers x except 0.The graph has no

y-intercept, because x can never equal 0. The graph has no x-intercept because theequation has no solution. Therefore, the graph of H will not cross ortouch either of the coordinate axes. Because

H is an even function, so its graph is symmetric with respect to the y-axis.

Table 9 shows the behavior of for selected positive numbers x. (We

will use symmetry to obtain the graph of H when .) From the first three rowsof Table 9, we see that, as the values of x approach (get closer to) 0, the values of

become larger and larger positive numbers, so H is unbounded in the positivedirection. We use limit notation, , read “the limit of as x

approaches zero equals infinity,” to mean that as .Look at the last four rows of Table 9. As , the values of approach 0

(the end behavior of the graph). In calculus, this is symbolized by writing. Figure 28 shows the graph. Notice the use of red dashed lines to

convey the ideas discussed above.limx:q

H1x2 = 0

H1x2x: qx: 0H1x2: q

H1x2limx:0

H1x2 = qH1x2 x 6 0

H1x2 = 1x2

H1-x2 = 11-x22 = 1x2 = H1x2

H1x2 = 0

H1x2 = 1x2

!Now Work P R O B L E M 3 3

!

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EXAMPLE 2 Graphing the Absolute Value Function DE GRAFICAS.pdffoundation for further graphing techniques. Constant Function f1x2 = b b is a real number x y f(x ) = b (0,b) Figure 31