introduction to analytic geometry · introduction to analytic geometry 1.1 the cartesian coordinate...

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Chapter 1

Introduction to Analytic Geometry

1.1 The Cartesian Coordinate System

1. Let (x2, y2) = (2, 4) and (x1, y1) = (5, 2). From the distance formula

d =√

(x2 − x1)2 + (y2 − y1)2

we get

d =√

(2− 5)2 + (4− 2)2 =√

(−3)2 + 22 =√

9 + 4 =√

13.

2. Let (x2 , y2) = (−3 , 2) and (x1 , y1) = (5 , −4). From the distance formula

d =

√(x2 − x1)

2+(y2 − y1)

2we get

d =

√(−3− 5)

2+[2− (−4)]

2=

√(−8)

2+(6)

2=√

64+36 = 10.

3. Let (x2, y2) = (−3,−6) and (x1, y1) = (5,−2). Then

d =√

(−3− 5)2 + [−6− (−2)]2 =√

(−8)2 + (−4)2 =√

64 + 16 =√

16 · 5 = 4√

5.

4. Let (x2 , y2) =(−√

5 , 2) and (x1 , y1) = (0 , 0).

Then d =

√(−√

5− 0)2+(2− 0)

2=√

5+4 = 3

5. Let (x2, y2) = (√

3, 4) and (x1, y1) = (0, 2). Then

d =

√(√

3− 0)2 + (4− 2)2 =√

3 + 4 =√

7.

6. d =

√(√

2−√

2)2+(√

5− 0)2=√

0+5 =√

5

7. d =√

[1− (−1)]2 + (−√

2− 0)2 =√

4 + 2 =√

6

8. d =

√[2− (−2)]

2+ (7− 3)2 =

√16 + 16 =

√2 · 16 = 4

√2

9. d =√

[−9− (−11)]2 + (−1− 1)2 =√

22 + (−2)2 =√

8 =√

2 · 4 = 2√

2

10. Distance from (0 , 0) to (4 , 3):

√(4− 0)

2+(3− 0)

2=√

16 + 9 = 5. Distance from (0 , 0) to

(6 , 0)= 6. Distance from (4 , 3) to (6 , 0):

√(6− 4)

2+(0− 3)

2=√

4 + 9 =√

13.

Perimeter = 5 + 6 +√

13 = 11 +√

13.

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n.Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig

Full file at https://TestbankDirect.eu/


https://TestbankDirect.eu/Solution-Manual-for-Technical-Calculus-with-Analytic-Geometry-5th-Edition-by-Kuhfittig


2 CHAPTER 1. INTRODUCTION TO ANALYTIC GEOMETRY

11b. x/y is negative whenever x and y have opposite signs: quadrants II and IV.

12. If x2 > x1, then x2 − x1 = P1P3 = |x2 − x1|. If x2 < x1, then x2 − x1 = −P1P3 and

P1P3 = |−P1P3| = |x2 − x1|.

13a. Any point on the y-axis has coordinates of the form (0, y).

14. Distance from (11, 2) to origin:√

(11− 0)2 + (2− 0)2 =√

125

Distance from (−5, 10) to origin:√

(−5− 0)2 + (10− 0)2 =√

125

Distance from (−1,−11) to origin:√

(−1− 0)2 + (−11− 0)2 =√

122

Answer: (−1,−11)

15. Let A = (−2,−5), B = (−4, 1) and C = (5, 4); then AB =√

[−2− (−4)]2 + (−5− 1)2 =√

40, AC =√

(−2− 5)2 + (−5− 4)2 =√

130, and BC =√

(−4− 5)2 + (1− 4)2 =√

90.

Since (AB)2 + (BC)2 = (AC)2, the triangle must be a right triangle.

16. Let A = (−1,−1), B = (−2, 3), and C = (6, 5). After calculating AB =√

17, BC =√

68, and

AC =√

85, we observe that (AB)2 + (BC)2 = (AC)2.

17. The points (12, 0), (−4, 8) and (−1,−13) are all 5√

5 units from (1,−2).

18. Distance from (−2 , 10) to (3 , −2): 13. Distance from (15 , 3) to (3 , −2): 13.

19. Distance from (−1,−1) to (2, 8):√(−1− 2)2 + (−1− 8)2 =

√9 + 81 =

√90 =

√9 · 10 = 3

√10

Distance from (2, 8) to (5, 17):√(5− 2)2 + (17− 8)2 =

√90 = 3

√10

Distance from (−1,−1) to (5, 17):√62 + 182 =

√360 = 6

√10

Total distance 6√

10 = 3√

10 + 3√

10, the sum of the other two distances.

20. Distance from (x , y) to (−1 , 2):

d =√

(x+ 1)2 + (y − 2)2 = 3

(x+ 1)2 + (y − 2)2 = (3)2 squaring both sides

x2 + 2x+ 1 + y2 − 4y + 4 = 9

x2 + y2 + 2x− 4y = 4

21. Distance from (x, y) to y-axis: x units

Distance from (x, y) to (2, 0):√

(x− 2)2 + (y − 0)2 =√

(x− 2)2 + y2

By assumption,√(x− 2)2 + y2 = x

(x− 2)2 + y2 = x2 squaring both sides

x2 − 4x+ 4 + y2 = x2

y2 − 4x+ 4 = 0

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Solution Manual for Technical Calculus with Analytic Geometry 5th Edition by Kuhfittig






1.2. THE SLOPE 3

22. Let (x1 , y1) = (−3 , −5) and (x2 , y2) = (−1 , 7). Then from the midpoint formula(x1 + x2

2,y1 + y2

2

)=

(−3 + (−1)

2,−5 + 7

2

)= (−2 , 1).

23. Let (x1, y1) = (−2, 6) and (x2, y2) = (2,−4). Then from the midpoint formula(x1 + x2

2,y1 + y2

2

)we get (

−2 + 2

2,

6 + (−4)

2

)= (0, 1).

24.

(x1 + x2

2,y1 + y2

2

)=

(−3 + (−2)

2,

5 + 9

2

)=

(−5

2,7

)25. Let (x1, y1) = (5, 0) and (x2, y2) = (9, 4). Then from the midpoint formula(

x1 + x22

,y1 + y2

2

)we get (

5 + 9

2,

0 + 4

2

)= (7, 2).

26.

(x1 + x2

2,y1 + y2

2

)=

(−4 + (−1)

2,

3 + (−7)

2

)=

(−5

2, − 2

)

27. The center is the midpoint:

(−2 + 6

2,−1 + 11

2

)= (2, 5).

28. Midpoint of given line segment: (2, 6). Midpoint of line segment from (2, 6) to (−2, 4): (0, 5).

1.2 The Slope

1. Let (x2, y2) = (1, 7) and (x1, y1) = (2, 6). Then, by formula (1.4),

m =y2 − y1x2 − x1

we get

m =7− 6

1− 2=

1

−1= −1.

2. Let (x2 , y2) = (−3 , −10) and (x1 , y1) = (−5, 2). Then by formula (1.4)

m =y2 − y1x2 − x1

=−10− 2

−3− (−5)=−12

2= −6.

3. Let (x1, y1) = (0, 2) and (x2, y2) = (−4,−4). Then

m =−4− 2

−4− 0=−6

−4=

3

2.

4. Let (x2 , y2) = (6 , −3) and (x1 , y1) = (4, 0). Then m =y2 − y1x2 − x1

=−3− 0

6− 4=−3

2= −3

2.

5. Let (x2, y2) = (7, 8) and (x1, y1) = (−3,−4). Then

m =8− (−4)

7− (−3)=

8 + 4

7 + 3=

12

10=

6

5.

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6. m =−4− 0

8− 0= −1

2

7. m =43− (−1)

−1− 1=

44

−2= −22

8. m =31− 1

0− (−1)= 30

9. m =−5− 4

3− 3=−9

0(undefined)

10. m =6− 6

5− (−3)=

0

8= 0

11. m =−3− (−3)

9− 5=

0

4= 0

12. m− 3− (−2)

4− 4=

5

0(undefined)

13. m =3− 2

12− (−2)=

1

14

14. (a) tan 0◦ = 0; (b) tan 30◦ =√33 ; (c) tan 150◦ = −

√33 ; (d) tan 90◦is undefined;

(e) tan 45◦ = 1; (f) tan 135◦ = −1

15. See answer section of book.

16. Slope of AB =0− 2

−2− (−1)=−2

−1= 2; of BC =

5

3; of AC = −3

4.

17. Slope of given line is1− (−5)

−7− 6=

6

−13= − 6

13. Slope of perpendicular is given by the negative

reciprocal and is therefore − 1

(−6/13)=

13

6.

18. Slope of line through (−4, 6) and (−1,−3):−3− 6

−1− (−4)= −3.

Slope of line through (−4, 6) and (1,−9):−9− 6

1− (−4)= −3.

Since the lines have the same slope and pass through (−4, 6), they must coincide.

19. Slope of line through (−4, 6) and (6, 10):6− 10

−4− 6=−4

−10=

2

5

Slope of line through (6, 10) and (10, 0):10− 0

6− 10=

10

−4= −5

2Since the slopes are negative reciprocals, the lines are perpendicular.

20. Slope of line through (−4 , 2) and (−1 , 8): 2;

Slope of line through (9 , 4) and (6 , −2): 2;

Slope of line through (−1 , 8) and (9 , 4): −2/5;

Slope of line through (−4 , 2) and (6 , −2): −2/5;

21. Slope of line through (0,−3) and (−2, 3):−3− 3

0− (−2)=−6

2= −3

Slope of line through (7, 6) and (9, 0):6− 0

7− 9=

6

−2= −3

Slope of line through (−2, 3) and (7, 6):3− 6

−2− 7=−3

−9=

1

3

Slope of line through (0,−3) and (9, 0):−3− 0

0− 9=

1

3

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1.3. THE STRAIGHT LINE 5

Since −3 and 13 are negative reciprocals, adjacent sides are perpendicular and opposite sides

are parallel.

22. Midpoint of line segment:

(−2 + 8

2,−4 + 8

2

)= (3, 2). Point: (6 , −4), m =

−4− 2

6− 3= −2

23. Midpoint:

(−3 + 9

2,−2 + 0

2

)= (3,−1)

Slope of line through (5, 6) and (3,−1):6− (−1)

5− 3=

7

2

24. Let (x , y) be the other end of the diameter. Since the center is the midpoint, we getx+ 4

2= 1

andy − 3

2= 2; solving, x = −2, y = 7.

25. tan θ =rise

run=

10.0 ft

160 ft= 0.0625

θ = 3.6◦

26. tan θ =rise

run=

2.5 m

19.0 m≈ 0.1316; θ = 7.5◦

27. Slope of line through (−1,−1) and (3,−5):−1− (−5)

−1− 3=

4

−4= −1

Slope of line through (x, 2) and (4,−6):2 + 6

x− 4=

8

x− 4Since the two slopes must be equal, we have:

8

x− 4= −1

8 = −x+ 4 multiplying both sides by x− 4

x = −4

28. Slope of line segment (2 , −1) to (−3 , 2): −3

5; Slope of other line segment:

−2 + 7

x− 4; so

5

x− 4=

5

3(negative reciprocals); solving, x = 7.

1.3 The Straight Line

1. Since (x1, y1) = (−7, 2) and m = 1/2, we get

y − 2 = 12 (x+ 7) y − y1 = m(x− x1)

2y − 4 = x+ 7 clearing fractions

0 = x− 2y + 7 + 4

x− 2y + 11 = 0

2. Since (x1 , y1) = (0 , 3) and m = −4, we get

y − 3 = −4(x− 0) y − y1 = m(x− x1)

4x+ y − 3 = 0

3. y − y1 = m(x− x1)

y + 4 = 3(x− 3) (x1, y1) = (3,−4); m = 3

y + 4 = 3x− 9

3x− y − 13 = 0

4. By (1.8), y = 2.Not For Sale

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5. y − y1 = m(x− x1)

y − 0 = − 13 (x− 0) (x1, y1) = (0, 0); m = −1/3

3y = −xx+ 3y = 0

6. By (1.9), x = −3.

7. The line y = 1 = 0x+ 1 has slope 0.

y − y1 = m(x− x1)

y − 0 = 0(x+ 4) (x1, y1) = (−4, 0); m = 0

y = 0 x-axis

8. First determine the slope: m = y2−y1‘

x2−x1= 2−6−3−7 = 2

5

Let (x1 , y1) = (−3 , 2), then

y − 2 = 25 (x+ 3)

5y − 10 = 2x+ 6 multiplying by 5

2x− 5y + 16 = 0

9. First determine the slope using m =y2 − y1x2 − x1

to get m =4− (−6)

−3− 3=

10

−6= −5

3.

Then let (x1, y1) = (−3, 4) to get

y − 4 = − 53 (x+ 3) y − y1 = m(x− x1)

3y − 12 = −5x− 15 multiplying by 3

5x+ 3y + 3 = 0

10. m = 6−3−4−2 = − 1

2

y − 3 = − 12 (x− 2) (x1, y1) = (2, 3)

2y − 6 = −x+ 2 multiplying by 2

x+ 2y − 8 = 0

11. m =−4− 0

9− 5= −1

y − 0 = −1(x− 5) choosing (x1, y1) = (5, 0)

x+ y − 5 = 0

12. m = 5−7−3−1 = 1

2

y − 5 = 12 (x+ 3) (x1, y1) = (−3, 5)

2y − 10 = x+ 3 multiplying by 2

x− 2y + 13 = 0

13. 6x+ 2y = 5

2y = −6x+ 5

y = −3x+ 52 y = mx + b

m = −3, y-intercept = 52 ; see graph in answer section of book.

14. Solving for y, we get y = x− 1. Slope:1, y-intercept :−1

15. Since 2x = 3y, y = 23x. From the form y = mx+ b, m = 2

3 and b = 0. The line passes through

the origin and has slope 23 . See graph in answer section of book.

16. From y = −4x+ 12, we get m = −4 and b = 12.

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17. 2y − 7 = 0

y = 0x+ 72 y = mx + b

m = 0, y-intercept = 72 ; see graph in answer section of book.

18. Solving for y: y = 14x−

32 slope: 1

4 ; y-intercept :− 32

19. 2x− 3y = 1 4x− 6y + 3 = 0

−3y = −2x+ 1 −6y = −4x− 3

y = 23x−

13 y = 4

6x+ 36

y = 23x+ 1

2

From the form y = mx+ b, m = 23 in both cases, so that the lines are parallel.

20. 2x+ 4y + 3 = 0 y − 2x = 2

y = − 12x−

34 y = 2x+ 2

slope = − 12 slope = 2

Answer: The lines are perpendicular.

21. 3x− 4y = 1 3y − 4x = 3

−4y = −3x+ 1 3y = 4x+ 3

y = 34x−

14 y = 4

3x+ 1

The lines are neither parallel nor perpendicular.

22. 7x− 10y = 6 y − 4 = 0

y = 710x−

35 y = 4

slope = 710 slope = 0

Answer: neither

23. x+ 3y = 5 y − 3x− 2 = 0

3y = −x+ 5 y = 3x+ 2

y = − 13x+ 5

3

The slopes are − 13 and 3, respectively. Since the slopes are negative reciprocals, the lines are

perpendicular.

24. 2x+ 5y = 2 6x+ 15y = 1

y = − 25x+ 2

5 y = − 25x+ 1

15

Slope is −2

5in each case; the lines are parallel.

25. 3x− 5y = 6 9x− 15y = 4

−5y = −3x+ 6 −15y = −9x+ 4

y = 35x−

65 y = −9

−15x+ 4−15

y = 35x−

415

From the form y = mx+ b, the slope m is 35 in both cases; so the lines are parallel.

26. 4y − 3x+ 6 = 0 6x− 8y + 1 = 0

4y = 3x− 6 −8y = −6x− 1

y = 34x−

32 y = −6

−8x+ −1−8

y = 34x+ 1

8

slope = 34 slope = 3

4

Answer: the lines are parallel.Not For Sale

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27. 2y − 3x = 6 6y + 4x = 5

2y = 3x+ 6 6y = −4x+ 5

y = 32x+ 3 y = −4

6 x+ 56

y = − 23x−

56

The respective slopes are 32 and − 2

3 . Since the slopes are negative reciprocals, the lines are

perpendicular.

28. x− 4y − 2 = 0 2y + 8x− 7 = 0

−4y = −x+ 2 2y = −8x+ 7

y =1

4x− 1

2y = −4x+

7

2The respective slopes are 1

4 and −4. Since the slopes are negative reciprocals, the lines are

perpendicular.

29. 3y − 2x− 12 = 0 2x+ 3y − 4 = 0

3y = 2x+ 12 3y = −2x+ 4

y = 23x+ 4 y = − 2

3x+ 43

The lines are neither parallel nor pependicular.

30. 3x+ 4y − 4 = 0 6x− 8y − 3 = 0

4y = −3x+ 4 −8y = −6x+ 3

y = −3

4x+ 1 y =

3

4x− 3

8The lines are neither parallel nor perpendicular.

31. 3x+ 4y = 5 (given line)

y = − 34x+ 5

4 slope = − 34

y − y1 = m(x− x1) point-slope form

y − 1 = − 34 (x+ 2) point: (−2, 1)

4y − 4 = −3x− 6

3x+ 4y + 2 = 0

32. The given line can be written y = −3

4x+

5

4. So the slope is −3

4. Slope of perpendicular:

4

3.

y − 1 =4

3(x+ 2) y − y1 = m(x− x1)

3y − 3 = 4x+ 8

4x− 3y + 11 = 0

33. To find the coordinates of the point of intersection, solve the equations simultaneously:

2x − 4y = 1

3x + 4y = 4

5x = 5 adding

x = 1

From the second equation, 3(1) + 4y = 4, and y = 14 . So the point of intersection is (1, 14 ).

From the equation 5x+ 7y + 3 = 0, we get

7y = −5x− 3

y = − 57x−

37 slope= −5/7

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Thus (x1, y1) = (1, 14 ) and m = − 57 . The desired line is y− 1

4 = − 57 (x− 1). To clear fractions,

we multiply both sides by 28:

28y − 7 = −20(x− 1)

28y − 7 = −20x+ 20

20x+ 28y − 27 = 0

34. The first two lines are perpendicular: y =1

3x+ 1 and y = −3x− 25

4

35. See graph in answer section of book.

36. F = 3x, slope = 3, passing through the origin

37. From F = kx, we get 3 = k · 12 . Thus k = 6 and F = 6x.

38. y-intercept : initial value; t-intercept : the year the value becomes zero

39. F = mC + b

212 = m(100) + b F = 212, C = 100

32 = m(0) + b F = 32, C = 0

b = 32 second equation

212 = m(100) + 32 substituting into first equation

m =180

100=

9

5Solution: F = 9

5C + 32

40. C = 1800 + 500t

41. R = aT + b

51 = a · 100 + b R = 51, T = 100

54 = a · 400 + b R = 54, T = 400

−3 = −300a subtracting

a =−3

−300= 0.01

From the first equation, 51 = a · 100 + b, we get

51 = (0.01)(100) + b (a = 0.01)

b = 50

So the formula R = aT + b becomes R = 0.01T + 50.

42. P = kx; let P = 187.2 lb and x = 3.0 ft. Then

187.2 = k(3.0) and k =187.2

3.0= 62.4

So the relationship is P = 62.4x.

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1.4 Curve Sketching

In the answers below, the intercepts are given first, followed by symmetry, asymptotes, and extent.

2. y = 2, x = 12 ; none; none; all x

3. Intercepts. If x = 0, then y = −9. If y = 0, then

0 = x2 − 9

x2 = 9 solving for x

x = ±3 x = 3 and x = −3.

Symmetry. If x is replaced by −x, we get y = (−x)2 − 9, which reduces to the given equation

y = x2 − 9. The graph is therefore symmetric with respect to the y-axis. There is no other

type of symmetry.

Asymptotes. Since the equation is not in the form of a quotient with a variable in the denom-

inator, there are no asymptotes.

Extent. y is defined for all x.

Graph.

............

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

x

y

0

4. y = 1; y-axis; none; all x

5. Intercepts. If x = 0, then y = 1. If y = 0, then

0 = 1− x2

x2 = 1 solving for x

x = ±1 x = 1 and x = −1.

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1.4. CURVE SKETCHING 11

Symmetry. If x is replaced by −x, we get y = 1− (−x)2, which reduces to the given equation

y = 1 − x2. The graph is therefore symmetric with respect to the y-axis. There is no other

type of symmetry.




Graph.

............

............

.................................................................................................................................................................................................................................................................................................................................................................................................................................

...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

x

y

0

6. y = 5, x = ±√

5; y-axis; none; all x

7. Intercepts. If x = 0, then y = 0, and if y = 0, then x = 0. So the only intercept is the origin.

Symmetry. If we replace x by −x, we get y2 = −x, which does not reduce to the given equa-

tion. So there is no symmetry with respect to the y-axis.

If y is replaced by −y, we get (−y)2 = x, which reduces to y2 = x, the given equation. It

follows that the graph is symmetric with respect to the x-axis.

To check for symmetry with respect to the origin, we replace x by−x and y by−y: (−y)2 = −x.

The resulting equation, y2 = −x, does not reduce to the given equation. So there is no sym-

metry with respect to the origin.

Asymptotes. Since the equation is not in the form of a fraction with a variable in the denom-


Extent. Solving the equation for y in terms of x, we get

y = ±√x.

Note that to avoid imaginary values, x cannot be negative. It follows that the extent is x ≥ 0.Not For Sale

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Graph.

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................................................................................................................

................................................................................................................................

...................................................................

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

x

y

0

8. origin; x-axis; none; x ≥ 0

9. Intercepts. If x = 0, then y = ±1. If y = 0, then x = −1.

Symmetry. If we replace x by −x we get y2 = −x + 1, which does not reduce to the given

equation. So there is no symmetry with respect to the y-axis.

If y is replaced by −y, we get (−y)2 = x+ 1, which reduces to y2 = x+ 1, the given equation.

It follows that the graph is symmetric with respect to the x-axis.

The graph is not symmetric with respect to the origin.

Asymptotes. Since the equation is not in the form of a fraction with a variable in the denom-


Extent. Solving the equation for y, we get y = ±√x+ 1. To avoid imaginary values, we must

have x+ 1 ≥ 0 or x ≥ −1. Therefore the extent is x ≥ −1.

Graph.

............

............

...................................................................................................................................

....................................................................................................................................................

.......................................................................................................................................

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

x

y

0

10. y = ±√

2; x = 2; x-axis; none; x ≤ 2

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11. Intercepts. If x = 0, then y = (0− 3)(0 + 5) = −15. If y = 0, then

0 = (x− 3)(x+ 5)

x− 3 = 0 x+ 5 = 0

x = 3 x = −5.

Symmetry. If x is replaced by −x, we get y = (−x − 3)(−x + 5), which does not reduce to

the given equation. So there is no symmetry with respect to the y-axis. Similarly, there is no

other type of symmetry.




Graph.

............

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

•

x

y

0

(0,-15)

12. y = −24; x = −6, 4; none; none; all x


0 = x(x+ 3)(x− 2)

x = 0,−3, 2.

Symmetry. If x is replaced by −x, we get y = −x(−x + 3)(−x − 2), which does not reduce

to the given equation. So the graph is not symmetric with respect to the y-axis. There is no




Extent. y is defined for all x.Not For Sale

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Graph.

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......

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.......

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......

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.......

........

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.......

.......

........

.......

........

.......

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.......................................................................................................................................................................................................................................................................................................................................

......................................................................................................................................................................................................................................

x

y

0

14. y = 0; x = 0,1, 4; none; none; all x

15. Intercepts. If x = 0, y = 0; if y = 0, then

x(x− 1)(x− 2)2 = 0

x = 0, 1, 2.

Symmetry. If x is replaced by −x, we get y = −x(−x − 1)(−x − 2)2, which does not reduce

to the given equation. So there is no symmetry with respect to the y-axis. Similarly, there is

no other type of symmetry.

Asymptotes. None (the equation does not have the form of a fraction).


Graph.

............

...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..................................................................................................................................................................................................

x

y

0

16. y = 0; x = −2, 0, 3; none; none; all x

Not For Sale

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0 = x(x− 1)2(x− 2)

x = 0, 1, 2.

Symmetry. If x is replaced with −x, we get y = −x(−x− 1)2(−x− 2), which does not reduce

to the given equation. Therefore there is no symmetry with respect to the y-axis. There is no


Asymptotes. None (the equation does not have the form of a fraction).


Graph.

............

...............................................................................................................................................................................................................................................................................................................................................................................................................

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

x

y

0

18. y = 0;x = −2, 0, 3; none; none; all x

............

......................................................................................................................................................................................................................................................................................................................

........................................................................................................................................................................................................................................................................................................................................................................................................................................

....................................................................................................................................

x

y

19. Intercepts. If x = 0, y = 1; if y = 0, we have

0 =2

x+ 2.

This equation has no solution.

Symmetry. Replacing x by −x, we get

y =2

−x+ 2

which does not reduce to the given equation. So there is no symmetry with respect to the

y-axis. Similarly, there is no other type of symmetry.

Asymptotes. Setting the denominator equal to 0, we get

x+ 2 = 0 or x = −2.

It follows that x = −2 is a vertical asymptote. Also, as x gets large, y approaches 0. So the

x-axis is a horizontal asymptote.

Extent. To avoid division by 0, x cannot be equal to −2. So the extent is all x except x = −2.Not For Sale

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Graph.

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.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

x

y

0

20. y = −1; none; x = 3; y = 0; x 6= 3


2

(x− 1)2= 0.

This equation has no solution.

Symmetry. Replacing x by −x, we get

y =2

(−x− 1)2

which does not reduce to the given equation. There are no other types of symmetry.

Asymptotes. Setting the denominator equal to 0 gives

(x− 1)2 = 0 or x = 1.

It follows that x = 1 is a vertical asymptote. Also, as x gets large, y approaches 0. So the

x-axis is a horizontal asymptote.

Extent. To avoid division by 0, x cannot be equal to 1. So the extent is the set of all x except

x = 1.

Graph.

............

............

.......................................................................................................................................................................................................................................................

.................................................................................................................................................................................................................. .........................................................................................................................................................................................................................................................................................................................................................................................................................................................................

x

y

0

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22. y = − 14 ; none; none; x = −2; y = 0;x 6= −2

..........

..........

x

y

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... ...............................................................................................................................................................................................................................................................................

.....................................................................................................................................................................................................................................................................................................................−6 −4 −2 20


0 =x2

x− 1.

The only solution is x = 0.

Symmetry. Replacing x by −x yields

y =(−x)2

−x− 1=

x2

−x− 1

which is not the same as the given equation. So the graph is not symmetric with respect to

the y-axis. Replacing y by −y, we have

−y =x2

x− 1

which does not reduce to the given equation. So the graph is not symmetric with respect to

the x-axis.

Similarly, there is no symmetry with respect to the origin.

Asymptotes. Setting the denominator equal to 0, we get x − 1 = 0, or x = 1. So x = 1 is a

vertical asymptote. There are no horizontal asymptotes.

(Observation: for very large x the 1 in the denominator becomes insignificant. So the graph

gets ever closer to y =x2

x= x; the line y = x is a slant asymptote.)

Extent. To avoid division by 0, x cannot be equal to 1. So the extent is all x except x = 1.

Graph.

............

............

.......................................................................................................................................................................................................................................................................................................

......................................................................................................................

.............................................................................................................................................................................................................................................................................................................................................................................................................................

x

y

0

24. y = 0; x = 0; none; x = −2; y = 1;x 6= −2

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25. Intercepts. If x = 0, then y = −1/2. If y = 0, then

0 =x+ 1

(x− 1)(x+ 2).

The only solution is x = −1.

Symmetry. Replacing x by −x yields

y =−x+ 1

(−x− 1)(−x+ 2)

which is not the same as the given equation. There are no types of symmetry.

Asymptotes. Setting the denominator equal to 0, we get (x − 1)(x + 2) = 0. So x = 1

and x = −2 are the vertical asymptotes. As x gets large, y approaches 0, so the x-axis is a

horizontal asymptote.

Extent. To avoid division by 0, the extent is all x except x = 1 and x = −2.

Graph.

............

............

............................................................................................................................................................................................................................................................................................................................................................................................

...............................................................................................................................................................................................................................................................................................................................................................................................................................

...................................................................................................................................................................................................................................................................................................................................................................................................................................................................

x

y

0

26. y = 0, x = 1, 0; none; x = −1, 2, y = 1;x 6= −1, 2


x2 − 4

x2 − 1= 0

x2 − 4 = 0 multiplying by x2 − 1

x = ±2. solution

Symmetry. Replacing x by −x reduces to the given equation. So there is symmetry with

respect to the y-axis. There is no other type of symmetry.

Asymptotes. Vertical: setting the denominator equal to 0, we have

x2 − 1 = 0 or x = ±1.

Horizontal: dividing numerator and denominator by x2, the equation becomes

y =1− 4

x2

1− 1x2

.

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As x gets large, y approaches 1. So y = 1 is a horizontal asymptote.

Extent. All x except x = ±1 (to avoid division by 0).

Graph.

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....................................................................................................................................

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.............................................................................................................................................................................................................................................................................................................

x

y

0

28. y = 14 , x = ±1; y-axis; x = ±2, y = 1;x 6= ±2

..........

..........

x

y

.....................................................................................................................................................................................................................................................................

.......

........

......

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.......

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..............................................................................................................................................................................................................................................................................................................................................................................................................

.....................................................................................................................................................................................................................................................................

−3−2−1 1 2 3

−3

−2

−1

1

2

3

0

29. Intercepts. If x = 0, then y2 = −4−1 = 4, or y = ±2. If y = 0, then

0 =x2 − 4

x2 − 1

which is possible only if x2 − 4 = 0, or x = ±2.

Symmetry. The even powers on x and y tell us that if x is replaced by −x and y is replaced by

−y, the resulting equation will reduce to the given equation. The graph is therefore symmetric

with respect to both axes and the origin.

Asymptotes. Vertical: setting the denominator equal to 0, we get

x2 − 1 = 0 or x = ±1.

Horizontal: dividing numerator and denominator by x2, we get

y2 =1− 4

x2

1− 1x2

.

The right side approaches 1 as x gets large. Thus y2 approaches 1, so that y = ±1 are the

horizontal asymptotes.Not For Sale

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Extent. From

y = ±√x2 − 4

x2 − 1

we conclude thatx2 − 4

x2 − 1=

(x− 2)(x+ 2)

(x− 1)(x+ 1)≥ 0.

Since the signs change only at x = 2,−2, 1, and −1, we need to use arbitrary “test val-

ues”between these points. The results are summarized in the following chart.

test

values x− 2 x− 1 x+ 1 x+ 2(x− 2)(x+ 2)

(x− 1)(x+ 1)

x > 2 3 + + + + +

1 < x < 2 3/2 − + + + −−1 < x < 1 0 − − + + +

−2 < x < −1 −3/2 − − − + −x < −2 −3 − − − − +

Note that the fraction is positive only when x > 2, −1 < x < 1 and x < −2. Since y = 0

when x = ±2, the extent is x ≥ 2, −1 < x < 1, x ≤ −2.

Graph.

............

............

............................................................................................................................................................................................................................................................................................................................................

......................................................................................................................................................................................................

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

....................................................................................................................................................................................................................................................................................................................

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...............................................................................................................................................................................................................................................

x

y

0

30. y = ± 12 , x = ±1; both axes; x = ±2, y = ±1;x < −2,−1 ≤ x ≤ 1, x > 2

31. Intercepts. If x = 0, y2 = (−3)(5) = −15, or y = ±√

15 j, which is a pure imaginary number.

If y = 0,

(x− 3)(x+ 5) = 0

x = 3,−5.

Symmetry. Replacing y by −y, we get (−y)2 = (x − 3)(x + 5), which reduces to the given

equation. Hence the graph is symmetric with respect to the x-axis.

Asymptotes. None (no fractions).

Extent. From y = ±√

(x− 3)(x+ 5), we conclude that (x− 3)(x+ 5) ≥ 0. If x ≥ 3,

(x − 3)(x + 5) ≥ 0. If x ≤ −5, (x − 3)(x + 5) ≥ 0, since both factors are negative (or zero).

If −5 < x < 3, (x − 3)(x + 5) < 0. [For example, if x = 0, we get (−3)(5) = −15.] These

observations are summarized in the following chart.

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test

values x− 3 x+ 5 (x− 3)(x+ 5)

x > 3 4 + + +

−5 < x < 3 0 − + −x < −5 −6 − − +

Extent: x ≤ −5, x ≥ 3

Graph.

............

............

..................................................................

..................................................................................................................................................................................................................................................................................................................................................................................................................

..........................................................................................................................................

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..........................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

• • x

y

0(-5,0) (3,0)

32. y = 0, x = 0;x-axis; x = −2, y = ±1;x < −2, x ≥ 0

33. Intercepts. If x = 0, y = 0; if y = 0, x = 0.

Symmetry. Replacing y by −y leaves the equation unchanged. So there is symmetry with

respect to the x-axis. There is no other type of symmetry.

Asymptotes. Vertical: setting the denominator equal to 0, we get

(x− 3)(x− 2) = 0 or x = 3, 2.

Horizontal: as x gets large, y approaches 0 (x-axis).

Extent. From

y = ±√

x

(x− 3)(x− 2)

we conclude thatx

(x− 3)(x− 3)≥ 0.

Since signs change only at x = 0, 2 and 3, we need to use “test values”between these points.

The results are summarized in the following chart.

test

values x x− 2 x− 3x

(x− 3)(x− 2)

x < 0 −1 − − − −0 < x < 2 1 + − − +

2 < x < 3 5/2 + + − −x > 3 4 + + + +

So the inequality is satisfied for 0 < x < 2 and x > 3. In addition, y = 0 when x = 0.Not For Sale

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So the extent is 0 ≤ x < 2 and x > 3.

Graph.

............

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.....................................................................................................................................

..................................................................................

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..................................................................................................................................................................................................................................................................................

..............................................................................................................................................................................................................................................................................................

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..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......................................................................................x

y

0

34. x = −1; x-axis; x = −2, 1, y = 0;−2 < x ≤ −1, x > 1

35. C =10−2C1

C1 + 10−2, C1 ≥ 0

The only intercept is the origin. Dividing numerator and denominator by C1, the equation

becomes

C =10−2

1 + 10−2/C1.

As C1 gets large, C approaches 10−2; so C = 10−2 is a horizontal asymptote.

See graph in answer section of book.

37. Intercepts. If t = 0, S = 0; if S = 0, we get

0 = 60t− 5t2

0 = 5t(12− t)or t = 0, 12.

Symmetry. None.

Asymptotes. None.

Extent. t ≥ 0 by assumption.

Graph. See graph in answer section of book.

38.

39. Extent L ≥ 0.

See graph in answer section of book.

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1.5. CURVES WITH GRAPHING UTILITIES 23

40. 2000 units (x = 2)

1.5 Curves with Graphing Utilities

Graphs are from the answer section of the book.

1. If y = 0, then

x2(x− 1)(x− 2) = 0.

Setting each factor equal to 0, we get

x = 0, 1, 2.

..............................................................................................................................................................................................................................................................................................................................................................

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

[−1, 3] by [−2, 2]

2. −2, 0, 1

[−4, 2] by [−4, 2]Not For Sale

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4. −2,−1, 0, 1

[−3, 3] by [−0.5, 0.5]

5. x4 − 2x3 = 0

x3(x− 2) = 0

x = 0, 2..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

................................................................................................................................................................................................................................................................................................................

[−1, 3] by [−2, 2]

6. ±√

6

2

[−2, 2] by [−15, 15]

8. 0,1

2

[−1, 1] by [−0.2, 0.2]

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introduction to analytic geometry · introduction to analytic geometry 1.1 the cartesian coordinate...

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