lesson 14.3 skills...

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Chapter 14 Skills Practice 771

14

LESSON 14.3 Skills Practice

Name meme Date Datete

It’s Time to Focus

Parabolas as Conics

Vocabulary

Write a de!nition for each term in your own words.

1. parabola

The set of all points in a plane that are equidistant from a fixed point and a fixed line.

2. locus of points

A locus of points is a set of points that share a property. The locus of points that defines

a parabola is the set of points that are equidistant from a fixed point and a fixed line.

3. focus of a parabola

The directrix of a parabola is the fixed point that, along with the directrix, determines the locus

of points that define a parabola. The focus is located in the interior of the parabola.

4. directrix of a parabola

The directrix of a parabola is the fixed line that, along with the focus, determines the locus

of points that define a parabola. The directrix is located in the exterior of the parabola.

5. concavity of a parabola

The concavity of a parabola describes the orientation of the curvature of the parabola. It is

concave up or down if there is an x 2 -term in its equation; it is concave right or left if there is

a y 2 -term in its equation.

6. conic form of a parabola

The conic form of a parabola is an equation written using the vertex (h, k) and a value p, which

is the distance from the vertex to the focus. It is written in the form (x 2 h) 2 5 4p(y 2 k) when the

parabola is oriented vertically, and in the form (y 2 k) 2 5 4p(x 2 h) when the parabola is oriented

horizontally. If the vertex of the parabola is the origin (0, 0), the conic forms become x 2 5 4py

and y 2 5 4px.

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772 Chapter 14 Skills Practice

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LESSON 14.3 Skills Practice page 2

Problem Set

For the graph of each parabola centered at the origin, write its equation in conic form. Show your work for

deriving the equation. Then, identify the symmetry and concavity of the parabola.

1.

28 26 24 22

24

22

20 4

(0, 0)

(0, 25)

6 8

28

26

8

6

4

2

x

y

y 5 5

Equation: x 2 5 220y

x 2 5 4py

x 2 5 4(25)y

x 2 5 220y

Axis of Symmetry: x 5 0

Concavity: concave down

2.

28 26 24 22

24

22

0 42

(0, 0) (2, 0)

6 8

28

26

8

6

4

2

x

y

x 5 22

Equation: y 2 5 8x

y 2 5 4px

y 2 5 4(2)x

y 2 5 8x

Axis of Symmetry: y 5 0

Concavity: concave right

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3.

28 26 24 22

24

22

20 4

(0, 0)(23, 0)

6 8

28

26

8

6

4

2

x

y

x 5 3

Equation: y 2 5 212x

y 2 5 4px

y 2 5 4(23)x

y 2 5 212x


Concavity: concave left

4.

24 23 22 21

22

21

10 2

(0, 0)

3 4

24

23

4

3

2

1

x

y

y 5 21

2

10, 21

2

Equation: x 2 5 2y

x 2 5 4py

x 2 5 4 ( 1 __ 2 ) y

x 2 5 2y


Concavity: concave up

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5.

216 212 28 24

28

24

40 8

(0, 0)(216, 0)

12 16

216

212

16

12

8

4

x

y

x 5 16

Equation: y 2 5 264x

y 2 5 4px

y 2 5 4(216)x

y 2 5 264x


Concavity: concave left

6.

28 26 24 22

24

22

20 4

(0, 0)

(0, 21)

6 8

28

26

8

6

4

2

x

y

y 5 1

Equation: x 2 5 24y

x 2 5 4py

x 2 5 4(21)y

x 2 5 24y


Concavity: concave down

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Write the conic form equation of each parabola. Identify the vertex and axis of symmetry. Show your

work for deriving the equation.

7.

28 26 24 22

24

22

20 4

(5, 0)

6 8

28

26

8

6

4

2

x

y

x 5 1

Vertex: (3, 0)


p-value: p 5 2

Equation: y 2 5 8(x 2 3)

(y 2 k) 2 5 4p(x 2 h)

(y 2 0) 2 5 4(2)(x 2 3)

y 2 5 8(x 2 3)

8.

28 26 24 22

24

22

20

4 6 8

28

26

8

6

4

2

x

y Vertex: (0, 21)


p-value: p 5 1

Equation: x 2 5 4(y 1 1)

(x 2 h) 2 5 4p(y 2 k)

(x 2 0) 2 5 4(1)(y 2 (21))

x 2 5 4(y 1 1)

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9.

28 26 24 22

24

22

20

4 6 8

8

6

4

2

12

10

x

y Vertex: (1, 3)


p-value: p 5 24

Equation: (y 2 3) 2 5 216(x 2 1)

(y 2 k) 2 5 4p(x 2 h)

(y 2 3) 2 5 4(24)(x 2 1)

(y 2 3) 2 5 216(x 2 1)

10.

242526 23 22 21

22

21

10

2

24

23

25

26

2

1

x

y Vertex: (22, 21)


p-value: p 5 21

Equation: (x 1 2) 2 5 28(y 1 1)

(x 2 h) 2 5 4p(y 2 k)

(x 2 (22)) 2 5 4(21)(y 2 (21))

(x 1 2) 2 5 24(y 1 1)

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11.

24 22

24

22

20

4 6 8 10 12

28

26

212

210

4

2

x

y Vertex: (3, 22)


p-value: p 5 3

Equation: (y 1 2) 2 5 12(x 2 3)

(y 2 k) 2 5 4p(x 2 h)

(y 2 (22)) 2 5 4(3)(x 2 3)

(y 1 2) 2 5 12(x 2 3)

12.

28210212214 26 24 22

22

20

8

6

4

2

12

10

14

x

y Vertex: (21, 6)


p-value: p 5 22

Equation: (y 2 6) 2 5 28(x 1 1)

(y 2 k) 2 5 4p(x 2 h)

(y 2 6) 2 5 4(22)(x 2 (21))

(y 2 6) 2 5 28(x 1 1)

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Calculate the equation for the directrix of each parabola using the given information.

Explain your reasoning.

13. A parabola has a vertex at (21, 2) and a focus at (21, 6).

The equation for the directrix is y 5 22.

The vertex is (21, 2) and the value of p is 4. The equation for the directrix of a parabola oriented

vertically is y 5 k 2 p. So, the equation for the directrix of this parabola is y 5 2 2 4, or y 5 22.


The equation for the directrix is x 5 1.


horizontally is x 5 h 2 p. So, the equation for the directrix of this parabola is x 5 3 2 2, or x 5 1.



The vertex is (22, 23) and the value of p is 23. The equation for the directrix of a parabola

oriented vertically is y 5 k 2 p. So, the equation for the directrix of this parabola is

y 5 23 2 (23), or y 5 0.


The equation for the directrix is x 5 9.


horizontally is x 5 h 2 p. So, the equation for the directrix of this parabola is x 5 4 2 (25), or x 5 9.

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vertically is y 5 k 2 p. So, the equation for the directrix of this parabola is y 5 7 2 11, or y 5 24.




vertically is y 5 k 2 p. So, the equation for the directrix of this parabola is y 5 8 2 (29), or y 5 17.

Calculate the coordinates of the focus of each parabola using the given information. Explain your reasoning.

19. A parabola has a vertex at (21, 21) and a directrix at y 5 3.

The coordinates of the focus are (21, 25).

The vertex is (21, 21) and the value of p is 24. The coordinates of the focus for a parabola

oriented vertically are (h, k 1 p). So, the coordinates of the focus of this parabola are

(21, 21 1 (24)), or (21, 25).

20. A parabola has a vertex at (2, 0) and a directrix at x 5 21.


The vertex is (2, 0) and the value of p is 3. The coordinates of the focus for a parabola oriented

horizontally are (h 1 p, k). So, the coordinates of the focus of this parabola are (2 1 3, 0), or (5, 0).

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vertically are (h, k 1 p). So, the coordinates of the focus of this parabola are (23, 7 1 1), or (23, 8).

22. A parabola has a vertex at (25, 25) and a directrix at x 5 23.


The vertex is (25, 25) and the value of p is 22. The coordinates of the focus for a parabola

oriented horizontally are (h 1 p, k). So, the coordinates of the focus of this parabola are

(25 1 (22), 25), or (27, 25).




vertically are (h, k 1 p). So, the coordinates of the focus of this parabola are (3, 1 1 (24)), or (3, 23).




vertically are (h, k 1 p). So, the coordinates of the focus of this parabola are (210, 0 1 1), or (210, 1).

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LESSON 14.4 Skills Practice


But I Regress . . .

Regression

Problem Set

Graph the points and analyze the data to determine the best regression model for the data. Explain your

reasoning. Then, use a graphing calculator to determine the regression equation.

1. The average salary in thousands of dollars, s(x), of a company’s presidents for various years since

1960 is displayed in the table.

Time Since 1960

(years)0 5 10 15 16 17 18

Average Salary

(thousands of dollars)170 325 750 1900 2000 2200 2600

0 2 4

1600

2000

1200

6

Time Since 1960 (years)

8 10 12 14 16 18

400

0

800

2800

2400

3600

3200

Sala

ry (th

ousand

s o

f d

olla

rs)

x

s(x)

The data increases over the given domain, but not at a constant rate, so I can use an

exponential regression to best model the data. The exponential regression equation is

s(x) 5 161.4299 ? 1.1692 x .

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2. The number of units produced, x, of a product and the pro!t associated with making that many units,

P(x), in thousands of dollars, are shown in the table.

Number of Units

Produced1280 1350 1500 1725 1960 2400 2650 2800

Profit

(thousands of

dollars)

1180 1170 1280 1560 1720 1960 1940 1800

0 400

800

1000

600

1200

Number of Units Produced

2000 2800 3600

200

0

400

1400

1200

1800

1600

Pro

!t

(tho

usand

s o

f d

olla

rs)

x

P(x)

The data increases and then decreases over the given domain, so you can use a

quadratic regression to best model the data. The quadratic regression equation is

P(x) 5 20.0005132x 2 1 2.5971x 2 1382.4928.

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3. The table shows the number of oriental rugs a company sells, n(x), and the price of each rug, x.

Price

(dollars)160 180 200 220 240 260 280

Number of Rugs 125 100 85 75 60 40 20

0 40

80

100

60

120

Price of a Rug (dollars)

200 280 36080 160 240 320

20

0

40

140

120

180

160

Num

ber

of

Rug

s S

old

x

n(x)

The data decreases at a constant rate over the given domain, so you can use a linear regression

to best model the data. The linear regression equation is n(x) 5 20.8214x 1 252.8571.

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4. A company conducts a study to see if the number of weekly TV ads that they buy, x, is related to the

sales of the product, s(x), in hundreds of dollars. The results are recorded in the table.

Number of Weekly

TV Ads Purchased 5 6 7 8 9 10

Sales

(hundreds of dollars)1000 1200 1400 1520 1000 950

0 1

800

1000

600

3

Number of Weekly TV Ads Purchased

5 7 92 4 6 8

200

0

400

1400

1200

1800

1600

Sale

s (hund

red

s o

f d

olla

rs)

x

s(x)



s(x) 5 273.75 x 2 1 1085.3929x 2 2598.5714.

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5. A college recorded the number of hours students studied for a math entrance exam, x, and their

scores on the exam, s(x). The results are shown in the table.

Study Time

(hours)1 3 4 9 10 12 13 14 16

Exam Score 350 400 490 580 650 600 700 730 770

0 2

400

500

300

6

Study Time (hours)

10 14 184 8 12 16

100

0

200

700

600

900

800

Exam

Sco

re

x

s(x)

The data increases at a constant rate over the given domain, so you can use a linear regression

to best model the data. The linear regression equation is s(x) 5 26.9219x 1 340.2668.

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6. A 75-gallon tank begins to leak and lose water. The table shows the amount of water left in the

tank a(x) after x minutes since the leak started.

Time Since

Leak Started

(minutes)

0 5 10 15 20 25 30

Amount of Water

in the Tank

(gallons)

75 45 27 18 13 8 7

0 5

40

50

30

15

Time Since Leak Started (minutes)

25 35 4510 20 30 40

10

0

20

70

60

90

80

Am

ount

of

Wate

r in

the T

ank (g

allo

ns)

x

a(x)

The data decreases over the given domain, but not at a constant rate, so you can use an


a(x) 5 66.4553 ? 0.9225 x .

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Use the given tables to answer the questions.

7. A car dealership studies the average fuel economy of one of its vehicles, f(x), in miles per gallon,

for various speeds, x, in miles per hour. The table shows the results.

Speed

(miles per hour)20 25 30 35 40 45 50 55 60 65

Fuel Economy

(miles per gallon)25.6 27.8 29.1 29.2 30.1 29.8 30.1 29.9 28.5 26.1

A new customer wants to average 28.5 miles per gallon. Write a regression equation

that best models the data, and use the equation to predict the fuel economy for a speed of

40 miles per hour.



f(x) 5 20.008136x 2 1 0.7107x 1 14.7879.

The predicted fuel economy for a speed of 40 miles per hour is approximately 30.2 miles per hour.

f(2) 5 20.008136(40 ) 2 1 0.7107(40) 1 14.7879 ¯ 30.2

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8. A small business conducts a study to see if the amount of money spent on monthly advertising,

x, in hundreds of dollars, is related to the amount of monthly sales s(x), in hundreds of dollars. The

table shows the results.

Monthly Advertising

Expenditures

(hundreds of dollars)

5.6 5.8 5.9 6.1 6.2 6.3 6.4 6.5 6.6

Monthly Sales

(hundreds of dollars)105 110 114 113 117 116 121 120 125

The company wants to have sales of $12,500. Write a regression equation that best models the data,

and use the equation to predict the amount of money the company will have to spend for advertising

each month in order to generate $12,500 in sales.

The data increases at a constant rate over the given domain, so you can use a linear regression

to best model the data. The linear regression equation is s(x) 5 17.2537x 1 9.4606.

The predicted amount of monthly advertising expenditures is $669.65.

s(x) 5 17.2537x 1 9.4606

125 5 17.2537x 1 9.4606

115.5394 5 17.2537x

6.6965 5 x

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9. The amount of medicine in the body, m(x), in milligrams/milliliter, x minutes after taking it, is shown in

the table.

Time Since

Taking Medicine

(minutes)

0 1 4 7 10 13 16 19

Amount of Medicine

(milligrams/milliliter)10.5 9.8 8.4 7.3 6.3 5.5 4.8 4.6

Write a regression equation that best models the data, and use the equation to predict the

number of minutes that the medicine has been in the system if there is 7.5 milligrams/milliliter left.

The data decreases over the given domain, but not at a constant rate, so I can use an


m(x) 5 10.1534 ? 0.9560 x .

The predicted amount of time that the medicine has been in the system is approximately

6.7 minutes.

m(x) 5 10.1534 ? 0.9560 x

7.5 5 10.1534 ? 0.9560 x

0.7387 ¯ 0.9560 x

log(0.7387) ¯ log (0.9560) x

log(0.7387) ¯ x log(0.9560)

log(0.7387)

___________ log(0.9560)

¯ x

6.7307 ¯ x

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10. A hospital notices that !u cases are on the rise and begins to track the number of !u patients.

The table shows the number of !u cases, f(x), and the time in days since the hospital started tracking

the data, x.

Time (days) 0 2 4 6 8 10

Number of

Flu Cases1500 1610 1650 1590 1550 1490

Write a regression equation that best models the data, and use the equation to predict the

number of !u cases the hospital can expect 12 days after it started tracking.



f(x) 5 25.2232x 2 1 48.0893x 1 1516.07143.

The predicted number of flu cases after 12 days is approximately 1341.

f(x) 5 25.2232x 2 1 48.0893x 1 1516.07143

f(12) 5 25.2232(12) 2 1 48.0893(12) 1 1516.07143

f(12) 5 1341.0022

lesson 14.3 skills...

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