Solution manual for calculus for the life sciences 2nd edition by greenwell

c Yes; because the data points are either on or very close to the horizontal line y = 1, it seems that the data should have a strong linear relationship.. a Using a TI-84 Plus graphing

1.1 Lines and Linear Functions

1. Find the slope of the line through (4, 5) and

The slope is undefined; the line is vertical

4. Find the slope of the line through (1, 5) and

Using the slope-intercept form, y=mx+b,

we see that the slope is 1

10 The x-axis is the horizontal line y =0

Horizontal lines have a slope of 0

11. y = 8This is a horizontal line, which has a slope of

y= x− with the slope, m, being 0

13. Find the slope of a line parallel to

Let m be the slope of any line perpendicular to

the given line Then 4 1 1

4

⋅ = − ⇒ = −

Section 1.1 LINES AND LINEAR FUNCTIONS 29

15 The line goes through (1, 3), with slope

18. The line goes through ( 8, 1),− with undefined

slope Since the slope is undefined, the line is

vertical The equation of the vertical line

passing through ( 8, 1)− is x = −8

19 The line goes through (4, 2) and (1, 3) Find

the slope, then use point-slope form with

either of the two given points

20. The line goes through (8, 1)− and (4, 3) Find

the slope, then use point-slope form with

either of the two given points

43

2 6

23

2762

( 2)2132

3 21[ ( 2)]

− − − which is undefined This

is a vertical line; the value of x is always –8

The equation of this line is x = −8

24 The line goes through ( 1, 3)− and (0, 3)

This is a horizontal line; the value of y is

always 3 The equation of this line is y = 3

25 The line has x-intercept –6 and y-intercept –3

Two points on the line are ( 6, 0)− and (0, 3).− Find the slope; then use slope-intercept form

3

m b

26 The line has x-intercept –2 and y-intercept 4

Two points on the line are (–2, 0) and (0, 4)

Find the slope; then use slope-intercept form

27. The line is vertical, through (–6, 5)

The line has an equation of the form x = k,

where k is the x-coordinate of the point In this

case, k = –6, so the equation is x = –6

28. The line is horizontal, through (8, 7)

The line has an equation of the form y = k,

where k is the y-coordinate of the point In this

case, k = 7, so the equation is y = 7

29. Write an equation of the line through ( 4, 6),−

3

6 62

32

30 Write the equation of the line through (2, 5),−

parallel to 2x−y= −4 Rewrite the equation

The slope of this line is 2

Use 2m = and the point (2, 5)− in the

The slope of this line is 1.− To find the slope

of a perpendicular line, solve

The slope of this line is 2

3 To find the slope

of a perpendicular line, solve

3m= − ⇒ = −m 2Use m = −32 and ( 2, 6)− in the point-slope form

3

23

23

2332

perpendicular line will be 5 If the y-intercept

is 4, then using the slope-intercept form we have

Section 1.1 LINES AND LINEAR FUNCTIONS 31

34 Write the equation of the line with x-intercept

The slope of this line is 2 Since the lines are

perpendicular, the slope of the needed line is

1

2

− The line also has an x-intercept of −23

Thus, it passes through the point ( 2 )

3, 0

−Using the point-slope form, we have

35 Do the points (4, 3), (2, 0), and ( 18, 12)− − lie

on the same line?

Find the slope between (4, 3) and (2, 0)

Since these slopes are not the same, the points

do not lie on the same line

36 (a) Write the given line in slope-intercept

form

223

This line has a slope of −23 The desired

line has a slope of 2

x x k

k k

2 ( 1)4

The slope of the line through ( 7 )

38 Two lines are perpendicular if the product of

The product of the slopes is (1)( 1)− = − so 1,

the diagonals are perpendicular

39 The line goes through (0, 2) and ( 2, 0)−

The correct choice is (a)

40 The line goes through (1, 3) and (2, 0)

The correct choice is (f)

41 The line appears to go through (0, 0) and

43 (a) See the figure in the textbook Segment

MN is drawn perpendicular to segment

PQ Recall that MQ is the length of

(c) Triangles MPQ, PNQ, and MNP are right

triangles by construction In triangles

MPQ and MNP, angle M =angle ,M and

in the right triangles PNQ and MNP,

angle angle N= N

Since all right angles are equal, and since

triangles with two equal angles are

similar, triangle MPQ is similar to

triangle MNP and triangle PNQ is similar

to triangle MNP Therefore, triangles MNQ and PNQ are similar to each other

(d) Since corresponding sides in similar

triangles are proportional,

1

MQ QN

=From (a) and (b), m1=MQ and

m m

=

−Multiplying both sides by m2, we have

a b

Section 1.1 LINES AND LINEAR FUNCTIONS 33 (c) If the equation of a line is written as

1

a+b = , we immediately know the

intercepts of the line, which are a and b

45. y= −x 1

Three ordered pairs that satisfy this equation

are (0, 1), (1, 0),− and (3, 2) Plot these points

and draw a line through them

46 y=4x+5

Three ordered pairs that satisfy this equation

are ( 2, 3),− − ( 1, 1),− and (0, 5) Plot these

points and draw a line through them

47. y= −4x+9

Three ordered pairs that satisfy this equation

are (0, 9), (1, 5), and (2, 1) Plot these points

and draw a line through them

48. y= −6x+12

Three ordered pairs that satisfy this equation

are (0, 12), (1, 6), and (2, 0) Plot these points

and draw a line through them

so the y-intercept is 4.−

Plot the ordered pairs (6, 0) and (0, 4)− and draw a line through these points (A third point may be used as a check.)

so the y-intercept is 9

Plot the ordered pairs ( 3, 0)− and (0, 9) and draw a line through these points (A third point may be used as a check.)

51. 3y−7x= −21

Find the intercepts

If y =0, then 3(0) 7+ x= −21⇒ −7x= −21⇒ =x 3

may be used as a check.)

− for any value of x The graph of this

equation is the horizontal line with y-intercept

2

−

54. x =4

For any value of y, the x-value is 4 Because

all ordered pairs that satisfy this equation have

the same first number, this equation does not

represent a function The graph is the vertical

line with x-intercept 4

55. x +5=0This equation may be rewritten as x = −5 For any value of y, the x-value is 5.− Because all ordered pairs that satisfy this equation have the same first number, this equation does not represent a function The graph is the vertical line with x-intercept 5.−

58. y= −5x

Three ordered pairs that satisfy this equation are (0, 0), ( 1, 5),− and (1, 5).− Use these points to draw the graph

Section 1.1 LINES AND LINEAR FUNCTIONS 35

59. x+4y=0

If y =0, then x =0, so the x-intercept is 0 If

0,

x = then y =0, so the y-intercept is 0

Both intercepts give the same ordered pair, (0,

0) To get a second point, choose some other

value of (or ).x y For example if x =4, then

x = then y =0, so the y-intercept is 0

Both intercepts give the same ordered pair

(0, 0) To get a second point, choose some

other value of (or ).x y For example, if x =5,

16 36+ =52 l =0.7(220 16) 143− ≈beats per minute

72 (a) The line goes through (4, 0.17) and

(7, 0.33)

0.33 0.17

7 40.160.0533

0.16

30.33 0.053 0.3730.053 0.043

10.2

t t t

73. Let x = 0 correspond to 1900 Then the “life

expectancy from birth” line contains the points

given by the equation y=0.297x+46

The “life expectancy from age 65” line

contains the points (0, 76) and (108, 83.8)

given by the equation y=0.072x+76

Set the two expressions for y equal to

determine where the lines intersect At this

point, life expectancy should increase no

=

≈

Determine the y-value when x ≈133.3 Use

the first equation

0.297(133.3) 46 85.6

Thus, the maximum life expectancy for

humans is about 86 years

74 Let x represent the force and y represent the

speed The linear function contains the points

The pony switches from a trot to a gallop at

approximately 4.3 meters per second

75 (a) The number of years of healthy life is

increasing linearly at a rate of about 28 million years every 10 years, so the slope

of the line is m =1028=2.8 Because 35 million years of healthy life was lost in

1900, it follows that y = 35 when t = 0

So, the y-intercept is b = 35 Therefore,

2.8 35

y=mt+ =b t+ is the number of years (in millions) of healthy life lost

globally to tobacco t years after 1900

(b) The number of years lost to diarrhea is

declining linearly at a rate of 22 million years every 10 years, so the slope of the line is m= −1022= −2.2 Because 100 million years of healthy life was lost in

1990, it follws that y = 100 when t = 0

So, the y-intercept is b = 100 Therefore,

2.2 100

y= − t+ is the number of years

lost (in millions) to diarrhea t years after

1990

(c) 2.8 35 2.2 100

5.0 6513

t t

=

=The amount of healthy life lost to tobacco will exceed the amount of healthy life lost

to diarrhea 13 years after 1990, or in

2003

76 (a) The two ordered pairs representing the

given information are (63, 0.109) and (243, 0.307) The slope of the line through these points is

(b) From part (a), b = 0.0397

This means that when the calf is at rest, it

is still expending energy at a rate of 0.0397 kilocalories per kg per minute

(c) Let y = 0.36 and solve for x

0.36 0.0011 0.03970.3203 0.0011

291

x x x

=

≈The calf is galloping about 291 meters per minute

Section 1.1 LINES AND LINEAR FUNCTIONS 37

77 (a) The function is of the form

( )

y= f t =mt+ and contains the point b

(5, 8.2) and (17, 33.34) The slope is

So, the number of male alates in an ant

colony that is t years old is given by

t

t t

=

Assuming the linear function continues to

be accurate, we would expect a colony to

be about 20 years old before it has

approximately 40 male aletes

78 (a) Let t = 0 correspond to 1900 Then the

“area of the ice” line contains the points

slope form to obtain the equation of the

t t t

1900 + 125 = 2025

79 (a) If the temperature rises 0.3C° per decade,

it rises 0.03C° per year Therefore, 0.03

m =

15,

b = since a point is (0, 15) If T is the

average global temperature in degrees Celsius, we have T =0.03t+15

(b) Let T =19. Find t

19 0.03 15

4 0.03133.3 133

t t t

=

So, 1970 133 2103.+ =The temperature will rise to 19°C in about the year 2103

80. Use the formulas derived in Example 14 in this section of the textbook

93255329

C = 37.5; find F

937.5 32 67.5 32 99.55

The range is between 97.7°F and 99.5°F

81 The cost to use the first thermometer on x patients is y = 2x + 10 dollars The cost to use the second thermometer on x patients is

0.75 120

y= x+ dollars If these two costs are equal, then

2 10 0.75 1201.25 110

110881.25

x x

Let M = the total cost to use the Amalgamated

Medical Supplies machine

Note that P=20x+40, 000 and

30 32, 000

M = x+ Set M = P and solve for x

(continued on next page)

(continued)

30 32, 000 20 40, 000

10 8000800

x x

=

=The total cost to use the machines will be

equal if each is used 800 times

83 (a) The line (for the data for men) goes

(c) Since 0.137>0.117, women seem to

have the faster increase in median age at

=

≈The median age at first marriage for men

will reach 30 in the year

The median age at first marriage for

women will be 28.2 when the median age

for men is 30 (The answer will be 28.3 if

the year t = 46 is used as the answer for

249,187 13, 223.97( 50)249,187 13, 223.97 661,198.5

Truncate the y-intercept because it

represents the number of immigrants admitted to the United states in 1900 There can’t be 0.5 a person

(b) The year 2015 corresponds to t =115

13, 223.97 115 412, 0111,108, 746

y y

≈The number of immigrants admitted to the United States in 2015 will be about 1,108,746

(c) The equation y=13, 223.97t−412, 011

has –412,011 for the y-intercept,

indicating that the number of immigrants admitted in the year 1900 was –412,011 Realistically, the number of immigrants cannot be a negative value, so the equation cannot be used for valid predicted values

85 (a) The line goes through (90, 88) and

88 1.55( 90)

88 1.55 139.51.55 227.5

121

t t t

≈The mortality rate will drop to 40 or below in the year 1900 121 2021+ =

Section 1.2 THE LEAST SQUARES LINE 39

86 Use the formula derived in Example 14 in this

section of the textbook

5(26) 14.49

5( 20 32)9

5( 52) 28.99

9

325

1.2 The Least Squares Line

1. The correlation coefficient measures the degree

to which two variables are linearly related A

positive correlation coefficient does not

necessarily mean that an increase in one of the

quantities causes the other to increase also

There are two reasons why this is the case:

• We don’t know the direction of the cause

Does X cause Y or does Y cause X?

• Another variable, or variables, may be

involved that is responsible for the change

A positive correlation coefficient means that as one quantity increases, the other quantity also increases

2 For the set of points (1, 4), (2, 5), and (3, 6),

Y = x + 3 For the set (4, 1), (5, 2), and (6, 3),

y m x b

Section 1.2 THE LEAST SQUARES LINE 41

The point (9, 20)− is an outlier that has a

strong effect on the least squares line and the correlation coefficient

(c) Yes; because the data points are either on

or very close to the horizontal line y = 1, it

seems that the data should have a strong

linear relationship The correlation

coefficient does not describe well a linear

relationship if the data points fit a

(c) No; a correlation coefficient of 0 means

that there isn’t a linear relationship

between the x and y values A parabola (a

quadratic relationship) seems to fit the

given data points

0.4875

Section 1.2 THE LEAST SQUARES LINE 43

Yes, the points lie in a linear pattern

(b) Using a calculator’s STAT feature, the

correlation coefficient is found to be

0.959

r ≈ This indicates that the

percentage of successful hunts does trend

to increase with the size of the hunting

party

(c) Y =3.98x+22.7

12 (a) Using a TI-84 Plus graphing calculator, the

linear regression function gives the

coefficient of correlation as r ≈0.9929

(b) Using a TI-84 Plus graphing calculator, the

linear regression function gives the least

squares line as Y =1.973x+6.96

The least squares line fits the data well at

the beginning, but seems to diverge

slightly at the end

(d) When x = 18, the linear model gives

( )

1.973 18 6.696 42.2

This is less than the actual score of 43.6

13 (a) Using a TI-84 Plus graphing calculator, the

linear regression function gives the coefficient of correlation as r ≈0.9940

(b) Using a TI-84 Plus graphing calculator, the

linear regression function gives the least squares line as Y =1.3525x−2.51

Yes, the line accurately fits the data

(d) The slope of the least squares line is

m = 1.3525, so the fetal stature is increasing by about 1.3525 cm each week

(e) When x = 45, the linear model gives

( )

1.3525 45 2.51 58.35 cm

14 (a)

The data are mostly linear

(b) Using a TI-84 Plus graphing calculator, the

least squares line is Y =1.06x+32.6

(c) Using a TI-84 Plus graphing calculator, the

coefficient of correlation is 0.972 Yes, it agrees with the estimate of the fit in part (b)

15 (a) Using a TI-84 Plus graphing calculator, the

linear regression function gives the least squares line as Y =0.212x−0.309

≈When the crickets are chirping 18 times

per second, the temperature is about

86.4°F

(d) Using a TI-84 Plus graphing calculator, the

linear regression function gives the

coefficient of correlation as r ≈0.835

16 (a) Using a TI-84 Plus graphing calculator, the

linear regression function gives the least

=

≈The women’s record will catch up with the

men’s record in 1900 137,+ or in the year

2037

(d) There have been no improvements in the

women’s record since 1983, so the linear

regression equation may not closely

represent the data It is possible that the

women’s record will never catch up with

the men’s record

with negative slope

Answers will vary

17 (a) Skaggs’s average speed was

(d) Using a graphing calculator, r ≈0.9971

Yes, the least squares line is a very good fit to the data

(e) A good value for Skaggs’ average speed

would be the slope of the least squares line,

or m = 4.317 miles per hour This value is

faster than the average speed found in part (a) The value 4.317 miles per hour is most likely the better value

2 7(1, 818, 667, 092) (108, 210)

Section 1.2 THE LEAST SQUARES LINE 45

( 2) ( )2

2

7(1, 942, 595) (105)(108, 210)7(2275) (105)456.35

The slope suggests that the taller the

student, the shorter the ideal partner’s

There is no linear relationship among all

10 data pairs However, there is a linear

relationship among the first five data pairs

(female students) and a separate linear

relationship among the second five data

pairs (male students)

20 (a) Using a graphing calculator, we have

0.0067 14.75

(b) Let 420;x = find Y

0.0067(420) 14.7511.936 12

(c) Let 620;x = find Y

0.0067(620) 14.7510.596 11

(e) There is no linear relationship between a

student’s math SAT and mathematics placement test scores

( )12.02 0.3657142857(17.5)

70.803

0.0769 5.91

b

n b

The predicted number of points expected

when a team is at the 50 yard line is 2.07

points

1.3 Properties of Functions

1.3 Exercises

1. The x-value of 82 corresponds to two y-values,

93 and 14 In a function, each value of x must

correspond to exactly one value of y

The rule is not a function

2. Each x-value corresponds to exactly one

y-value The rule is a function

3. Each x-value corresponds to exactly one

y-value.The rule is a function

4. 9 corresponds to 3 and 3,− 4 corresponds to 2

and 2,− and 1 corresponds to 1 and 1

The rule is not a function

5.

y=x +

Each x-value corresponds to exactly one y

-value The rule is a function

6. y= x

Each x-value corresponds to exactly one y

-value The rule is a function

7. | |x= y

Each value of x (except 0) corresponds to two

y-values The rule is not a function

8. x= y2+ 4Solve the rule for y

y = −x y= ± x−Each value of x (greater than 4) corresponds to two y-values

−Pairs: ( 2, 1),− − (–1, 1), (0, 3), (1, 5), (2, 7), (3, 9)

Pairs: (–2, 15), (–1, 12), (0, 9), (1, 6), (2, 3), (3, 0)

Section 1.3 PROPERTIES OF FUNCTIONS 47

Pairs: ( 2,− 4), ( 1,− 1), (0, 0), (1, 1), (2, 4), (3, 9) Range: {0, 1, 4, 9}

16. y= −4x2

x y

Pairs: ( 2,− −16), ( 1,− −4), (0, 0), (1, 4),− (2, 16− ), (3, 36− ) Range: { 36,− −16, − 0} 4,

17. ( )f x =2x

x can take on any value, so the domain is the set

of real numbers, (−∞ ∞ , )

18 f x( )=2x+ 3

x can take on any value, so the domain is the set

of real numbers, which is written (−∞ ∞ , )

(continued on next page)

(continued)

Values in the interval ( 2,− 2) satisfy the

inequality; 2x= and x= − also satisfy the 2

inequality The domain is [ 2,− 2]

Thus, the domain is any real number except

6 or 6.− In interval notation, the domain is

(−∞ −, 6)∪( 6, 6)− ∪(6, ).∞

( 4)( 4)16

would produce a negative radicand and

(x−4) (⋅ x+4)=0 would lead to division by

zero

(x−4)(x+4)=0⇒ =x 4 or x= − 4

Use the values 4− and 4 to divide the number line into 3 intervals, (−∞ −, 4), ( 4,− 4) and (4, ).∞ Only the values in the intervals (−∞ −, 4) and (4, )∞ satisfy the inequality The domain is (−∞ −, 4)∪(4, ).∞

28.

2

5( )

36

f x

x

= −+

x can take on any value No choice for x will produce a zero in the denominator Also, no choice for x will produce a negative number under the radical The domain is (−∞,∞ )

29. f x( )= x2−4x− =5 (x−5)(x+1)See the method used in Exercise 21

(x−5)(x+1)≥0 when x≥ and when 51

Solve (3x−1) (x+1)=0

3 1 0 or1or3

x x

− =

1 01

x x

+ =

= −(continued on next page)

Section 1.3 PROPERTIES OF FUNCTIONS 49

(continued)

Use the values 1− and 1

3 to divide the number line into 3 intervals, (−∞ −, 1),

33. By reading the graph, the domain is all numbers

greater than or equal to 5− and less than 4 The

range is all numbers greater than or equal to 2−

and less than or equal to 6

Domain: [ 5,− 4); range: [ 2,− 6]

34. By reading the graph, the domain is all numbers

greater than or equal to 5.− The range is all

numbers greater than or equal to 0

Domain: [ 5, )− ∞ range: [0, )∞

35. By reading the graph, x can take on any value,

but y is less than or equal to 12

Domain: (−∞ ∞ range: (, ); −∞, 12]

36. By reading the graph, both x and y can take on

any values

Domain: (−∞ ∞ range: (, ); −∞ ∞ , )

37. The domain is all real numbers between the end

points of the curve, or [ 2, 4].−

The range is all real numbers between the

minimum and maximum values of the function

38 The domain is all real numbers between the end

points of the curve, or [–2, 4]

The range is all real numbers between the minimum and maximum values of the function

f ⎛ ⎞

= −

⎜ ⎟

⎝ ⎠

(d) From the graph, ( ) 1f x = when x=2.5

40. The domain is all real numbers between the end points of the curve, or [–2, 4]

The range is all real numbers between the minimum and maximum values of the function,

m m

( ) 1( 3)( 4) 1

12 1

13 0

1 1 522

23.140 or 4.140

9 1

2 2

m m m

f m

m m m

5

x x

2 4 4

m

m m m

+

−

− +

x

−

=+

= −

Section 1.3 PROPERTIES OF FUNCTIONS 51

45. f x( )=6x2− 2

2 2 2 2

h h h

x h

=+

h

h x h h

58 A vertical line drawn anywhere through the graph will intersect the graph in only one place The graph represents a function

59. Avertical line drawn through the graph may intersect the graph in two places The graph does not represent a function

60. Avertical line drawn through the graph may intersect the graph in two or more places The graph does not represent a function

61. A vertical line drawn anywhere through the graph will intersect the graph in only one place The graph represents a function

62. Avertical line is not the graph of a function since the one x-value in the domain corresponds

to more than one, in fact, infinitely many yvalues The graph does not represent a function

Section 1.3 PROPERTIES OF FUNCTIONS 53

x x

x

x x x x

=+

67 (a) The curve in the graph crosses the point

with x-coordinate 17:37 and y-coordinate

of approximately 140 So, at time 17 hours, 37 minutes the whale reaches a depth of about 140 m

(b) The curve in the graph crosses the point

with x-coordinate 17:39 and y-coordinate

of approximately 240 So, at time 17 hours, 39 minutes the whale reaches a depth of about 250 m

( ( )) (0.454 ) 19.7(0.454 )

19.7(0.454 )10.9

z z

=

≈

69 (a) (i) By the given function f, a muskrat

weighing 800 g expends

0.88(800) 0.01(800)

3.6, or approximately

≈3.6 kcal/km when swimming at the surface

of the water

(ii) A sea otter weighing 20,000 g

expends

0.88(20, 000) 0.01(20, 000)

(b) If z is the number of kilograms of an

animal’s weight, then x=g z( ) 1000= z is

the number of grams since 1 kilogram

equals 1000 grams

0.88 0.88 0.88 0.88

( ( )) (1000 )

0.01(1000 )0.01(1000 )4.4

z z z

(d) Answers will vary

71 (a) Let w = the width of the field and

let l = the length

The perimeter of the field is 6000 ft, so

(d) Answers will vary

72 (a) From the graph, it appears that the energy

consumption in 2015 for the United States will be about 100 quadrillion Btu, about

125 quadrillion Btu for China, and about

28 quadrillion Btu for India

(b) China will first consume 150 quadrillion

Btu in about 2023

(c) The energy consumption of China equaled

that of the United States in about 2009

73 (a) The independent variable is the year

(b) The dependent variable is the number of

1 If a ≥ 1, then the graph of y=ax2 becomes

narrower as the value of a increases If 0 < a

<1, then the graph of y=ax2 becomes wider

as the value of a decreases

2. If a < 0, then the graph of y=ax2 is the same

as the graph of y= a x2, but reflected across

the x-axis

3. The graph of y=x2−3is the graph of y=x2

translated 3 units downward

This is graph D

4. The graph of y=(x−3)2is the graph of y=x2

translated 3 units to the right

This is graph F

Section 1.4 QUADRATIC FUNCTIONS; TRANSLATION AND REFLECTION 55

5 The graph of y=(x−3)2+2 is the graph

of y=x2 translated 3 units to the right and 2

units upward

This is graph A

6. The graph of y=(x+3)2+2 is the graph

of y=x2translated 3 units to the left and

2 units upward

This is graph B

7. The graph of y= −(3−x)2+2is the same as the

graph of y= −(x−3)2+2 This is the graph of

2

y=x reflected in the x-axis, translated 3 units

to the right and 2 units upward

This is graph C

8. The graph of y= −(x+3)2+2is the graph

of y=x2 reflected in the x-axis, translated

three units to the left and two units upward

The x-intercepts are 3− and 2.− Set x =0 to

find the y-intercept

2

0 5(0) 66

y y

The x-intercepts are 5− and 1

Set x = 0 to find the y-intercept

2 2

6 6 4( 3)(4)2( 3)

The axis is x = –1, the vertical line through the

vertex

17. y=2x2+8x− 8Let y =0

2 2 2

The y-intercept is –8

(continued on next page)

Section 1.4 QUADRATIC FUNCTIONS; TRANSLATION AND REFLECTION 57

(continued)

The x-coordinate of the vertex is

82

The axis is x =3

19 f x( )=2x2−4x+ 5Let ( )f x =0

x-intercepts

Let 0.x =

22(0) 4(0) 55

y y

=The y-intercept is 5

b x a

22(1) 4(1) 5 2 4 5 3

21

2

12 12 4(1)(48)2(1)

12 144 1922

x-intercepts Let x =0

21(0) 6(0) 24 242

The y-intercept is 24

(continued on next page)