1 - 1

Chapter 1

Linear Functions

1 - 2

Section 1.1

Slopes and Equations of

Lines

Figure 1

Figure 2

Figure 3

Your Turn 1

Find the slope of the line through (1,5) and (4,6).

1 1 2 2 Let ( , ) (1,5) andSol ( ,ut )ion: (4,6). x y x y

Use the definition of the slope .y

mx

6 54 1

13

Figure 4

Your Turn 2

Find the equation of the line with x-intercept − 4 and

y-intercept 6.

Solution: Notice that b = 6.

To find m, use the definition of the slope after writing the

x-intercept as (− 4, 0) and y-intercept as (0,6).

Substituting these values into y = mx + b, we have

0 6 6 34 0 4 2

m

36.

2y x

Your Turn 3

Find the slope of the line whose equation is 8 3 5.x y

To find the slope, solve the equatiSolution on f: or y.

8 3 5x y

Subtract 8 from3 8 5 both sides. y x x

Divide both si8 5

3

des b 33

y . y x

The coefficient of is 8 / 3, which is the slope of the line. x

Your Turn 4

Find the equation of the line through (2,9) and (5,3).

Put your answer in slope-intercept form. Begin by using the definition of slope of the line that

passes through the give

So

n

lution

poi

:

nts.3 9 6

25 2 3

m

Either (2,9) or (5,3) can be used in the point-slope form with 2.m

1 1

1 1

1 1

If ( , ) (2,9), then

( )

9 2( 2)

9, 2, 2 y x

x y

y y m x x

y m x

9 2 4 Distributiv e Propertyy x

2 13 Add 9 b ot h sidesy x

Figure 5

Your Turn 5

Find the equation of the line that passes through the point (4,5)

and is parallel to the line 3 6 7.x y The slope of 3 6 7 can be found by writing the

equation in slope-int

Solutio

ercept form.

n

: x y

subtract 3 from both sides a

3 6 7

6 3 7

3 7

nd

and divid 6 6

e both sides by 6.

1 72 6

x

x y

y x

y x

y x

The result shows that the slope is 1/ 2. Since the lines

are parallel, 1/ 2 is also the slope of the line whose equation

we want. cont inued

Your Turn 5 continued

1 1

1 1

This line passes through (4,5). Substituting 1/ 2,

4 and 5 into the point-slope form give

(

s

)y

m

x y

y m x x

1 1

1 1

This line passes through (4,5). Substituting 1/ 2,

4 and 5 into the point-slope form giv s

( ).

e

y x

x

y

m

m x

y

12 5 Add 5 to both sid e .

2sy x

13

2y x

Your Turn 6

Find the equation of the line that passes through the point (3,2)

and is perpendicular to the line 2 3 4.x y To find the slope, write 2 3 4 in

slope-int

Solut

erce

ion

pt r

:

m:

fo

x y

subtract 2 from both sides

2 3 4

3 2 4

2 4

3 3

and

and divide both sides by 3.

x y

y x

y

x

x

The slope is 2 / 3. Since the lines are perpendicular, if line

2has slope , then 1

33 / 2. continued

L

m m

m

Your Turn 6 continued

1 1

1 1

Now subtitute 3 / 2, 3 and 2 into the point-slope

( )

form.

y y m

m x y

x x

32 ( 3)

2y x

3 92 Distri

2butive propert

2yy x

Add 2 to both sides and ge3 9

+2 2 23 5

t a

common de 2 2

nominator.

y x

y x

Figure 6

Figure 7

Figure 8

Figure 9

Figure 10

1 - 28

Section 1.2

Linear Functions and Applications

Your Turn 1

Let ( ) 4 5. Find ( 5).g x x g

To find ( 5), substituteSolution: 5 for .g x

( 5) 4( 5) 5 g

( 5) 20 5 25g

Your Turn 2(a) Suppose that Greg Tobin, manager of a giant supermarket

chain, has studied the supply and demand for watermelons. He

has noticed that the demand increases as the price decreases.

He has determined that the quantity (in thousands) demanded

weekly, q, and the price (in dollars) per watermelon, p, are

related by the linear function

(a) Find the quantity of watermelons demanded at a price of $3.30 per watermelon.

( ) 9 0.75 . Demand functionp D q q

Subtract 9 from both sides.

3.30 9 0.75

5.70 0.75

7.6 Divide both sides by 0.7

Thus, at

Sol

a

u

price $3.30,

t

ion:

the

5.

q

q

q

quantity demanded is 7600 watermelons.

Your Turn 2(b)

Greg also noticed that the quantity of watermelons supplied

decreased as the price decreased. Price p and supply q are

related by the linear function

(b) Find the quantity of watermelons supplied at a price of $3.30 per watermelon.

( ) 0.75 . Supply functionp S q q

Divide both sides by 0.75.

3.30 0.75

4.4

Thus, at a price $3.30, the quantity supplied is 4400 watermelons

S

.

olution: q

q

Your Turn 3 Find the equilibrium quantity and price for the watermelons

using the demand equation

and the supply equation

Solution: The equilibrium quantity is found when the prices

from both supply and demand are equal. Set the two

expressions for p equal to each other and solve.

The equilibrium quantity is 8000 watermelons.

The equilibrium price can be found by plugging the value of q = 8

into either the demand or the supply function.

Continued

( ) 10 0.85D q q ( ) 0.4 .S q q

10 0.85 0.4

10 1.25 Add 0.85 to both sides.

8

q q

q

q

q

Your Turn 3 continued

The equilibrium price can be found by plugging the value of q = 8

into either the demand or the supply function.

Using the demand function,

The equilibrium price is $3.20.

(8) 10 0.85(8) 3.2.p D

Figure 11

Figure 12

Your Turn 4

The marginal cost to make x batches of a prescription medication

is $15 per batch, while the cost to produce 80 batches is $1930.

Find the cost function C(x), given that it is linear.

Solution: Since the cost function is linear, it can be expressed in the

form C(x) = mx+b. The marginal cost is $15 per batch, which gives

the value for m. Using x = 80 and C(x) = 1930 in the point-slope

form of the line gives

( ) 1930 15( 80)

( ) 1930 15 1200

( ) 15 730 Ad d 1930 to both sides.

C x x

C x x

C x x

Your Turn 5A firm producing poultry feed finds that the total cost C(x) in

dollars of producing x units is given by

Management plans to charge $58 per unit for the feed.

How many units must be sold to produce a profit of $8030?

Solution: Since R(x) = p x and p = 58, R(x) = 58x.

Use the formula for profit P(x) = R(x) – C(x).

( ) 35 250.C x x

Solve for8030 58 (35 250)

8030 58 35 250

8030 23

250

.xx x

x x

x

8030 250 23

8280 23

Add 250 t

360

Sales of 360 units will produce $8030 profi

o both sides.

Divide both sides by 23.

t.

x

x

x

Figure 13