2

Functions and Their Graphs

- The Cartesian Coordinate System and

Straight lines - Equations of Lines
- Functions and Their Graphs
- The Algebra of Functions
- Linear Functions
- Quadratic Functions
- Functions and Mathematical Models

2.1

- The Cartesian Coordinate System and Straight lines

The Cartesian Coordinate System

- We can represent real numbers geometrically by

points on a real number, or coordinate, line

The Cartesian Coordinate System

- The Cartesian coordinate system extends this

concept to a plane (two dimensional space) by

adding a vertical axis.

4 3 2 1 1 2 3 4

The Cartesian Coordinate System

- The horizontal line is called the x-axis, and the

vertical line is called the y-axis.

y

4 3 2 1 1 2 3 4

x

The Cartesian Coordinate System

- The point where these two lines intersect is

called the origin.

y

4 3 2 1 1 2 3 4

Origin

x

The Cartesian Coordinate System

- In the x-axis, positive numbers are to the right

and negative numbers are to the left of the

origin.

y

4 3 2 1 1 2 3 4

Positive Direction

Negative Direction

x

The Cartesian Coordinate System

- In the y-axis, positive numbers are above and

negative numbers are below the origin.

y

4 3 2 1 1 2 3 4

Positive Direction

x

Negative Direction

The Cartesian Coordinate System

- A point in the plane can now be represented

uniquely in this coordinate system by an ordered

pair of numbers (x, y).

y

( 2, 4)

4 3 2 1 1 2 3 4

(4, 3)

x

(3, 1)

(1, 2)

The Cartesian Coordinate System

- The axes divide the plane into four quadrants as

shown below.

y

4 3 2 1 1 2 3 4

Quadrant I (, )

Quadrant II (, )

x

Quadrant IV (, )

Quadrant III (, )

Slope of a Vertical Line

- Let L denote the unique straight line that passes

through the two distinct points (x1, y1) and (x2,

y2). - If x1 x2, then L is a vertical line, and the

slope is undefined.

y

L

(x1, y1)

(x2, y2)

x

Slope of a Nonvertical Line

- If (x1, y1) and (x2, y2) are two distinct points

on a nonvertical line L, then the slope m of L is

given by

y

L

(x2, y2)

y2 y1 ?y

(x1, y1)

x2 x1 ?x

x

Slope of a Nonvertical Line

- If m gt 0, the line slants upward from left to

right.

y

L

m 1

?y 1

?x 1

x

Slope of a Nonvertical Line

- If m gt 0, the line slants upward from left to

right.

y

L

m 2

?y 2

?x 1

x

Slope of a Nonvertical Line

- If m lt 0, the line slants downward from left to

right.

y

m 1

?x 1

?y 1

x

L

Slope of a Nonvertical Line

- If m lt 0, the line slants downward from left to

right.

y

m 2

?x 1

?y 2

x

L

Examples

- Sketch the straight line that passes through the

point (2, 5) and has slope 4/3.

- Solution
- Plot the point (2, 5).
- A slope of 4/3 means that if x increases by 3,

y decreases by 4. - Plot the resulting point (5, 1).
- Draw a line through the two points.

y

6 5 4 3 2 1

?x 3

(2, 5)

?y 4

(5, 1)

x

1 2 3 4 5 6

L

Examples

- Find the slope m of the line that goes through

the points (1, 1) and (5, 3). - Solution
- Choose (x1, y1) to be (1, 1) and (x2, y2) to be

(5, 3). - With x1 1, y1 1, x2 5, y2 3, we find

Examples

- Find the slope m of the line that goes through

the points (2, 5) and (3, 5). - Solution
- Choose (x1, y1) to be (2, 5) and (x2, y2) to be

(3, 5). - With x1 2, y1 5, x2 3, y2 5, we find

Examples

- Find the slope m of the line that goes through

the points (2, 5) and (3, 5). - Solution
- The slope of a horizontal line is zero

y

6 4 3 2 1

(3, 5)

(2, 5)

L

m 0

x

2 1 1 2 3 4

Parallel Lines

- Two distinct lines are parallel if and only if

their slopes are equal or their slopes are

undefined.

Example

- Let L1 be a line that passes through the points

(2, 9) and (1, 3), and let L2 be the line that

passes through the points ( 4, 10) and (3, 4).

- Determine whether L1 and L2 are parallel.
- Solution
- The slope m1 of L1 is given by
- The slope m2 of L2 is given by
- Since m1 m2, the lines L1 and L2 are in fact

parallel.

2.2

- Equations of Lines

Equations of Lines

- Let L be a straight line parallel to the y-axis.
- Then L crosses the x-axis at some point (a, 0) ,

with the x-coordinate given by x a, where a is

a real number. - Any other point on L has the form (a, ), where

is an appropriate number. - The vertical line L can therefore be described as
- x a

y

L

(a, )

(a, 0)

x

Equations of Lines

- Let L be a nonvertical line with a slope m.
- Let (x1, y1) be a fixed point lying on L, and let

(x, y) be a variable point on L distinct from

(x1, y1). - Using the slope formula by letting (x, y) (x2,

y2), we get - Multiplying both sides by x x1 we get

Point-Slope Form

- An equation of the line that has slope m and

passes through point (x1, y1) is given by

Examples

- Find an equation of the line that passes through

the point (1, 3) and has slope 2. - Solution
- Use the point-slope form
- Substituting for point (1, 3) and slope m 2, we

obtain - Simplifying we get

Examples

- Find an equation of the line that passes through

the points (3, 2) and (4, 1). - Solution
- The slope is given by
- Substituting in the point-slope form for point

(4, 1) and slope m 3/7, we obtain

Perpendicular Lines

- If L1 and L2 are two distinct nonvertical lines

that have slopes m1 and m2, respectively, then L1

is perpendicular to L2 (written L1 - L2) if and

only if

Example

- Find the equation of the line L1 that passes

through the point (3, 1) and is perpendicular to

the line L2 described by - Solution
- L2 is described in point-slope form, so its slope

is m2 2. - Since the lines are perpendicular, the slope of

L1 must be - m1 1/2
- Using the point-slope form of the equation for L1

we obtain

Crossing the Axis

- A straight line L that is neither horizontal nor

vertical cuts the x-axis and the y-axis at, say,

points (a, 0) and (0, b), respectively. - The numbers a and b are called the x-intercept

and y-intercept, respectively, of L.

y

y-intercept

(0, b)

x-intercept

x

(a, 0)

L

Slope-Intercept Form

- An equation of the line that has slope m and

intersects the y-axis at the point (0, b) is

given by - y mx b

Examples

- Find the equation of the line that has slope 3

and y-intercept of 4. - Solution
- We substitute m 3 and b 4 into y mx b

and get - y 3x 4

Examples

- Determine the slope and y-intercept of the line

whose equation is 3x 4y 8. - Solution
- Rewrite the given equation in the slope-intercept

form. - Comparing to y mx b, we find that m ¾ and

b 2. - So, the slope is ¾ and the y-intercept is 2.

Applied Example

- Suppose an art object purchased for 50,000 is

expected to appreciate in value at a constant

rate of 5000 per year for the next 5 years. - Write an equation predicting the value of the art

object for any given year. - What will be its value 3 years after the

purchase? - Solution
- Let x time (in years) since the object was

purchased - y value of object (in dollars)
- Then, y 50,000 when x 0, so the y-intercept

is b 50,000. - Every year the value rises by 5000, so the slope

is m 5000. - Thus, the equation must be y 5000x 50,000.
- After 3 years the value of the object will be

65,000 - y 5000(3) 50,000 65,000

General Form of a Linear Equation

- The equation
- Ax By C 0
- where A, B, and C are constants and A and B are

not both zero, is called the general form of a

linear equation in the variables x and y.

General Form of a Linear Equation

- An equation of a straight line is a linear

equation conversely, every linear equation

represents a straight line.

Example

- Sketch the straight line represented by the

equation - 3x 4y 12 0
- Solution
- Since every straight line is uniquely determined

by two distinct points, we need find only two

such points through which the line passes in

order to sketch it. - For convenience, lets compute the x- and

y-intercepts - Setting y 0, we find x 4 so the x-intercept

is 4. - Setting x 0, we find y 3 so the y-intercept

is 3. - Thus, the line goes through the points (4, 0) and

(0, 3).

Example

- Sketch the straight line represented by the

equation - 3x 4y 12 0
- Solution
- Graph the line going through the points (4, 0)

and (0, 3).

y

L

1 1 2 3 4

(4, 0)

x

1 2 3 4 5 6

(0, 3)

Equations of Straight Lines

- Vertical line x a
- Horizontal line y b
- Point-slope form y y1 m(x x1)
- Slope-intercept form y mx b
- General Form Ax By C 0

2.3

- Functions and Their Graphs

Functions

- A function f is a rule that assigns to each

element in a set A one and only one element in a

set B. - The set A is called the domain of the function.
- It is customary to denote a function by a letter

of the alphabet, such as the letter f. - If x is an element in the domain of a function f,

then the element in B that f associates with x is

written f(x) (read f of x) and is called the

value of f at x. - The set B comprising all the values assumed by y

f(x) as x takes on all possible values in its

domain is called the range of the function f.

Example

- Let the function f be defined by the rule
- Find f(1)
- Solution

Example

- Let the function f be defined by the rule
- Find f( 2)
- Solution

Example

- Let the function f be defined by the rule
- Find f(a)
- Solution

Example

- Let the function f be defined by the rule
- Find f(a h)
- Solution

Applied Example

- ThermoMaster manufactures an indoor-outdoor

thermometer at its Mexican subsidiary. - Management estimates that the profit (in dollars)

realizable by ThermoMaster in the manufacture and

sale of x thermometers per week is - Find ThermoMasters weekly profit if its level of

production is - 1000 thermometers per week.
- 2000 thermometers per week.

Applied Example

- Solution
- We have
- The weekly profit by producing 1000 thermometers

is - or 2,000.
- The weekly profit by producing 2000 thermometers

is - or 7,000.

Determining the Domain of a Function

- Suppose we are given the function y f(x).
- Then, the variable x is called the independent

variable. - The variable y, whose value depends on x, is

called the dependent variable. - To determine the domain of a function, we need to

find what restrictions, if any, are to be placed

on the independent variable x. - In many practical problems, the domain of a

function is dictated by the nature of the problem.

Applied Example Packaging

- An open box is to be made from a rectangular

piece of cardboard 16 inches wide by cutting away

identical squares (x inches by x inches) from

each corner and folding up the resulting flaps.

x

10 10 2x

x

16 2x

x

x

16

Applied Example Packaging

- An open box is to be made from a rectangular

piece of cardboard 16 inches wide by cutting away

identical squares (x inches by x inches) from

each corner and folding up the resulting flaps. - Find the expression that gives the volume V of

the box as a function of x. - What is the domain of the function?

- The dimensions of the resulting box are

x

10 2x

16 2x

Applied Example Packaging

- Solution
- a. The volume of the box is given by multiplying

its dimensions (length ? width ? height), so

x

10 2x

16 2x

Applied Example Packaging

- Solution
- b. Since the length of each side of the box must

be greater than or equal to zero, we see that - must be satisfied simultaneously. Simplified
- All three are satisfied simultaneously provided

that - Thus, the domain of the function f is the

interval 0, 5.

More Examples

- Find the domain of the function
- Solution
- Since the square root of a negative number is

undefined, it is necessary that x 1 ? 0. - Thus the domain of the function is 1,?).

More Examples

- Find the domain of the function
- Solution
- Our only constraint is that you cannot divide by

zero, so - Which means that
- Or more specifically x ? 2 and x ? 2.
- Thus the domain of f consists of the intervals (

?, 2), (2, 2), (2, ?).

More Examples

- Find the domain of the function
- Solution
- Here, any real number satisfies the equation, so

the domain of f is the set of all real numbers.

Graphs of Functions

- If f is a function with domain A, then

corresponding to each real number x in A there is

precisely one real number f(x). - Thus, a function f with domain A can also be

defined as the set of all ordered pairs (x, f(x))

where x belongs to A. - The graph of a function f is the set of all

points (x, y) in the xy-plane such that x is in

the domain of f and y f(x).

Example

- The graph of a function f is shown below

y

y

(x, y)

Range

x

x

Domain

Example

- The graph of a function f is shown below
- What is the value of f(2)?

y

4 3 2 1 1 2

x

1 2 3 4 5 6 7 8

(2, 2)

Example

- The graph of a function f is shown below
- What is the value of f(5)?

y

4 3 2 1 1 2

(5, 3)

x

1 2 3 4 5 6 7 8

Example

- The graph of a function f is shown below
- What is the domain of f(x)?

y

4 3 2 1 1 2

x

1 2 3 4 5 6 7 8

Domain 1,8

Example

- The graph of a function f is shown below
- What is the range of f(x)?

y

4 3 2 1 1 2

Range 2,4

x

1 2 3 4 5 6 7 8

Example Sketching a Graph

- Sketch the graph of the function defined by the

equation - y x2 1
- Solution
- The domain of the function is the set of all real

numbers. - Assign several values to the variable x and

compute the corresponding values for y

x y

3 10

2 5

1 2

0 1

1 2

2 5

3 10

Example Sketching a Graph

- Sketch the graph of the function defined by the

equation - y x2 1
- Solution
- The domain of the function is the set of all real

numbers. - Then plot these values in a graph

y

10 8 6 4 2

x y

3 10

2 5

1 2

0 1

1 2

2 5

3 10

x

3 2 1 1 2 3

Example Sketching a Graph

- Sketch the graph of the function defined by the

equation - y x2 1
- Solution
- The domain of the function is the set of all real

numbers. - And finally, connect the dots

y

10 8 6 4 2

x y

3 10

2 5

1 2

0 1

1 2

2 5

3 10

x

3 2 1 1 2 3

Example Sketching a Graph

- Sketch the graph of the function defined by the

equation - Solution
- The function f is defined in a piecewise fashion

on the set of all real numbers. - In the subdomain ( ?, 0), the rule for f is

given by - In the subdomain 0, ?), the rule for f is given

by

Example Sketching a Graph

- Sketch the graph of the function defined by the

equation - Solution
- Substituting negative values for x into

, while - substituting zero and positive values into

we get

x y

3 3

2 2

1 1

0 0

1 1

2 1.41

3 1.73

Example Sketching a Graph

- Sketch the graph of the function defined by the

equation - Solution
- Plotting these data and graphing we get

y

x y

3 3

2 2

1 1

0 0

1 1

2 1.41

3 1.73

3 2 1

x

3 2 1 1 2 3

The Vertical Line Test

- A curve in the xy-plane is the graph of a

function y f(x) if and only if each vertical

line intersects it in at most one point.

Examples

- Determine if the curve in the graph is a function

of x - Solution
- The curve is indeed a function of x, because

there is one and only one value of y for any

given value of x.

y

x

Examples

- Determine if the curve in the graph is a function

of x - Solution
- The curve is not a function of x, because there

is more than one value of y for some values of x.

y

x

Examples

- Determine if the curve in the graph is a function

of x - Solution
- The curve is indeed a function of x, because

there is one and only one value of y for any

given value of x.

y

x

2.4

- The Algebra of Functions

The Sum, Difference, Product and Quotient

of Functions

- Consider the graph below
- R(t) denotes the federal government revenue at

any time t. - S(t) denotes the federal government spending at

any time t.

y

2000 1800 1600 1400 1200 1000

y R(t)

y S(t)

S(t)

Billions of Dollars

R(t)

t

1990 1992 1994 1996 1998 2000

t

Year

The Sum, Difference, Product and Quotient

of Functions

- Consider the graph below
- The difference R(t) S(t) gives the budget

deficit (if negative) or surplus (if positive) in

billions of dollars at any time t.

y

2000 1800 1600 1400 1200 1000

y R(t)

y S(t)

S(t)

Billions of Dollars

D(t) R(t) S(t)

R(t)

t

1990 1992 1994 1996 1998 2000

t

Year

The Sum, Difference, Product and Quotient

of Functions

- The budget balance D(t) is shown below
- D(t) is also a function that denotes the federal

government deficit (surplus) at any time t. - This function is the difference of the two

functions R and S. - D(t) has the same domain as R(t) and S(t).

y

400 200 0 200 400

y D(t)

t

Billions of Dollars

t

1992 1994 1996 1998 2000

D(t)

Year

The Sum, Difference, Product and Quotient

of Functions

- Most functions are built up from other, generally

simpler functions. - For example, we may view the function f(x) 2x

4 as the sum of the two functions g(x) 2x and

h(x) 4.

The Sum, Difference, Product and Quotient of

Functions

- Let f and g be functions with domains A and B,

respectively. - The sum f g, the difference f g, and the

product fg of f and g are functions with domain A

n B and rule given by - (f g)(x) f(x) g(x) Sum
- (f g)(x) f(x) g(x) Difference
- (fg)(x) f(x)g(x) Product
- The quotient f/g of f and g has domain A n B

excluding all numbers x such that g(x) 0 and

rule given by - Quotient

Example

- Let and g(x) 2x 1.
- Find the sum s, the difference d, the product p,

and the quotient q of the functions f and g. - Solution
- Since the domain of f is A 1,?) and the

domain of g is B ( ?, ?), we see

that the domain of s, d, and p is A

n B 1,?). - The rules are as follows

Example

- Let and g(x) 2x 1.
- Find the sum s, the difference d, the product p,

and the quotient q of the functions f and g. - Solution
- The domain of the quotient function is 1,?)

together with the restriction x ? ½ . - Thus, the domain is 1, ½) U ( ½,?).
- The rule is as follows

Applied Example Cost Functions

- Suppose Puritron, a manufacturer of water

filters, has a monthly fixed cost of 10,000 and

a variable cost of - 0.0001x2 10x (0 ? x ? 40,000)
- dollars, where x denotes the number of filters

manufactured per month. - Find a function C that gives the total monthly

cost incurred by Puritron in the manufacture of x

filters.

Applied Example Cost Functions

- Solution
- Puritrons monthly fixed cost is always 10,000,

so it can be described by the constant function - F(x) 10,000
- The variable cost can be described by the

function - V(x) 0.0001x2 10x
- The total cost is the sum of the fixed cost F and

the variable cost V - C(x) V(x) F(x)
- 0.0001x2 10x 10,000 (0 ? x ? 40,000)

Applied Example Cost Functions

- Lets now consider profits
- Suppose that the total revenue R realized by

Puritron from the sale of x water filters is

given by - R(x) 0.0005x2 20x (0 x 40,000)
- Find
- The total profit function for Puritron.
- The total profit when Puritron produces 10,000

filters per month.

Applied Example Cost Functions

- Solution
- The total profit P realized by the firm is the

difference between the total revenue R and the

total cost C - P(x) R(x) C(x)
- ( 0.0005x2 20x) ( 0.0001x2 10x

10,000) - 0.0004x2 10x 10,000
- The total profit realized by Puritron when

producing 10,000 filters per month is - P(x) 0.0004(10,000)2 10(10,000) 10,000
- 50,000
- or 50,000 per month.

The Composition of Two Functions

- Another way to build a function from other

functions is through a process known as the

composition of functions. - Consider the functions f and g
- Evaluating the function g at the point f(x), we

find that - This is an entirely new function, which we could

call h

The Composition of Two Functions

- Let f and g be functions.
- Then the composition of g and f is the function

ggf (read g circle f ) defined by - (ggf )(x) g(f(x))
- The domain of ggf is the set of all x in the

domain of f such that f(x) lies in the domain of

g.

Example

- Let
- Find
- The rule for the composite function ggf.
- The rule for the composite function fgg.
- Solution
- To find ggf, evaluate the function g at f(x)
- To find fgg, evaluate the function f at g(x)

Applied Example Automobile Pollution

- An environmental impact study conducted for the

city of Oxnard indicates that, under existing

environmental protection laws, the level of

carbon monoxide (CO) present in the air due to

pollution from automobile exhaust will be

0.01x2/3 parts per million when the number of

motor vehicles is x thousand. - A separate study conducted by a state government

agency estimates that t years from now the number

of motor vehicles in Oxnard will be 0.2t2 4t

64 thousand. - Find
- An expression for the concentration of CO in the

air due to automobile exhaust t years from now. - The level of concentration 5 years from now.

Applied Example Automobile Pollution

- Solution
- Part (a)
- The level of CO is described by the function
- g(x) 0.01x2/3
- where x is the number (in thousands) of motor

vehicles. - In turn, the number (in thousands) of motor

vehicles is described by the function - f(t) 0.2t2 4t 64
- where t is the number of years from now.
- Therefore, the concentration of CO due to

automobile exhaust t years from now is given by - (ggf )(t) g(f(t)) 0.01(0.2t2 4t 64)2/3

Applied Example Automobile Pollution

- Solution
- Part (b)
- The level of CO five years from now is
- (ggf )(5) g(f(5)) 0.010.2(5)2 4(5)

642/3 - (0.01)892/3 0.20
- or approximately 0.20 parts per million.

2.5

- Linear Functions

Linear Function

- The function f defined by
- where m and b are constants, is called a linear

function.

Applied Example Linear Depreciation

- A Web server has an original value of 10,000 and

is to be depreciated linearly over 5 years with a

3000 scrap value. - Find an expression giving the book value at the

end of year t. - What will be the book value of the server at the

end of the second year? - What is the rate of depreciation of the server?

Applied Example Linear Depreciation

- Solution
- Let V(t) denote the Web servers book value at

the end of the tth year. V is a linear function

of t. - To find an equation of the straight line that

represents the depreciation, observe that V

10,000 when t 0 this tells us that the line

passes through the point (0, 10,000). - Similarly, the condition that V 3000 when t 5

says that the line also passes through the point

(5, 3000). - Thus, the slope of the line is given by

Applied Example Linear Depreciation

- Solution
- Using the point-slope form of the equation of a

line with point (0, 10,000) and slope m 1400,

we obtain the required expression - The book value at the end of the second year is

given by - or 7200.
- The rate of depreciation of the server is given

by the negative slope of the depreciation line m

1400, so the rate of depreciation is 1400 per

year.

Applied Example Linear Depreciation

- Solution
- The graph of V is

V

(0, 10,000)

10,000

(5, 3000)

3000

t

1 2 3 4 5 6

Cost, Revenue, and Profit Functions

- Let x denote the number of units of a product

manufactured or sold. - Then, the total cost function is
- C(x) Total cost of manufacturing x units of

the product - The revenue function is
- R(x) Total revenue realized from the sale of x

units of the product - The profit function is
- P(x) Total profit realized from manufacturing

and selling x units of the product

Applied Example Profit Function

- Puritron, a manufacturer of water filters, has a

monthly fixed cost of 20,000, a production cost

of 20 per unit, and a selling price of 30 per

unit. - Find the cost function, the revenue function, and

the profit function for Puritron. - Solution
- Let x denote the number of units produced and

sold. - Then,

Finding the Point of Intersection

- Suppose we are given two straight lines L1 and L2

with equations - y m1x b1 and y m2x b2
- (where m1, b1, m2, and b2 are constants) that

intersect at the point P(x0, y0). - The point P(x0, y0) lies on the line L1 and so

satisfies the equation y m1x b1. - The point P(x0, y0) also lies on the line L2 and

so satisfies y m2x

b2 as well. - Therefore, to find the point of intersection

P(x0, y0) of the lines L1 and L2, we solve for x

and y the system composed of the two equations - y m1x b1 and y m2x b2

Example

- Find the point of intersection of the straight

lines that have equations - y x 1 and y 2x 4
- Solution
- Substituting the value y as given in the first

equation into the second equation, we obtain - Substituting this value of x into either one of

the given equations yields y 2. - Therefore, the required point of intersection is

(1, 2).

Example

- Find the point of intersection of the straight

lines that have equations - y x 1 and y 2x 4
- Solution
- The graph shows the point of intersection (1, 2)

of the two lines

y

5 4 3 2 1

L1

(1, 2)

x

1 1 2 3 4 5

L2

Applied Example Break-Even Level

- Prescott manufactures its products at a cost of

4 per unit and sells them for 10 per unit. - If the firms fixed cost is 12,000 per month,

determine the firms break-even point. - Solution
- The revenue function R and the cost function C

are given respectively by - Setting R(x) C(x), we obtain

Applied Example Break-Even Level

- Prescott manufactures its products at a cost of

4 per unit and sells them for 10 per unit. - If the firms fixed cost is 12,000 per month,

determine the firms break-even point. - Solution
- Substituting x 2000 into R(x) 10x gives
- So, Prescotts break-even point is 2000 units of

the product, resulting in a break-even revenue of

20,000 per month.

2.6

- Quadratic Functions

Quadratic Functions

- A quadratic function is one of the form
- where a, b, and c are constants and a ? 0.
- For example, the function
- is quadratic, with a 2, b 4, and c 3.

Quadratic Functions

- Below is the graph of the quadratic function
- The graph of a quadratic function is a curve

called a parabola that opens upward or downward.

y

10 8 6 4 2

Parabola

x

2 1 1 2 3 4

Quadratic Functions

- The parabola is symmetric with respect to a

vertical line called the axis of symmetry. - The axis of symmetry also passes through the

lowest or highest point of the parabola, which is

called the vertex of the parabola.

Axis of symmetry

y

10 8 6 4 2

Parabola

Vertex (1, 1)

x

2 1 1 2 3 4

Quadratic Functions

- We can use these properties to help us sketch the

graph of a quadratic function. - Suppose we want to sketch the graph of
- If we complete the square in x, we obtain
- Note that (x 1)2 is nonnegative it equals to

zero when x 1 and is greater than zero if x ?

1. - Thus, we see that f(x) ? 2 for all values of

x. - This tells us the vertex of the parabola is the

point (1, 2).

Quadratic Functions

- We know the vertex of the parabola is the point

(1, 2) and that it is the minimum point of the

graph, since f(x) ? 2 for all values of x. - Thus, the graph of f(x) 3x2 6x 1 looks as

follows

y

4 2 2

x

2 2 4

Vertex (1, 2)

Properties of Quadratic Functions

- Given f(x) ax2 bx c (a ? 0)
- The domain of f is the set of all real numbers.
- If a gt 0, the parabola opens upward, and if a lt

0, it opens downward. - The vertex of the parabola is
- The axis of symmetry of the parabola is
- The x-intercepts (if any) are found by solving

f(x) 0. The y-intercept is f(0)

c.

Example

- Given the quadratic function f(x) 2x2 5x

2 - Find the vertex of the parabola.
- Find the x-intercepts (if any) of the parabola.
- Sketch the parabola.
- Solution
- Here a 2, b 5, and c 2. therefore, the

x-coordinate of the vertex of the parabola is - The y-coordinate of the vertex is therefore

given by - Thus, the vertex of the parabola is the point

Example

- Given the quadratic function f(x) 2x2 5x

2 - Find the vertex of the parabola.
- Find the x-intercepts (if any) of the parabola.
- Sketch the parabola.
- Solution
- For the x-intercepts of the parabola, we solve

the equation - using the quadratic formula with a 2, b 5,

and c 2. - We find
- Thus, the x-intercepts of the parabola are 1/2

and 2.

Example

- Given the quadratic function f(x) 2x2 5x

2 - Find the vertex of the parabola.
- Find the x-intercepts (if any) of the parabola.
- Sketch the parabola.
- Solution
- The sketch

y

Vertex

1 1 2

x-intercepts

x

1 1 2

y-intercept

Some Economic Models

- Peoples decision on how much to demand or

purchase of a given product depends on the price

of the product - The higher the price the less they want to buy of

it. - A demand function p d(x) can be used to

describe this.

Some Economic Models

- Similarly, firms decision on how much to supply

or produce of a product depends on the price of

the product - The higher the price, the more they want to

produce of it. - A supply function p s(x) can be used to

describe this.

Some Economic Models

- The interaction between demand and supply will

ensure the market settles to a market

equilibrium - This is the situation at which quantity demanded

equals quantity supplied. - Graphically, this situation occurs when the

demand curve and the supply curve intersect

where d(x) s(x).

Applied Example Supply and Demand

- The demand function for a certain brand of

bluetooth wireless headset is given by - The corresponding supply function is given by
- where p is the expressed in dollars and x is

measured in units of a thousand. - Find the equilibrium quantity and price.

Applied Example Supply and Demand

- Solution
- We solve the following system of equations
- Substituting the second equation into the first

yields - Thus, either x 400/9 (but this is not

possible), or x 20. - So, the equilibrium quantity must be 20,000

headsets.

Applied Example Supply and Demand

- Solution
- The equilibrium price is given by
- or 40 per headset.

2.7

Functions and Mathematical Models

Mathematical Models

- As we have seen, mathematics can be used to solve

real-world problems. - We will now discuss a few more examples of

real-world phenomena, such as - The solvency of the U.S. Social Security

trust fund - Global warming

Mathematical Modeling

- Regardless of the field from which the real-world

problem is drawn, the problem is analyzed using a

process called mathematical modeling. - The four steps in this process are

Real-world problem

Mathematical model

Formulate

Solve

Test

Solution of real-world problem

Solution of mathematical model

Interpret

Modeling With Polynomial Functions

- A polynomial function of degree n is a function

of the form - where n is a nonnegative integer and the numbers

a0, a1, . an are

constants called the coefficients of the

polynomial function. - Examples
- The function below is polynomial function of

degree 5

Modeling With Polynomial Functions

- A polynomial function of degree n is a function

of the form - where n is a nonnegative integer and the numbers

a0, a1, . an are

constants called the coefficients of the

polynomial function. - Examples
- The function below is polynomial function of

degree 3

Applied Example Global Warming

- The increase in carbon dioxide (CO2) in the

atmosphere is a major cause of global warming. - Below is a table showing the average amount of

CO2, measured in parts per million volume (ppmv)

for various years from 1958 through 2007

Year 1958 1970 1974 1978 1985 1991 1998 2003 2007

Amount 315 325 330 335 345 355 365 375 380

Applied Example Global Warming

Year 1958 1970 1974 1978 1985 1991 1998 2003 2007

Amount 315 325 330 335 345 355 365 375 380

- Below is a scatter plot associated with these

data

y (ppmv)

380 360 340 320

t (years)

10 20 30 40 50

Applied Example Global Warming

Year 1958 1970 1974 1978 1985 1991 1998 2003 2007

Amount 315 325 330 335 345 355 365 375 380

- A mathematical model giving the approximate

amount of CO2 is given by

y (ppmv)

380 360 340 320

t (years)

10 20 30 40 50

Applied Example Global Warming

Year 1958 1970 1974 1978 1985 1991 1998 2003 2007

Amount 315 325 330 335 345 355 365 375 380

- Use the model to estimate the average amount of

atmospheric CO2 in 1980 (t 23). - Assume that the trend continued and use the model

to predict the average amount of atmospheric CO2

in 2010.

Applied Example Global Warming

Year 1958 1970 1974 1978 1985 1991 1998 2003 2007

Amount 315 325 330 335 345 355 365 375 380

- Solution
- The average amount of atmospheric CO2 in 1980 is

given by - or approximately 338 ppmv.
- Assuming that the trend will continue, the

average amount of atmospheric CO2 in 2010 will be

Applied Example Social Security Trust Fund

- The projected assets of the Social Security trust

fund (in trillions of dollars) from 2008 through

2040 are given by - The scatter plot associated with these data is

Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040

Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0

y (trillion)

6 4 2

t (years)

5 10 15 20 25 30

Applied Example Social Security Trust Fund

- The projected assets of the Social Security trust

fund (in trillions of dollars) from 2008 through

2040 are given by - A mathematical model giving the approximate value

of assets in the trust fund (in trillions of

dollars) is

Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040

Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0

y (trillion)

6 4 2

t (years)

5 10 15 20 25 30

Applied Example Social Security Trust Fund

Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040

Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0

- The first baby boomers will turn 65 in 2011. What

will be the assets of the Social Security trust

fund at that time? - The last of the baby boomers will turn 65 in

2029. What will the assets of the trust fund be

at the time? - Use the graph of function A(t) to estimate the

year in which the current Social Security system

will go broke.

Applied Example Social Security Trust Fund

Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040

Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0

- Solution
- The assets of the Social Security fund in 2011 (t

3) will be - or approximately 3.18 trillion.
- The assets of the Social Security fund in 2029

(t 21) will be - or approximately 5.59 trillion.

Applied Example Social Security Trust Fund

Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040

Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0

- Solution
- The graph shows that function A crosses the

t-axis at about t 32, suggesting the system

will go broke by 2040

y (trillion)

6 4 2

t (years)

5 10 15 20 25 30

Rational and Power Functions

- A rational function is simply the quotient of two

polynomials. - In general, a rational function has the form
- where f(x) and g(x) are polynomial functions.
- Since the division by zero is not allowed, we

conclude that the domain of a rational function

is the set of all real numbers except the zeros

of g (the roots of the equation g(x) 0)

Rational and Power Functions

- Examples of rational functions

Rational and Power Functions

- Functions of the form
- where r is any real number, are called power

functions. - We encountered examples of power functions

earlier in our work. - Examples of power functions

Rational and Power Functions

- Many functions involve combinations of rational

and power functions. - Examples

Applied Example Driving Costs

- A study of driving costs based on a 2007

medium-sized sedan found the following average

costs (car payments, gas, insurance, upkeep, and

depreciation), measured in cents per mile - A mathematical model giving the average cost in

cents per mile is - where x (in thousands) denotes the number of

miles the car is driven in 1 year.

Miles/year, x 5000 10,000 15,000 20,000

Cost/mile, y () 83.8 62.9 52.2 47.1

Applied Example Driving Costs

Miles/year, x 5000 10,000 15,000 20,000

Cost/mile, y () 83.8 62.9 52.2 47.1

- Below is the scatter plot associated with this

data

y ()

140 120 100 80 60 40 20

C(x)

x (?1000 miles/year)

5 10 15 20 25

Applied Example Driving Costs

Miles/year, x 5000 10,000 15,000 20,000

Cost/mile, y () 83.8 62.9 52.2 47.1

- Using this model, estimate the average cost of

driving a 2007 medium-sized sedan 8,000 miles per

year and 18,000 miles per year. - Solution
- The average cost for driving a car 8,000 miles

per year is - or approximately 68.8/mile.

Applied Example Driving Costs

Miles/year, x 5000 10,000 15,000 20,000

Cost/mile, y () 83.8 62.9 52.2 47.1

- Using this model, estimate the average cost of

driving a 2007 medium-sized sedan 8,000 miles per

year and 18,000 miles per year. - Solution
- The average cost for driving a car 18,000 miles

per year is - or approximately 48.95/mile.

Constructing Mathematical Models

- Some mathematical models can be constructed using

elementary geometric and algebraic arguments. - Guidelines for constructing mathematical

models - Assign a letter to each variable mentioned in the

problem. If appropriate, draw and label a figure. - Find an expression for the quantity sought.
- Use the conditions given in the problem to write

the quantity sought as a function f of

one variable. - Note any restrictions to be placed on the domain

of f by the nature of the

problem.

Applied Example Enclosing an Area

- The owner of the Rancho Los Feliz has 3000 yards

of fencing with which to enclose a rectangular

piece of grazing land along the straight portion

of a river. - Fencing is not required along the river.
- Letting x denote the width of the rectangle, find

a function f in the variable x giving the area of

the grazing land if she uses all of the fencing.

Applied Example Enclosing an Area

- Solution
- This information was given
- The area of the rectangular grazing land is A

xy. - The amount of fencing is 2x y which must equal

3000 (to use all the fencing), so - 2x y 3000
- Solving for y we get
- y 3000 2x
- Substituting this value of y into the expression

for A gives - A x(3000 2x) 3000x 2x2
- Finally, x and y represent distances, so they

must be nonnegative, so x ? 0 and y 3000 2x ?

0 (or x ? 1500). - Thus, the required function is
- f(x) 3000x 2x2 (0 ? x ? 1500)

Applied Example Charter-Flight Revenue

- If exactly 200 people sign up for a charter

flight, Leisure World Travel Agency charges 300

per person. - However, if more than 200 people sign up for the

flight (assume this is the case), then each fare

is reduced by 1 for each additional person. - Letting x denote the number of passengers above

200, find a function giving the revenue realized

by the company.

Applied Example Charter-Flight Revenue

- Solution
- This information was given.
- If there are x passengers above 200, then the

number of passengers signing up for the flight is

200 x. - The fare will be (300 x) dollars per passenger.
- The revenue will be
- R (200 x)(300 x)
- x2 100x 60,000
- The quantities must be positive, so x ? 0 and 300

x ? 0 (or x ? 300). - So the required function is
- f(x) x2 100x 60,000 (0 ? x ? 300)

End of Chapter