1.1 Functions

• This section deals with the topic of functions, one of the most important topics in all of mathematics. Let’s discuss the idea of the Cartesian coordinate system first.

Cartesian Coordinate System

• The Cartesian coordinate system was named after Rene Descartes. It consists of two real number lines which meet at a point called the origin. The two number lines which meet at a right angle divide the plane into four areas called quadrants.

• The quadrants are numbered using Roman numerals as shown. Each point in the plane corresponds to one and only one ordered pair of numbers (x , y). Two ordered pairs are shown.

x

y

III

III IV

(3,1)

(-1,-1)

Graphing an equation

• To graph an equation in x and y, we need to find ordered pairs that solve the equation and plot the ordered pairs on a grid.

For example, let’s plot the graph of the equation y = x2 + 2

Making a table of ordered pairs

• Let’s make a table of ordered pairs that satisfy the equation y = x2 + 2

210-1-2-3

yx2( 3) 2− +

2( 2 ) 2 6− + =2( 1) 2 3− + =

2(0) 2 2+ =2(1) 2 3+ =

2(2) 2 6+ =

Plotting the points

• Next, plot the points and connect them with a smooth curve. You may need to plot additional points to see the pattern formed.

Function

• The previous graph is the graph of a function. The idea of a function is this: a relationship between two sets D and R

• such that for each element of the first set, D, there corresponds one and only one element of the second set, R.

For example, the cost of a pizza (C) is related to the size of the pizza. A 10 inch diameter pizza costs 9.00 while a 16 inch diameter pizza costs 12.00.

Function definition• You can visualize a function by the following diagram which shows

a correspondence between two sets, D, the domain of the functionand R, the range of the function. The domain gives the diameter of pizzas and the range gives the cost of the pizza.

10

12 169.00

12.0010.00

domainrange

Functions specified by equations

• Consider the previous equation that was graphed

-2

2

Input x = -2

Process: square (–2) then subtract 2

Output: result is 2 (-2,2) is an ordered pair of the function.

Function Notation

• The following notation is used to describe functions The variable y will now be called

• This is read as “ f of x” and simply means the y coordinate of the function corresponding to a given x value.

Our previous equation

can now be expressed as

Function evaluation

• Consider our function

• What does mean? Replace x with the value –3 and evaluate the expression

• The result is 11 . This means that the point (-3,11) is on the graph of the function.

Some Examples

• 1.

( ) 3( ) 2af a= −

( ) 3( ) 2 18 3 2

1 3

6

6

6h hf h

h

= − = + −

= +

+ +

Domain of a Function

• Consider

• which is not a real number. Question: for what values of x is the function defined?

( ) 3 2f x x= −

0

(0) ?

( ) 3( ) 20 2

f

f

=

= − = −

Domain of a function

• Answer:

• is defined only when the radicand (3x-2)• is greater than or equal to zero. This implies that

3x-2 0or

≥23

x≥

Domain of a function

• Therefore, the domain of our function is the set of real numbers that are greater than or equal to

• Examples. Find the domain of the following functions.

• Answer:

23

1( ) 42

f x x= −

{ }8 , [8, )x x ≥ ∞

More examples

• Find the domain of

• In this case, the function is defined for all values of x except where the denominator of the fraction is zero. This means all real numbers x except

1( )3 5

f xx

=−

53

Mathematical modeling

• The price-demand function for a company is given by•

• where P(x) represents the price of the item and x represents the number of items. Determine the revenue function and find the revenue generated if 50 items are sold.

( ) 1000 5 , 0 100p x x x= − ≤ ≤

Solution

• Revenue = price x quantity so • R(x)= p(x)*x =• When 50 items are sold, x = 50 so we will evaluate the

revenue function at x = 50

• The domain of the function has already been specified. We are told that

(1000 5 )x x− i

(50) (1000 5(50)) 50 37,500R = − =i

0 100x≤ ≤

1.2 Elementary Functions; Graphs and Transformations

• In this presentation, you will be given an equation of a function and asked to draw its graph. You should be able to state how the graph is related to a “standard” function. It is not important that you plot a great many points for each graph. It IS important that you recognize the general shape of the graph. You can verify your answers using a graphing calculator, but only after you have attempted to construct the graph by hand.

Problem 1

• Construct the graph of

Solution

0

2

4

6

8

10

12

-4 -2 0 2 4

Series1

Problem 2

• Now, sketch the related graph given by the equation below and explain, in words, how it is related to the first function you graphed.

Solution: Problem 2

• The graph has the same shape as the original function. The difference is that the original graph has been translated two units to the right on the x-axis. Conclusion: The graph of the function f(x-2) is the graph of f(x) shifted horizontally two units to the right on the x-axis.

• Notice that replacing x by x-2 shifts the graph horizontally to the right and not the left.

Problem 3

• Now, graph the following “standard”function: Complete the table:

23

10-1-2-3

Solution to problem 3

-30

-20

-10

0

10

20

30

-4 -2 0 2 4Series1

Problem 4

• Now, graph the following related function:

Solution to problem 4

-30

-20

-10

0

10

20

30

-4 -2 0 2 4Series1

Problem 4 solution

• The graph of

• is obtained from the graph of

• by translating the graph of the original function up one unit vertically on the positive y-axis.

Problem 5

• Graph:

• What is the domain of this function?

Solution to problem 5

• The domain is all non-negative real numbers. Here is the graph:

0

1

2

3

4

5

6

0 5 10 15 20 25 30

Series1

Problem 6

• Graph:

• Explain, in words, how it compares to problem 5.

Problem 6 solution(Notice that the graph lies entirely within the fourth

quadrant)

-6

-5

-4

-3

-2

-1

00 5 10 15 20 25 30

Series1

Graph of –f(x)

• The graph of the function –f(x) is a reflection of the graph of f(x) across the x-axis. That is, if the graphs of f(x) and –f(x) are folded along the x-axis, the two graphs would coincide.

Cube root function

• Sketch the graph of the cube root function. Complete the table of ordered pairs:

8

-8

27

10-1

-27yx

Variation of cube root function

• Sketch the following variation of the cube root function:

Same graph as graph of cube root function. Shifted horizontally to the left one unit.

Graph of f(x+c) compared to graph of f(x):

• The graph of f(x+c) has the same shape as the graph of f(x) with the exception that the graph of f(x+c) is translated horizontally to the left c units when c >0 and is translated horizontally to the right c units when c < 0.

Absolute Value function

• Now, graph the absolute value function. Be sure to choose x values that are both positive and negative as well as zero.

( )a x x=

Graph of absolute value function

Notice the symmetry of the graph.

Variation of absolute value function

( ) 21 1 2x xa − = + −+

Shift absolute value graph to the left one unit and down two units on the vertical axis.

Linear functions and Straight Lines

Linear Functions

• The equation f(x) = mx+b m and b are real numbers is the equation of a linear function. The domain is the set of all real numbers. The graph of a linear function is a straight line. Some examples of graphs will follow in the next few slides.

1( ) 84

f x x= +

f(x)= -2x+3

More examples

( ) 2f x = −

Graphing

• Graph

• using a table of values for x and y

3( ) 24

f x x= +

840-4

yx

Solution:

Graphing using intercepts

• Graph 5x+6y = 30 using the x and y intercepts:

• 1. Set x = 0 and solve for y • 5(0) + 6y = 30 • y = 5 . • 2. Now, let y =0 and solve for x: • 5x + 6(0) = 30, x = 6 • 3. Plot the two ordered pairs (0, 5) and (6,0) and connect the points

with a straight line.

solution

Special cases

• 1. The graph of x=k is the graph of a vertical line k units from the y-axis.

• 2. The graph of y=k is the graph of the horizontal line k units from the x-axis.

Some examples:

• 1. Graph x=-7

• 2. Graph y = 3

solutionsX=7

Y=3

Slope of a line

• Slope of a line: m =

• = riserun

2 1

2 1

y yx x−−

Rise

run

( )2 2,x y

( )1 1,x y

Slope-intercept form

• The equation

• is called the slope-intercept form of an equation of a line .

• The letter m represents the slope and brepresents the y intercept.

y mx b= +

Find the slope and intercept from an equation of a line

• 1. Find the slope and y intercept of the line whose equation is

• 5x – 2y = 10 Solution: Solve the equation for y in

terms of x. Identify the coefficient of x as the slope and the y-intercept is the constant term.

Therefore: the slope is 5/2 and the y intercept is -5

-

5 2 102 5 105 10 5 52 2 2

x yy xxy x

− =− =− +−

= + = −− −

Point-slope form• The point- slope form of

the equation of a line is as follows:

• It is derived from the definition of the slope of a line:

1 1( )y y m x x− = −

2 1

2 1

y y mx x−

=−

Examples

• Find the equation of the line through the points (-5, 7) and (4, 16) :

• Solution:

( )5,7 (4,16)16 7 9 1

4 ( 5) 916 1( 4) 4 16 12

m

y x y x x

−

−= = =

− −− = − → = − + = +

Applications • Office equipment was purchased for $20,000 and will

have a scrap value of $2,000 after 10 years. If its value is depreciated linearly , find the linear equation that relates value (V) in dollars to time (t) in years:

Solution: when t = 0 , V = 20,000 and when t = 10, V = 2,000. Thus, we have two ordered pairs (0, 20,000) and (10, 2000). We find the slope of the line using the slope formula. The y-intercept is already known (when t = 0, V = 20,000, so the y-intercept is 20,000) . The slope is : (2000-20,000)/(10 – 0) = -1,800.

Therefore, our equation is V(t)= - 1,800t + 20,000

Quadratic functions

• If a, b, c are real numbers with a not equal to zero, then the function

• is a quadratic function and its graph is a parabola.

2( )f x ax bx c= + +

Vertex form of the quadratic function

• It is convenient to convert the general form of a quadratic equation

• to what is known as the vertex form.

2( )f x ax bx c= + +

2( ) ( )f x a x h k= − +

Completing the square to find the vertex of a quadratic function

• The example below illustrates the procedure:

Consider 2( ) 3 6 1f x x x= − + −Complete the square to find the vertex:

2

_____

2

2

( ) 3( 2 ) 1

( ) 3 1)( 2( ) 3( 1)

14

3

f x x x

f x x xf x x

+ −

= − − + −

= − −

= − − −

−

Completing the square, continued

• The vertex is (1 , 2)• The quadratic function opens down since

the coefficient of the x squared term is negative (-3) .

Intercepts of a quadratic function

• Find the x-intercepts of.

• Set f(x) = 0

• Use the quadratic formula: X =

2( ) 3 6 1f x x x= − + −

20 3 6 1x x= − + −

2 42

b b aca

− ± −

Intercepts of a quadratic function

• X=

•

26 6 4( 3)( 1) 6 24 0.184,1.8162( 3) 6

− ± − − − − ±= ≈

− −

Find the y-intercept : Let x = 0 and solve for y: We have (0, -1)

2( ) 3 6 1f x x x= − + −2( ) 3(0) 6(0) 1f x = − + −

Generalization

• Summary:

• where a is not equal to zero.

2( ) ( )f x a x h k= − +

Graph of f is a parabola: if a > 0, the graph opensupward if a < 0 , the graph opens downward.

Generalization, continued• Vertex is (h , k)• Line or axis of symmetry: x = h • f(h) = k is the minimum if a > 0, otherwise, f(h) = k is

the maximum • Domain : set of all real numbers

• Range: if a < 0. If a > 0, the range is{ }y y k≤

{ }y y k≥

Application of Quadratic Functions

• A Macon Georgia peach orchard farmer now has 20 trees per acre. Each tree produces, on the average, 300 peaches. For each additional tree that the farmer plants, the number of peaches per tree is reduced by 10. How many more trees should the farmer plant to achieve the maximum yield of peaches? What is the maximum yield?

•

Solution

• Solution: Yield= number of peaches per tree x number of trees

• Yield = 300 x 20 = 6000 ( currently) • Plant one more tree: Yield = ( 300 – 1(10))

* ( 20 + 1) = 290 x 21 = 6090 peaches. • Plant two more trees: • Yield = ( 300 – 2(10)* ( 20 + 2) = 280 x 22

= 6160

Solution, continued

• Let x represent the number of additional trees. Then Yield =( 300 – 10x) (20 + x)=

• Y(x)=• To find the maximum yield, note that the Y(x) function is

a quadratic function opening downward. Hence, the vertex of the function will be the maximum value of the yield.

•

210 100 6000x x− + +

Solution, continued • Complete the square to find the vertex of the parabola: •• Y(x) =

• (we have to add 250 on the outside since we multiplied • –10 by 25 = -250. The equation is unchanged, then.

210( 10 25) 6000 250x x− − + + +

Solution,continued

• Y(x)=

• Thus, the vertex of the quadratic function is ( 5 , 6250) . So the farmer should plant 5 additional trees and obtain a yield of 6250 peaches. We know this yield is the maximum of the quadratic function since the the value of a is -10. The function opens downward so the vertex must be the maximum.

•

210( 5) 6250x− − +