Elementary Functions andTransformations

Section 4

c©csun Fall 2011 v12

Part I

• A beginning library of elementary functions

• Graphs of elementary functions

• Shifts and stretches

• Piecewise defined functions

• Quadratic functions and parabolas

• Rational functions

68

Beginning Library

• identity function f(x) = x

• absolute value function f(x) = |x|

• square function f(x) = x2

• square-root function f(x) =√

x

• piecewise defined functions

69

Identity and Absolute value functions

Identity function Absolute-value function

Expression f(x) = x Expression f(x) = |x| or abs (x)

Domain: all numbers (−∞,∞) Domain: all numbers (−∞,∞)

Range: all numbers (−∞,∞) Range: y ≥ 0, [0,∞)

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

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Square and Square-root functions

Square function Square-root functionExpression:f(x) = x2 Expression: f(x) =

√x

Domain: all numbers (−∞,∞) Domain: x ≥ 0, [0,∞)Range: y ≥ 0, [0,∞) Range: y ≥ 0, [0,∞)

-5 -4 -3 -2 -1 1 2 3 4 5

5

10

15

20

25

-5 5 10 15 20 25

-1

1

2

3

4

5

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Building Business Functions

Adding together multiples of x2 and x and constants we buildfunctions that model cost, revenue and profit.The graphs of cost revenue and profit retain the form of thoseabove: Lines and parabolas.

Why do we need square root and absolute value?

• Using the square root function we can find break even pointsand find intervals of profitability.

• Using the absolute value we can measure sizes of changes incost, revenue and profit.

72

Vertical shifting/stretching lines and

parabolas

How does the graph of y = x2 + 2 differ from that of y = x2?

• The point (1,1) on the graph ofy = x2 will correspond to (1,4)on the graph of y = x2 + 3.Why?

• The point (10,100) on thegraph of y = x2 will correspondto (10, ) on the graph ofy = x2 + 3.

• The point (x, x2) on y = x2

will correspond to the point(x, x2 + 3) on y = x2 + 3. Avertical shift up 3 units.

!3 !2 !1 0 1 2 3

3

6

9

12

73

Transformations of lines and

parabolas

• vertical translations (shift)

• vertical stretch

• horizontal translation (shift)

74

Vertical shift

Vertical shift 5 units up

-4 -3 -2 -1 1 2 3 4

-10

-8

-6

-4

-2

2

4

6

8

10

The graph of f(x) is blue(dark line).

The graph of f(x) + 5is red (light line).

The vertical distance be-tween the curves is 5.

75

Vertical stretch

Vertical stretch by a factor of 2:

-4 -3 -2 -1 1 2 3 4

-10

-8

-6

-4

-2

2

4

6

8

10

The graph of f(x) is blue(dark line).

The graph of 2f(x) is red(light line).

The vertical distance fromthe x-axis of the graph of2f(x) is twice that of f(x).

76

Vertical flip

Vertical flip:

!5000 5000 10 000 15 000 20 000 25 000

!1 000 000

!500 000

500 000

1 000 000

The graph ofy = .01x2 − 200x + 1000is blue (dark line).

The graph ofy = −(.01x2 − 200x + 1000) ory = −.01x2 + 200x− 1000is red (dashed).

The graph ofy = −.01x2 + 200x− 1000is identical to that ofy = .01x2 − 200x + 1000except it is flipped across the x-axis.

77

Example

Here are the graphs of y = x2 and y = −3x2 − 5.

Notice: The second equation’sgraph is the same basic shape as thefirst: a parabola• stretched up by by a factor of 3

flipped across the x-axisshifted down by a factor of 5

• Does the order matter?• What is the vertex?

!3 !2 !1 1 2 3

!30

!25

!20

!15

!10

!5

5

78

Practice

Graph y = −2x2 + 3 andfind the domain and range of f(x) = −2x2 + 3.

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

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Horizontal shift

Suppose P1(t) is a function that gives annual profits for a lampmanufacturer, Illuminous, where t is measured in years and t = 0corresponds to the year 1990.Thus P1(0) gives the profits from what year?

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20years since 19000

5000

10 000

15 000

20 000

25 000

30 000

profit in dollars

The marketing department in the year 2011, decides that theprofit information from 1990 to 2000 is no longer relevant andwants to re-graph profits starting from 2000. So now P2(t),should measure annual profits with P2(0) giving the profits inthe year 2000. How would the graph change?

Horizontal shift

The graphical way to think about this is to think of just shiftingthe vertical axis to the right by 10 units.

!10!9 !8 !7 !6 !5 !4 !3 !2 !1 0 1 2 3 4 5 6 7 8 9 10years since 2000

5000

10 000

15 000

20 000

25 000

30 000

profit in dollars

Finding relationship algebraically between P1(t) and P2(t):If t = 10, P1(10) should give the profits in what year?

This is equal to P2( )

Thus P2(0) = P1(10).Similarly since P1(15) would give profits in the year 2005:

P2( ) = P1(15)

In general P2(t) = P1(t + 10).Corresponds to shifting the vertical axis to the right 10 units.

81

Beta: There are some negative values of t shown in thegraph of P2(t). What do those mean?

Alpha: You’re asking what P2(−4) stands for, for example

Beta: Yes, that’s what I want to know.

Alpha: I’ll answer with a question. P2(+4) stands for theprofit four years after 2000. So what is P2(−4)?

Beta: Could it be the profit four years before ?

Alpha: That’s it!

Beta: So, P2(−4) is the profit in the year .

Alpha: Right. And what was the profit in 1996?

Beta: The height of the graph ofat t = −4 is .So the profit in 1996 was $15,000.

Alpha: Yes.

82

Caution

Suppose we were so foolish as to graph y = P1(t) and y = P2(t)on the same pair of axes.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20years since ????0

5000

10 000

15 000

20 000

25 000

30 000

profit in dollars

Notice that it appears that the graph of y = P2(t) = P1(t + 10)is just a horizontal shift to the left of the graph of y = P1(t).

83

Practice

Graph look back at the graph of y = −2x2 + 3graph y = −2(x + 1)2 + 3 and find the domain and range ofg(x) = −2(x + 1)2 + 3.

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

84

Practice: Each function corresponds to geomet-ric description

f(x− 5) horizontal shift of the x-axis 5 units to the left

f(x) + 7

3f(x)

f(x + 3)

vertical shift 2 units up

vertical shrink by a factor of 1/2

horizontal shift of the x-axis 4 units to left

85

Piecewise defined functions, an

example

A car rental agency charges $30 per day (or partial day) or $150per week, whichever is least. What is the rental cost C(x) for x

days?

Here are the costs for various numbers of days, x. Fill in the twomissing costs.

x 1.0 2.0 2.6 3.0 3.1 4.0 4.2 5.0 6.0 7.0 7.1C(x) 60 90 90 120 150 150 150 150 180

86

A car rental agency charges $30 per day (or partial day) or $150per week, whichever is least. Graph the cost function C(x).

1 2 3 4 5 6 7 8 9 10Days

30

60

90

120

150

180

210

240

270Cost

87

Example from business T (x) is the tax on taxable income of x.

The federal income tax rate is

Between But Not Over Base Tax Rate Of the

Amount Over

$0 $7,550 0 10% $0.00

$7,550 $30,650 $755.00 15% $7,550

$30,650 $74,200 $4,220.00 25% $30,650

$74,200 $154,800 $15,107.50 28% $74,200

$154,800 $336,550 $37,675.50 33% $154,800

$336,550 $97,653.00 35% $336,550

If you have a taxable income of x = $110,000, your tax is

T (110,000) = Base Tax + (Rate×Amount Over)

= 15,107.50 + [.28× (110,000− 74,200)]

= 15,107.50 + [.28× 35,800]

= 15,107.50 + 10,024.00

= 25,131.50

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Between But Not Over Base Tax Rate Of the

Amount Over

$0 $7,550 0 10% $0.00

$7,550 $30,650 $755.00 15% $7,550

$30,650 $74,200 $4,220.00 25% $30,650

$74,200 $154,800 $15,107.50 28% $74,200

$154,800 $336,550 $37,675.50 33% $154,800

$336,550 $97,653.00 35% $336,550

Suppose you have a taxable income of x =$50,000. What isyour tax?

89

The graph of T (x):

!!

!

!

!

!!

!

!

!

100 000 200 000 300 000 400 000Income

20 000

40 000

60 000

80 000

100 000

120 000

Tax

• What is the slope of the line segment between 74,200 and154,800?

• What is the height of the function when income is $74,200?

90

The graph of T (x):

!!

!

!

!

!!

!

!

!

100 000 200 000 300 000 400 000Income

20 000

40 000

60 000

80 000

100 000

120 000

Tax

• What are the coordinates of the point shown at the upperright of the graph?

• Why is it important that the graph has no ”jumps” in it?

91

Tax for incomes between $74,200 and $154,800

The equations for T (x):

Between But Not Over Base Tax Rate Of the

Amount Over

$0 $7,550 0 10% $0.00

$7,550 $30,650 $755.00 15% $7,550

$30,650 $74,200 $4,220.00 25% $30,650

$74,200 $154,800 $15,107.50 28% $74,200

$154,800 $336,550 $37,675.50 33% $154,800

$336,550 $97,653.00 35% $336,550

For income between $74,200 and $154,800:Line 4 in the table.74200 ≤ x ≤ 154800: T (x) =

92

Income at the end of an income bracket

What is the tax on an income of exactly $74,200? Which line inthe table should be used–line 3 or line 4?

Between But Not Over Base Tax Rate Of the

Amount Over

$0 $7,550 0 10% $0.00

$7,550 $30,650 $755.00 15% $7,550

$30,650 $74,200 $4,220.00 25% $30,650

$74,200 $154,800 $15,107.50 28% $74,200

$154,800 $336,550 $37,675.50 33% $154,800

$336,550 $97,653.00 35% $336,550

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Quadratic Functions and Parabolas

• Parabolas

• Quadratic equations and functions

• Graphs of quadratic functions

• Applications

94

Quadratic Functions and Expressions

The standard form for a quadratic function is

f(x) = ax2 + bx + c.

The graph of a quadratic function is a parabola.

The parabola opens up The parabola opens downif a > 0 if a < 0.

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Vertex-axis form for a quadratic function

• f(x) = ax2 + bx + c (standard form)

• f(x) = a(x− h)2 + k (vertex-axis form)

Since

a(x− h)2 + k = ax2 − 2ahx + ah2 + k,

b = −2ah, or h = −b/2a.

Once you know the x-coordinate (called h) of the vertex, they-coordinate (called k) is

k = f(h).

96

Vertex of the parabola, maximum and minimum

f(x) = ax2 + bx + c.

The vertex of the parabola is found at the point where x = −b/2a.

Often the vertex is denoted (h, k). In this case, h = −b/2a andk can be determined by the equation: k = f(h).

Example:

What are the coordinates of the vertex of the graph off(x) = 3x2 + 6x− 5? (h, k) = ( , )

97

Example and Exercise:

Remember: h = −b/2a.

Find the coordinates of the vertex of the graph of

f(x) = 2x2 + 12x + 13.

a = 2, b = 12, c = 13, so h = −3 and thenk = f(−3) = 2(−3)2 + 12(−3) + 13 = −5.

So the coordinates of the vertex are (−3,−5).

Next find the vertex-axis form of the quadratic function f(x).

The vertex-axis form is

f(x) = a(x− h)2 + k = 2(x + 3)2 − 5.

Check it:

2(x + 3)2 − 5 = 2(x2 + 6x + 9)− 5 = 2x2 + 12x + 13.

98

Graphing a quadratic function

y = 2x2 + 12x + 13

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

The vertex is the point atThe axis of symmetry is the vertical line

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Exercise:

Find the coordinates of the vertex, convert the standard forminto the vertex-axis form and graph:f(x) = −x2 + 6x− 5

Opens:

Vertex:

y-intercept:

100

Exercise:

Graph the parabola:f(x) = 3x2 + 6x + 1

Opens:

Vertex:

y-intercept:

101

The quadratic formula

The general quadratic function:

f(x) = ax2 + bx + c

The quadratic formula tells you the solutions to f(x) = 0,which is the same as locating the x-intercepts on the graph:

x =−b ±

√b2 − 4ac

2a.

102

The quadratic formula

x =−b ±

√b2 − 4ac

2a.

Example: Solve

2x2 − 5x− 3 = 0,

for x.

a = 2, b = −5, c = −3

x =5±

√52 − 4(2)(−3)

2(2)=

5±√

49

4=

5± 7

4.

So x = 3 and x = −1/2 are the solutions.

103

The quadratic formula

x =−b ±

√b2 − 4ac

2a.

Exercise: Graph y = x2−5x−6 and solve x2−5x−6 = 0, forx.

a = , b = , c =

104

Summary: Quadratic Functionsf(x) = ax2 + bx + c,

where a is not equal to zero

• if a > 0, the graph opens

• if a < 0, the graph opens

• x-coordinate of vertex h =

• f(h) = k is minimum if a > 0

• f(h) = k is maximum if a < 0

• Domain: set of all real numbers

• Range:(−∞, k], if a < 0[k,∞) , if a > 0

105

Rational Functions

• Introduction: Average Cost

• graphs

• asymptotes

106

Application: average cost

The cost to produce a gardening book is $8000 plus $10.00per copy. So the cost function is

C(x) = 8000 + 10x,

where x is the number of books produced.

What is the average cost per book, if x = 2000 copies are pro-duced?

C(2000) = 28000 So it costs $28,000 to produce 2000 books.That’s an average of

$28000

2000=

The average cost to produce x books is

C̄(x) = C(x)x =

which is a rational function.

107

Application: average cost

C̄(x) =8000 + 10x

x=

8000

x+ 10.

So what happens to C̄(x) as x gets large (i.e. goes to infinity)?

Symbolically mathematicians write: limx→∞ C̄(x) =

What does this limit mean in the context of this example?

0 2000 4000 6000 8000 10 000Books0

10

20

30

40

Average cost

108

Rational functions

A rational function is a function of the form

f(x) =n(x)

d(x),

where both the numerator, n(x), and the denominator, d(x) arepolynomials.

These are rational functions:

f(x) =1

x

C̄(x) =8000 + 10x

x

E(p) =p

40− p

f(x) =29x19 + 77x2 − 89

x16 − 55x9 + 44

f(x) =x + 26

4x2 − 7x + 22We will focus mostly on examples similar to the first three.

109

Simple rational functions

The graph of y = 1x is:

!10!9!8!7!6!5!4!3!2!1 1 2 3 4 5 6 7 8 910

!10!9!8!7!6!5!4!3!2!1

12345678910

In this case as x gets large y = 1x goes to .

(Compare to Average cost.)

110

Simple rational functions

The line that f(x) = ax+bcx+d approaches as x goes to infinity is

called the horizontal asymptote of the graph of y = f(x).What is the horizontal asymptote of:

1. y = 1x

2. y = 8000+10xx = 8000

x + 10

3. y = p40−p

111

Simple rational functions

The graph of y = 1x is:

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

In this case as x gets close to zero y = 1x goes to infinity.

(Compare to Average cost.)

112

Simple rational functions

The vertical line that f(x) = ax+bcx+d approaches as x approaches

−dc is called the vertical asymptote of the graph of y = f(x).

To find it set the denominator equal to zero and solve for x.What are the vertical asymptotes of:

1. y = 1x

2. y = 8000+10xx = 8000

x + 10

3. y = p40−p

113

The graph of f(x) =4x + 1

2x

!4 !2 2 4

!4

!2

2

4

As x→ ±∞, f(x)→ 2

limx→±∞

f(x) = 2

The graph of f(x) has ahorizontal asymptote aty = 2.

As x→ 0, f(x)→ ±∞

The graph of f(x) has avertical asymptote atx = 0.

114

Example:

f(x) =500 + 25x

x

115

Example:

E(p) =p

80− p

116