Section 3 Quadratic Functions - Intersg.inter.edu/mis/maec2140/ppt/bmb12e_ppt_2_3-1.pdf ·...

Chapter 2

Functions and

Graphs

Section 3

Quadratic Functions

2 Barnett/Ziegler/Byleen College Mathematics 12e

Learning Objectives for Section 2.3

Quadratic Functions

The student will be able to identify and define quadratic

functions, equations, and inequalities.

The student will be able to identify and use properties of

quadratic functions and their graphs.

The student will be able to solve applications of quadratic

functions.


Table of Content

Quadratic Functions, Equations

and Inequalities

Properties of Quadratic Functions

and their Graphs

Applications


Terms

quadratic function

quadratic formula

second-degree function

parabola

vertex form


Quadratic Functions

Is the degree of a linear equation is increased by one, we

obtain a second-degree function. Another basic function

that will be studied.

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


Quadratic Functions

y

x 0

1

2

3

4

5

6

7

8

9

10

1 3 5 7 9 11 13

h(x) = x^2


Quadratic Functions

y

x

f(x) = x^2 - 4

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

1 3 5 7 9 11 13

y

x 0

10

20

30

40

50

60

70

80

90

1 4 7 10 13 16 19

g(x) = 3x^2 – 12x - 14


Quadratic Functions

y

x

h(x) = 3 – 2x - x^2

-90

-80

-70

-60

-50

-40

-30

-20

-10

0

10

1 5 9 13 17 21 25 29 33 37


Quadratic Functions

Important Note.

The most important method for solving quadratic functions is

the quadratic formula. It says that if

Then


Example 1: Quadratic Functions

Graph, Intercepts.

(A) Sketch a graph of f(x) = -x2 + 5x + 3

(B) Find x and y intercepts algebraically to four decimal

places



Solution

(A) Hand-sketching a graph of f:

x y

-1 -3

0 3

1 7

2 9

3 9

4 7

5 3

6 -3

-4

-2

0

2

4

6

8

10

1 2 3 4 5 6 7 8

y

x

12

Barnett/Ziegler/Byleen College Mathematics 12e


(B) Finding intercepts algebraically:

y intercept:

x intercept: f(x) = 0


Example 2: Intercepts of a

Quadratic Function

Find the x and y intercepts of 2( ) 3 6 1f x x x


Example 2: Intercepts of a

Quadratic Function

Find the x and y intercepts of

x intercepts: Set f (x) = 0:

Use the quadratic formula:

x = =

2( ) 3 6 1f x x x

20 3 6 1x x

2 4

2

b b ac

a

26 6 4( 3)( 1) 6 240.184,1.816

2( 3) 6


Example 2: Intercepts of a Quadratic

Function

y intercept: Let x = 0. If x = 0, then y = –1, so (0, –1)

is the y intercept.

2( ) 3 6 1f x x x

1)0(6)0(3)0( 2 f


Properties of the Quadratic

Functions and their Graphs

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

Complete the square to find the vertex.

2( ) 3 6 1f x x x


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

Solution:

Factor the coefficient of x2 out of the first two terms:

f (x) = –3(x2 – 2x) –1


Completing the square

(continued)

The vertex is (1, 2)

The quadratic function opens down since the coefficient of the

x2 term is –3, which is negative.

Add 1 to complete the square inside the parentheses. Because

of the –3 outside the parentheses, we have actually added –3,

so we must add +3 to the outside.

f (x) = –3(x2 – 2x +1) –1+3

f (x) = –3(x – 1)2 + 2


Example 2: Completing the Square

to Find the Vertex of a QF

Consider the quadratic function given by

f (x) = –2x2 + 16x – 24



Example 2: Completing the Square

to Find the Vertex of a QF

Consider the quadratic function given by

f (x) = –2x2 + 16x – 24


Solution:

Factor the coefficient of x2 out of the first two terms:

f (x) = –2(x2 – 8x) – 24


Example 2: Completing the

square (continued)

If x = 4, then -2(x - 4) 2 = 0 y f (4) = 8. For any other value of

x, the negative number -2(x - 4) 2 is added to 8, making it

smaller. Therefore f (4) = 8 is the maximum value of f(x) for

all x.

Add 16 to complete the square inside the parentheses.

Because of the –2 outside the parentheses, we have actually

added –32, so we must add +32 to the outside.

f(x) = –2(x2 – 8x +16) – 24+ 32

f(x) = –2(x – 4)2 + 8



square (continued)

Therefore, the vertical line x = 4 is a line of symmetry.

Furthermore, if we choose any two x values that are the same

distance from 4, we will obtain the same value. For example,

x = 3 and x = 5, we have:

f (3) = –2(3 – 4)2 + 8 = 6

f (5) = –2(5 – 4)2 + 8 = 6



square (continued)

The graph shows that as x moves

from left to right, f(x) is

increasing on , and

decreasing on , and that f(x)

can assume no value greater than

8. Thus,

Range of f:

-4

-2

0

2

4

6

8

10

1 2 3 4 5 6 7 8

y

x


[Conceptual Insight]

Applying the graph transformation

properties discussed in Section 2-2

to the transformed equation,

f (x) = –2x2 + 16x – 24

= –2(x – 4)2 + 8

We see that the graph of f(x) is the

graph of g(x) = x2 vertically

stretched by a factor of 2, reflected

in the x axis, and shifted to the right

4 units and up 8 units.

-4

-2

0

2

4

6

8

10

1 2 3 4 5 6 7 8

y

x 0

2

4

6

8

10

1 4 7 10 13


Generalization

If a 0, then the graph of f is a parabola.

• If a > 0, the graph opens upward.

• If a < 0, the graph opens downward. Vertex is (h , k)

Axis of symmetry: x = h

f (h) = k is the minimum if a > 0, otherwise the maximum

Domain = set of all real numbers

Range: if a < 0. If a > 0, the range is y y k

2( ) ( )f x a x h k For

y y k


Generalization


Analyzing a Quadratic Function

Given the quadratic function

f(x) = 0.5x2 - 6x + 21

(A)Find the vertex form for f.

(B) Find the vertex and the maximum or minimum. State

the range of f.

(C) Describe how the graph of function f can be obtained

from the graph of g(x) = x2 using transformations.

(D) Sketch the graph of function f in a rectangular

coordinate system.


Answer:

(A)Complete the square to find the vertex form:

f(x) = 0.5x2 - 6x + 21

= 0.5(x2 - 12x + ?) + 21

= 0.5(x2 - 12x + 36) + 21 - 18

= 0.5(x – 6)2 + 3

(B) From the vertex form, we see that h = 6 and k = 3. Thus,

vertex: (6,3); minimum f(6) = 3; range: y ≥ 3 or [3,∞).

Analizing a Quadratic Function


Answer:

(C) The graph of

f(x) = 0.5(x – 6)2 + 3

is the same as the graph of

g(x) = x2 vertically shrunk by

a factor of 0.5, and shifted to

the right 6 units and up 3

units.

(D) Graph

Analizing a Quadratic Function

y

x 0

10

20

30

40

50

60

70

80

90

1 4 7 10 13 16 19


The financial department of a company that produces digital

cameras arrived at the following price-demand function and the

corresponding revenue function:

p(x) = 94.8 - 5x Price-demand function

R(x) = xp(x) = x(94.8 – 5x) Revenue function

Where p(x) is the whole price per camera at which x million

cameras can be sold and R(x) is the corresponding revenue (in

million dollars). Both have domain 1 < x < 15.

Application of Quadratic Functions:

Maximum Revenue


(A) Find the value of x to the nearest thousands cameras that will

generate the maximum revenue. What is the maximum

revenue to the nearest thousands dollars?

(B) What is the whole price per camera (to the nearest dollar)

that generates the maximum revenue?

(C) Sketch a graph of function f in a rectangular coordinate

system.


Maximum Revenue (continued)


(A) Algebraic solution:

R(x) = x(94.8 – 5x)

= -5x2 + 94.8x

= -5(x2 – 18.96x + ?)

= -5(x2 – 18.96x + 89.8704) + 449.352

= -5(x - 9.48)2 + 449.352

The maximum revenue of 449.352 million dollars

($449,352,000) occurs when x = 9.480 million cameras

(9,480,000 cameras).

Solution to Maximum Revenue


(B) Finding the wholesale price per camera. Use the price-

demand function for un output of 9,480,000 cameras:

p(x) = 94.8 - 5x

p(9.480) = 94.8 - 5(9.480)

= $47 per camera

C. Graph

Solution to Maximum Revenue

(continued)


The financial department of a company that produces digital

cameras has the revenue and cost functions for x million cameras

are as follows:

R(x) = x(94.8 – 5x)

C(x) = 156 + 19.7x. Both have domain 1 < x < 15

(A)Sketch the graphs of both functions in the same coordinate

system.

(B) Break-even points are the production levels at which R(x) =

C(x). Find the break-even points algebraically to the nearest

thousand cameras.


Break-Even Analysis


Solution to Break-Even Problem

(A) If we graph the cost and revenue functions on a same

coordinate system, we obtain the following graphs, showing

the two intersection points:


Solution to Break-Even Problem

(continued)

(B) Algebraic solution: Find x such as R(x) equal to C(x):

x(94.8 – 5x) = 156 + 19.7x

–5x2 + 75.1x – 156 = 0

𝑥 = −75.1 ± 75.1 − 4(5)(−156)

2(−5) =

−75.1 ± 75.1 − 3120

2(−5)=

−75.1 ± 2520.01

−10

𝑥 = −75.1 ± 2520.01

−10=

−75.1 ± 50.20

−10

x = 2.490 or 12.530

The company breaks even at x = 2.490 and 12.530 million

cameras.

Chapter 2

Functions and

Graphs

Section 3

Quadratic Functions

END Last Update: February 25/2013

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