Download - Chapter 3: Polynomial Functions


Chapter 3: Polynomial Functions

3.1 Complex Numbers3.2 Quadratic Functions and Graphs3.3 Quadratic Equations and Inequalities3.4 Further Applications of Quadratic Functions and Models3.5 Higher Degree Polynomial Functions and Graphs3.6 Topics in the Theory of Polynomial Functions (I)3.7 Topics in the Theory of Polynomial Functions (II)3.8 Polynomial Equations and Inequalities; Further

Applications and Models


3.2 Quadratic Functions and Graphs

• Quadratic Functions are polynomial functions, discussed later. • P is often used to represent a polynomial function.• A function of the form

with a 0 is called a quadratic function.• Recall

is the graph of stretched or shrunk and shifted horizontally and vertically.

• Example

khxaxg 2)()(2)( xxf

Figure 9 pg 3-13

cbxaxxP 2)(


3.2 Completing the Square

• Rewrite in the form 1642)( 2 xxxP .)()( 2 khxaxg

Completing The Square1. Divide both sides of the equation by a, so that the

coefficient of is 1.2. Add to both sides.3. Add to both sides the square of half the coefficient of

x, 4. Factor the right side as the square of a binomial and

combine terms on the left.5. Isolate the term involving P(x) on the left.6. Multiply both sides by a.

2x

ac

.22ab


3.2 Example of Completing the Square

1642)( 2 xxxP

Divide by 2 to make the coefficient of x2 equal to 1.

822

)( 2 xxxP

xxxP 282

)( 2

12182

)( 2 xxxP

Add 8 to both sides.

Add [½·2]2 to both sides to complete the square on the right.

2)1(92

)( xxP

Combine terms on the left; factor on the right.

9)1(2

)( 2 xxPSubtract 9 from both sides.

18)1(2)( 2 xxP Multiply both sides by 2.


3.2 Example of Completing the Square

• From we can determine several components of the graph of

,18)1(2)( 2 xxP.1642)( 2 xxxP

),1[on increasing-

]1,(on decreasing-

),18[:range),,( :domain-

1:symmetry of axis-

)18,1(),( :vertex-

18,1,2-

x

kh

kha


3.2 Graphs of Quadratic Functions

Transform into

86)( 2 xxxP

86)( 2 xxxP

2)3(1)( xxP

.)()( 2 khxaxP 86)( 2 xxxP

xxxP 68)( 2

9698)( 2 xxxP

1)3()( 2 xxP

1)3()( 2 xxP

- P has vertex (-3,1), so the graph of f (x) = x2 is shifted left 3 and up 1. - The coefficient of (x+3)2 is –1, so the graph opens downward. - y-intercept: (0,–8)- Axis of symmetry: line x = -3- Domain: (-,); Range: (-,1]- increasing: (-,-3]; decreasing: [-3,)


• One method to determine the coordinates of the vertex is to complete the square.

• Rather than go through the procedure for each individual function, we generalize the result for P(x) = ax² + bx + c.

3.2 Graph of P(x) = a(x-h)2 + k

The graph of

(a) is a parabola with vertex (h,k), and the vertical line x = h as axis of symmetry;

(a) opens upward if a > 0 and downward if a < 0;

(b) is broader than and narrower than

,0,)()( 2 akhxaxP

10 if 2 axy.1 if 2 axy


3.2 Vertex Formula for Parabola P(x) = ax² + bx + c (a 0)

abac

abxaxP

aacb

abx

ay

abx

aacb

ay

abx

abx

ab

ac

ay

xabx

ac

ay

acx

abx

ay

acbxaxyacbxaxxP

44

2)(

44

2

244

44

)0()0()(

22

2

22

2

2

2

2

22

2

2

2

2

2

2

Standard form

Replace P(x) with y to simplify notation.

Divide by a.

Subtract

Add

Combine terms on the left; factor on the right.

.a

c

.42

1

2

22

a

b

a

b

Get y-term alone on the left.

Multiply by a and write in the form .)()( 2

khxaxP

h

k


3.2 Vertex Formula

Example Use the vertex formula to find the coordinates of the vertex of the graph of

Analytic Solution – exact solution

Approximation Using a calculator, we find

The vertex of the graph of is the point

)0()( 2 acbxaxxP

.2

,2

abP

ab

.4265.)( 2 xxxP

.4)65(.2

22

)65(.2

265.

2 and

)65(.2

2

)65.(2

2

22

a

bPy

a

bx

.77.4 and 09.1 yx

so ,2,65. ba


3.2 Extreme Values

• The vertex of the graph of is the– lowest point on the graph if a > 0, or– highest point on the graph if a < 0.

• Such points are called extreme points (also extrema, singular: extremum).

cbxaxxP 2)(

For the quadratic function defined by

(a) if a > 0, the vertex (h,k) is called the minimum point of the graph. The minimum value of the function is P(h) = k.

(b) if a < 0, the vertex (h,k) is called the maximum point of the graph. The maximum value of the function is P(h) = k.

,)( 2 cbxaxxP


3.2 Identifying Extreme Points and Extreme Values

Example Give the coordinates of the extreme point and the corresponding maximum or minimum value for each function.(a) (b)1642)( 2 xxxP 86)( 2 xxxP

The vertex of the graph is (–1,–18). Since a > 0, the minimum point is (–1,–18), and the minimum value is –18.

The vertex of the graph is (–3,1). Since a < 0, the maximum point is (–3,1), and the maximum value is 1.


3.2 Finding Extrema with the Graphing Calculator

Let

• One technique is to use the fmin function. We get the x-value where the minimum occurs. The y-value is found by substitution.

.1642)( 2 xxxP

Figure 14 pg 3-20b


Example The table gives data for the percent increase (y) on hospital services in the years 1994 – 2001, where x is the number of years since 1990. The data are plotted in the scatter diagram.

A good model for the data is the function defined by

(a) Use f (x) to approximate the year when the percent increase was a minimum.The x-value of the minimum point is

(b) Find the minimum percent increase.The minimum value is .64 differing slightly from the data value of .5 in the table.

3.2 Applications and Modeling

Year x Percent increase y

Year Percent increase y

4 1.8 8 3.4

5 0.8 9 5.8

6 0.5 10 7.1

7 1.3 11 12.0

4.1 5.54 62 2(0.37)ba

2( ) .37 4.1 12f x x x

(5.54) .37(5.54) 4.1(5.54) 12 .64f


3.2 Height of a Propelled Object

• The coefficient of t ², 16, is a constant based on gravitational force and thus varies on different surfaces.

• Note that s(t) is a parabola, and the variable x will be used for time t in graphing-calculator-assisted problems.

Height of a Propelled Object

If air resistance is neglected, the height s (in feet) of an object propelled directly upward from an initial height s0 feet with initial velocity v0 feet per second is

where t is the number of seconds after the object is propelled.

00216)( stvtts


A ball is thrown directly upward from an initial height of 100 feet with an initial velocity of 80 feet per second.(a) Give the function that describes height in terms of time t.

(b) Graph this function.

(c) The cursor in part (b) is at the point (4.8,115.36). What does this mean?

3.2 Solving a Problem Involving Projectile Motion

1008016)( 2 ttts

After 4.8 seconds, the object will be at a height of 115.36 feet.



(d) After how many seconds does the projectile reach its maximum height?

(e) For what interval of time is the height of the ball greater than 160 feet?

ft. 200 ofheight a reaches projectile theseconds, 2.5After

.200100)5.2(80)5.2(16 and

5.2)16(2

80

2

vertexat the occurs maximum The

2

y

a

bx

Figure 19 pg 3-24

Using the graphs, t must be between .92 and 4.08 seconds.



(f) After how many seconds will the ball fall to the ground?

When the ball hits the ground, its height will be 0, so we need to find the positive x-intercept. From the graph, the x-intercept is about 6.04, so the ball will reach the ground 6.04 seconds after it is projected.

Figure 21 pg 3-25

Download - Chapter 3: Polynomial Functions

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