lect -6-38.pptx

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232 Mathematics

Polynomial Functions

Defn:

Polynomial function

In the form of: .

The coefficients are real numbers.

The exponents are non-negative integers.

The domain of the function is the set of all real numbers.

𝑓 (𝑥 )=5 𝑥 +2 𝑥2 −6 𝑥3+3

𝑔

(

𝑥

)

=

2

𝑥

−

4

𝑥

+

√

𝑥

−

2

h

(

𝑥

)

=

2

𝑥

(

4

𝑥

+

3

𝑥

)

Are the following functions polynomials?

yes no

yes

𝑘( 𝑥 )= 2 𝑥

3

+3

4 𝑥5+3 𝑥

Polynomial Functions

Definition:

Degree of a Function

The largest degree of the function represents the degree of the function.

The zero function (all coefficients and the constant are zero) does not have a degree.

𝑎

¿

𝑓

(

𝑥

)

5

𝑥

+

2

𝑥

2

−

6

𝑥

3

+

3

𝑔

(

𝑥

)

=

2

𝑥

−

4

𝑥

+

𝑥

−

2

h

(

𝑥

)

=

2

𝑥

(

4

𝑥

+

3

𝑥

)

3 5

8

𝑘( 𝑥 )=4 𝑥3+6 𝑥11− 𝑥10+ 𝑥12 12

1

Constant function:

Examples :

Is x= c (constant) is function ?

Linear Function :

• _{A Linear Function Is a function of the form} •

• _{where m and b are real numbers and m is the slope and}

b is the y - intercept.

• _{The x – intercept is}

• _{The domain and range of a linear}

function are all real numbers.

2

• _{Linear function y = x plotted as}

b

mx

x

f

(

)





m b

Example 1 : Graph using x and y intercepts 6x + 9y = 18

x-intercept( put y = 0)

6x = 18

x = 3

)

3

,

0

(

y-intercept(put x = 0)

9y = 18

y = 2

)

0

,

2

(

D: (0, 2)

C: (3, 0)

D

C

Quadratic

Function :

The Standard Form of a quadratic function

f

(

x

)



a

(

x



h

)



k, a



0

is in standard form. The graph of f is a parabola whose vertex is the point (h, k). The parabola is symmetric to the line x  h. If a > 0, the parabola opens upward; if a < 0, the parabola opens downward.

Copyright © by Houghton Mifflin Company, Inc. All rights reserved

. 10

Example: Graph f(x) = (x – 3)2 _{+ 2 and find the vertex and axis}

.

f(x) = (x – 3)2 _{+ 2 is the same shape as the graph of}

g(x) = (x – 3)2 _{shifted upwards two units}

.

g(x) = (x – 3)2 _{is the same shape as y = x}2 _{shifted to the right three units}

.

f(x) = (x – 3)2 _{+ 2}

g(x) = (x – 3)2

y = x2

Example

Solution:

Step 1 Determine how the parabola opens. Note that a, the coefficient of

x2_{, is -1. Thus, a} < _{0; this negative value tells us that the parabola opens}

downward.

Step 2 Find the vertex. We know the x-coordinate of the vertex is –b/2a. We identify a, b, and c to substitute the values into the equation for the

x-coordinate:

x = -b/(2a) = -6/2(-1) = 3.

The x-coordinate of the vertex is 3. We substitute 3 for x in the equation of the function to find the y-coordinate:

the parabola has its vertex at (3,7).

Graph the quadratic function

f

(

x

)





x



6

x

.

2

(3) 3 6(3) 2 9 18 2 7

Example

Graph the quadratic function

f

(

x

)





x



6

x

.

Step 3 Find the x-intercepts. Replace f(x) with 0 in f(x)  x2  6x  2. 0 =

x2  6x  2

a  1,b  6,c  2

x  b  b2  4ac

2a

 6 62  4(1)(2)

2(₁₎

 6 36  8 2

 6 28

₂  

6 2 7

Example

Graph the quadratic function

f

(

x

)





x



6

x

.

Step 4 Find the y-intercept. Replace x with 0 in f

(x)  x2  6x  2.

f(0)  02  6 • 0  2  

The y-intercept is –2. The parabola passes through (0, 2).

Step 5 Graph the parabola. -10 -8 -6 -4 -2 2 4 6 8 10

4

5

Exponential function :

The exponential function with base a is denoted by f(x) = ax , where a > 0, a

1 and x represents any real number.

Example 1: Graph using a table of values.

x 2x

0 1 1 2 2 4 -1 1/2 -2 1/4

x y

( )

Example 2: Graph using a table of values. X 1 2 0 -1 -2 1/3 1/9 1 3 9

( )

Example 3 : To graph f(x) =ex _{, you will need to use your calculator.}

Graph f(x) =ex _{using a table of values.}

x ex

1 e1 = 2.72

2 e2 = 7.39

0 e0 = 1

-1 e-1 = 0.37

-2 e-2 = 0.14

Text Example

Write each equation in its equivalent exponential form.

a. 2 = log₅ x b. 3 = log_b 64 c. log₃ 7 = y

Solution With the fact that y = log_b x means by = x,

c. log₃ 7 = y or y = log₃ 7 means 3y = 7.

a. 2 = log₅ x means 52 = x.

Logarithms are exponents. Logarithms are

exponents.

b. 3 = log_b 64 means b3 = 64.

Logarithms are exponents. Logarithms are

22

y = log₂ (

) y2 =

Examples: Write the equivalent exponential equation

and solve for y

.

1

=

5

y

y = log₅1

16

=

4y

y = log₄16

16

=

2y

y = log₂16

Solution Equivalent Exponential Equation Logarithmic Equation 16 = 24

 y = 4

=

2-1 _{y = –1}

16

=

42

 y = 2

1

=

50

 y = 0

23

x y

Graph f(x) = log₂x

Since the logarithm function is the inverse of the exponential function of the same base, its graph is the reflection of the exponential function in the line y = x

. 8 3 4 2 2 1 1 0 – 1 – 2 2x x

y = log₂ x y = x y = 2x

( 1 , 0 ) x-intercept horizontal asymptote y = 0

vertical asymptote

x = 0

4 1

24

Example: Graph the common logarithm function f(x) = log₁₀ x

. by calculator 1 0.602 0.301 0 – 1 – 2

f(x) = log₁₀ x

10 4 2 1 x y x 5 – 5

f(x) = log₁₀ x

Definitions of Trigonometric Functions For a Right Triangle

Fundamental Trigonometric Identities

• _{Quotient Identities}

• _{Reciprocal Identities}

• _{Pythagorean Identities}

Example : Find the sine, the cosine, and the tangent of angle A. Give a fraction and decimal answer (round to 4 places).

9

6

10.8

A

Now, figure out your ratios.

hyp

opp

A



sin

8

.

10

9





.

8333

hyp

adj

A



cos

8

.

10

6





.

5556

adj

opp

A



tan

6

9

Graphs of Trigonometric Functions

Sin x Cos x

Ex: 1

Figure out which ratio to use. Find x. Round to the

nearest tenth .

20

m

x

tan

20

5

5

)

Shrink yourself down and stand where the angle is

.

Now, figure out which trig ratio you have and set up the problem

.



55

( )

20

55

tan



x

m

6

.

28



x

( )

55



x

Ex: 2

Find the missing side. Round to the nearest

tenth .

80

ft

x

tan

80

7

2

=



(

)

)

Shrink yourself down and stand where the angle is

.

Now, figure out which trig ratio you have and set up the problem

.



72

( )

x

80

72

tan



ft

26



x

( )

72

80

tan



x

( )

(

tan

72

)

80



Example

32

Solving A Right Triangle Given an Angle and a Side

Solve the right triangle.

The third angle is 60, the complement of 30. Use the values of the

trigonometric functions of 30o_.

10 ₆₀○

30○

5 30○

5

Since = sin 30 = = , it follows that hyp = 10. hyp

opp

hyp

5

To get the last side, note that

= cos 30 = ;

therefore, adj = 5 3

33

Example :

A bridge is to be constructed across a small river from post A to post B. A surveyor walks 100 feet due south of post A. She sights on both posts from this location and finds that the angle between the posts is 73. Find the distance across the river from post A to post B.

It follows that x = 327.

The distance across the river from post A to post B is 327 feet. Use a calculator to find

tan 73o _{= 3.27.}

Post B Post A

100 ft.

x

73○

3.27 = tan 73= = adj opp

100

4. A closed circuit television camera is mounted on a wall 7.4 feet above a security desk in an office building. It is used to view an entrance door 9.3 feet from the desk. Find the angle of depression from the camera lens to the entrance

door.

7.4

feet

?

?

9.3

7. A train must climb at a constant gradient of 5.5 m for every 200 m of track. Find the angle of incline.

200 m

5.5

m

36

Inverse Trigonometric Functions on a Calculator

Example:

Find the acute angle  for which cos  = 0.25. Calculator keystrokes: (SHIFT) cos1 0.25 =

Inverse functions are often accessed by using a key that maybe be labeled SHIFT, INV, or 2nd. Check the

manual for your calculator.

Labels for sin1_{, cos}1_{, and tan}1 _{are usually written} above the sin, cos, and tan keys.

37

Example: Solving a Right Triangle Given Two Sides

hyp2 _{= 6}2 _{+ 5}2 _{= 61}

Subtract to calculate the third angle:

90 39.805571 = 50.194428. Solve for the hypotenuse:

Solve for :

Calculator Keystrokes: (SHIFT) tan1 _{( 5}  _{6 )}

Display: 39.805571 hyp = = 7.8102496

39.8○

6

5

50.2○

Solve the right triangle shown.

θ

5

38

Example 2:

A ship at sea is sighted by an observer at the edge of a cliff 42 m high. The angle of depression to the ship is 16. What is the distance from the ship to the base of the cliff?

The ship is 146 m from the base of the cliff.

line of sight

angle of depression

horizontal observer

ship cliff

42 m

16○

16○

d

d = = 146.47.



16 tan