Math 1111

Final Exam Review

1. Identify the type of function.

f(x) = 56

4

2

-2

-5 5

Constant

-2 2

2

1

-1

2. Identify the type of function.

4)(

2 x

xxf

Quotient of two polynomials.

RationalFunction

3. Identify the type of function.

1572)( 3 xxxf

PolynomialFunction

Highest Exponent 3

-10 -5 5 10

10

8

6

4

2

-2

-4

-6

-8

-10

-12

-14

-16

-18

-20

-22

4. Identify the type of function.

2847)( 2 xxxf

QuadraticFunction

Highest Exponent 2

-10 10

55

50

45

40

35

30

25

20

15

10

5

5. Identify the type of function.

23)( 1 xxf

ExponentialFunction

Variable in the Exponent

8

6

4

2

-2

-5 5

6. Identify the type of function.

6

4

2

-2

-5 5

Linear

3)( xxf

7. $1500 is invested at a rate of 6¼% compounded continuously. What is the balance at the end of two years?

• A = balance at the end of investment period

• P = Principal (Money invested)

• r = rate = 0.0625• t = time in years = 2

rtPeA)2)(0625.0(1500eA

)125.0(1500eA)13315.1(1500A

725.1699A $1699.73

Continuous Compounding

8. Evaluate the expression.

• Round your answer to 3 decimal places.

37.4 e

≈ 26.565

10. Evaluate the expression.

• When t = 15.

• Round your answer to 2 decimal places.

te 076.0300

≈ 95.95

)15(076.0300 e14.1300 e

11. What transformations are used to create the “child”?

xxf 2)( Parent

52)( xxgChild

6

4

2

-2

-5 5

4

2

-2

-5 5

1. Reflection of x-axis.

2. Vertical shift of 5 units.

12. What transformations are used to create the “child”?

xxf 2)( Parent

12)( 4 xxgChild

6

4

2

-2

-5 5

6

4

2

51. Reflection of y-axis.

2. Vertical shift up of 1 unit.

3. Horizontal shift right of 4 units.

13. $2100 is invested at a rate of 7% compounded monthly. What is the balance at the end of 10 years?

• A = balance at the end of investment period

• P = Principal (Money invested)

• r = rate = 0.07• n = number of periods

per year = 12• t = number of years

nt

n

rPA

1

$4220.29

)10(12

12

07.012100

A

12000583.12100A 00966.22100A

n Compoundings per year

15a. Given f(x) = x3 – 2x2 – 21x – 18answer the following questions.

• What is the degree of the polynomial?

3• According to the Fundamental Theorem of Algebra, how many zeros will

this polynomial have?

3• Use Descartes’ Rule of Signs to determine the number of possible positive

real zeros.

1 possible positive zero, one sign variation• Use Descartes’ Rule of Signs to determine the number of possible negative

real zeros.

f(-x) = (-x)3 – 2(-x)2 -21(-x) – 18 = -x3 – 2x2 + 21x - 18 2 or 0 possible negative zeros, two sign variations

15b. Given f(x) = x3 – 2x2 – 21x – 18answer the following questions.

• Use the Rational Root Test to list All possible rational zeros of the polynomial.

• Where p is a factor of -18 and q is a factor of 1

p

q

1; 2; 3; 6; 9; 18p

q

Given all of these candidates, how can you tell which is the actual root?

f(x) = x3 – 2x2 – 21x – 18

1; 2; 3; 6; 9; 18p

q

Synthetic Division

Graphing calculator

15c. Given f(x) = x3 – 2x2 – 21x – 18answer the following questions.

• Use the synthetic division to find all zeros of the polynomial.

1 -2 -21 -18-1

1

-1

-3

3

-18

18

0

)183))(1(( 2 xxx

)183)(1( 2 xxx)3)(6)(1( xxx

}6,1,3{ xEach zero has a multiplicity of one.

16a. Given f(x) = x4 + x3 – 11x2 + x – 12answer the following questions.

• What is the degree of the polynomial?

4• According to the Fundamental Theorem of Algebra, how many zeros will

this polynomial have?

4• Use Descartes’ Rule of Signs to determine the number of possible positive

real zeros.3 or 1 possible positive zero, three sign variations

• Use Descartes’ Rule of Signs to determine the number of possible negative real zeros.

f(-x) = (-x)4 + (-x)3 – 11(-x)2 + (-x) – 12 = x4 – x3 – 11x2 – x – 12

1 possible negative zero, one sign variation

16b. Given f(x) = x4 + x3 – 11x2 + x – 12answer the following questions.

• Use the Rational Root Test to list All possible rational zeros of the polynomial.

• Where p is a factor of -12 and q is a factor of 1

p

q

12;6;4;3;2;1 q

p

16c. Given f(x) = x4 + x3 – 11x2 + x – 12answer the following questions.1 1 -11 1 -123

1

3

4

12

1

3

4

0)44)(3( 23 xxxx

{ 4, 3, , }x i i Extract

the root.

12

0

0)]4(1)4()[3( 2 xxxx

0)1)(4)(3( 2 xxx

Factor by grouping

2( 1) 0x 2 1x

1x i

Each zero has a multiplicity of one.

17a. Find and plot the y-intercept. Write as an ordered pair.

5( )

3

xf x

x

Set x = 0

0 5

0 3y

5

3y

50,3

y-intercept

17b. Find and plot the zeros. Write as an ordered pair.

5( )

3

xf x

x

Set f(x) = y = 0

50

3

x

x

5x

5, 0

-5 5 10

8

6

4

2

-2

-4

-6

-8

(4, 9)17. Vertical Asymptote

50,3

5, 0

5( )

3

xf x

x

Set the denominator = 0

3 0x 3x

x = 3

17. Horizontal Asymptote

50,3

5, 0

-5 5 10

8

6

4

2

-2

-4

-6

-8

(4, 9)5

( )3

xf x

x

Since the degree of the two polynomials is the same find the ratio of the leading coefficient of the numerator divided by the leading coefficient of the denominator.

n

m

ayb

x = 3

1

1

1y y = 1

-5 5 10

8

6

4

2

-2

-4

-6

-8

(4, 9)17. Find function values to help you graph.

50,3

5, 0

5( )

3

xf x

x

x = 3

1

9( 6)f

y = 1

16,9

( 1)f 1

1, 1

(2)f 7

2, 7

(4)f 9

4,9

(6)f 113

116,3

-5 5 10

8

6

4

2

-2

-4

-6

-8

(4, 9)

50,3

5, 0

5( )

3

xf x

x

x = 3

y = 1

16,9

1, 1

2, 7

4,9

116,3

17. Find function values to help you graph.

18a. Find and plot the y-intercept. Write as an ordered pair.

2 2( )

3

x xf x

x

Set x = 0

20 0 2

0 3y

2

3y

2

0,3

y-intercept

18b. Find and plot the zeros. Write as an ordered pairs.

2 2( )

3

x xf x

x

Set f(x) = y = 0

2 20

3

x x

x

20 2x x

2, 0 (1,0)0 ( 2)( 1)x x

{ 2,1}x

-5 5 10

8

6

4

2

-2

-4

-6

-8

(4, 9)18. Vertical Asymptote

20,3

2, 0

2 2( )

3

x xf x

x

Set the denominator = 0

3 0x

3x x =

3

1, 0

18. Horizontal Asymptote 2 2

( )3

x xf x

x

Since the degree of the numerator is greater than the degree of the denominator there is no horizontal asymptote.

Slant Asymptote

y = x + 4

-20 -15 -10 -5 5 10 15 20

22

20

18

16

14

12

10

8

6

4

2

-2

-4

-6

18. Find function values to help you graph.

( 3)f 2

3

2 2( )

3

x xf x

x

2

0,3

2, 0

x = 3

1, 0

( 1)f 1

2(2)f 4(4)f 18

(6)f 13.3

y =x +

4

19a. Find and plot the y-intercept. Write as an ordered pair.

Set x = 0

y-intercept

43)(

2

xx

xxf

4)0(3)0(

02

y

4

0

y)0,0(

19b. Find and plot the zeros. Write as an ordered pairs.

Set f(x) = y = 0

x0

)0,0(

The Only Zero

43)(

2

xx

xxf

-6 -4 -2 2 4

5

4

3

2

1

-1

-2

-3

-4

-5

19. Vertical Asymptote Set the denominator = 0

x = -4

)0,0(

43)(

2

xx

xxf

430 2 xx)1)(4(0 xx

}1,4{x

x = 1

19. Horizontal Asymptote

Since the degree of the numerator is less than the degree of the denominator the horizontal asymptote is y = 0.

43)(

2

xx

xxf

-6 -4 -2 2 4

5

4

3

2

1

-1

-2

-3

-4

-5

19. Find function values to help you graph.

x = -4

)0,0(

)6(fx =

1

43)(

2

xx

xxf

y = 0

)3(f4

37

3

)1(f6

1

)2(f3

1

)4(f6

1

20. Find the Domain:

1)(

2

x

xxf 01x 1x

}1,|{ xxx

21. Find the Domain:

10

4)(

2

x

xxf 0102x

102 x

}10,|{ xxx

10x

22. Match the function with the graph:

A) f(x) = 4x – 5

B) f(x) = 4x + 5

C) f(x) = 4-x + 5

D) f(x) = 4-x – 5 -4 -2 2 4 6

8

7

6

5

4

3

2

1

-1

-2

23. Match the function with the graph:

A) f(x) = 3x-1

B) f(x) = 3x – 1

C) f(x) = 31- x

D) f(x) = 3-x – 1 -4 -2 2 4 6

8

7

6

5

4

3

2

1

-1

-2

24. Match the function with the graph:

A) f(x) = 5x+1 – 2

B) f(x) = 5x+2 – 1

C) f(x) = 5x-1+ 2

D) f(x) = 5x-2 + 1 -4 -2 2 4

8

7

6

5

4

3

2

1

-1

-2