Honors PreCalculus Summer Work 2015

• due date: third day of class • estimated time: 10 hours (for planning purposes only; work until you finish)

Dear Honors PreCalculus Students,

This assignment is designed for you to review all of your algebra skills and make sure you are well prepared for the start of Honors PreCalculus in the fall. The assignment is in two parts: written problems in this packet, and problems on iXL.com. See the next page for detailed instructions. We anticipate that the work will take about 10 hours to complete.

Both the written work and the online work are due on the third day of class. At that time you will have a short quiz with problems taken directly from both this packet and the iXL sections listed on the next page. A portion of the quiz grade will be based on the completeness of your summer work.

We are really looking forward to the upcoming school year and hope you are, too! Please email any of us if you have questions.

Barbara.Filler@stewardschool.org Todd.Serr@stewardschool.org

Jennifer.Maitland@stewardschool.org Karen.Hudson@stewardschool.org

Page 1 of 19

Assessment of Summer Work 1. On the third day of class, you will submit your written work, and your teacher will check

your online work. 2. You will have a quiz made up of problems taken directly from the on-line and written work. 3. Your written and online work will be evaluated for completeness; that grade will form part

of the quiz grade.

On-Line Work 1. Log in to iXL.com with the username and password you were given in your math class.

Students new to Steward: your parents will receive an email with your username and password.

2. Work the problems listed below, until you earn a score of at least 60% for each section. If you find that the name of the section does not match the numbers below, go by the

section name. 3. If you are having difficulty of any sort with iXL, reach out to a teacher at one of the email

addresses above.

1. click this tab 2. find these sections 3. work these problem sets

Algebra II

Complex numbers H.1 - Introduction to complex numbers H.3 - Complex conjugates

Quadratic functions I. 5 - Match quadratic functions and graphs I.9 - Complete the square

Logarithms R.1 - Convert between exponential and logarithmic form: rational bases

Exponential and logarithmic functions

S.1 - Domain and range of exponential and logarithmic functions

Written Work 1. Complete each problem on the following pages. 2. Do the problems on separate paper. Do the problems in order and number them clearly.

Exception: print the graphing problems (p.8-9)rather than drawing your own graphs.

3. Please show all of your work as was required of you throughout the entire school year. Remember: no work, no credit.

Page 2 of 19

Section 1: Radicals

Simplifying radicals involves two concerns. 1. No radicand (expression left under a radical) has a perfect square/cubic/etc factor. 2. There is no radical in a denominator (you must rationalize).

Examples: Simplify.

1. √12 = √4 ∙ 3 = 2√3

2. �95

= 3√5

= 3√5∙ √5√5

= 3√55

3. 1

−2−√6= 1

−2−√6∙ −2+√6−2+√6

= −2+√64−6

= −2+√6−2

= 2−√62

Some things to think about in the above examples:

1. You can use factor trees to help you if you want. 2. Examples 2 and 3 use the trick of creatively multiplying by 1. 3. Example 3 uses the concept of a conjugate. A conjugate occurs with binomials when the sign

between them changes.

Simplify.

1) √32

2) �(3𝑤𝑤)6

3) √−1285

4) �27𝑎𝑎4𝑏𝑏93

5) �1225

6) √363 ∗ √603

7) �√5 −√3�(√10 + √5)

8) �43

9) 1√2

10) 2

2 − √7

This word should match the root value of the radical. See problem 3 below.

Page 3 of 19

Section 2: Imaginary and Complex Numbers

Imaginary numbers allow us to “imagine” what an even root of a negative number might be, if it were allowed. Recall the imaginary number “i”: i = √−1 and 𝑖𝑖2 = √−1 ∗ √−1 = −1 Complex numbers are expressions that contain both a real and imaginary parts: a + bi Examples:

1. √−10 = √−1 ∗ 10 = 𝑖𝑖√10 2. 5𝑖𝑖(4 − 2𝑖𝑖) = 20𝑖𝑖 − 10𝑖𝑖2 = 20𝑖𝑖 − 10(−1) = 20𝑖𝑖 + 1

3. 5+𝑖𝑖3𝑖𝑖

= 5+𝑖𝑖3𝑖𝑖• 𝑖𝑖𝑖𝑖

= 5𝑖𝑖+𝑖𝑖2

3𝑖𝑖2= 5𝑖𝑖−1

−3= 1−5𝑖𝑖

3

Some things to think about in the above examples:

1. Treat “i” like any other variable when you +,−,×,÷ 2. BUT remember 𝑖𝑖2 = −1 3. Since i = √−1 it cannot be in a denominator

Simplify.

11) √−64

12) 2√−32

13) 3𝑖𝑖(2 + 4𝑖𝑖) − 6(7𝑖𝑖 + 1)

14) (3 − 6𝑖𝑖)2

15) (x + 2i)(x – 2i)

16) 3 − 4𝑖𝑖−2𝑖𝑖

Page 4 of 19

Section 3: Geometry

Recall these relationships for right triangles: • Pythagorean Theorem: 𝑎𝑎2 = 𝑏𝑏2 + 𝑐𝑐2 • 30° − 60° − 90° right triangles • 45° − 45° − 90° right triangles

Find the value of x.

17)

18)

Page 5 of 19

19)

Find the value of x and y. The triangles are 𝟒𝟒𝟒𝟒° − 𝟒𝟒𝟒𝟒° − 𝟗𝟗𝟗𝟗° .

20)

21)

Find the value of x and y. The triangles are 𝟑𝟑𝟗𝟗° − 𝟔𝟔𝟗𝟗° − 𝟗𝟗𝟗𝟗°.

22)

23)

Page 6 of 19

Section 4: Linear Equations

Slope Intercept Form: y = mx + b

Point-Slope Form: 𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1)

Standard Form: Ax + By = C

Vertical line: 𝑥𝑥 = 𝑎𝑎 (slope is undefined)

Horizontal line: 𝑦𝑦 = 𝑏𝑏 (slope = 0)

Slope: 𝑦𝑦2−𝑦𝑦1𝑥𝑥2−𝑥𝑥1

and −𝐵𝐵𝐴𝐴

24) State the slope and y-intercept of 3𝑥𝑥 − 2𝑦𝑦 = 10.

25) State the x- and y-intercepts of 𝑥𝑥 − 4𝑦𝑦 = 5.

26) Rewrite in standard form: 𝑦𝑦 = −23𝑥𝑥 + 1.

27) Write in slope-intercept form the equation of the line passing through (-2, 5) with a slope of -6.

28) Write in point slope form the equation of the line passing through (8, -13) and (4, 0).

29) Write in standard for the equation of the line with an x-intercept of 3 and a y-intercept of -9.

30) Write the equation of the line passing through (-1, 2) with an undefined slope.

Page 7 of 19

Section 5: Graphing

Print this section. Graph the equations and systems.

31) 3𝑥𝑥 − 4𝑦𝑦 = 12 32) �2𝑥𝑥 + 𝑦𝑦 = 4𝑥𝑥 − 𝑦𝑦 = 2

33) 𝑦𝑦 < −4𝑥𝑥 − 2 34) 𝑦𝑦 = |𝑥𝑥 + 1|

35) 𝑦𝑦 > |𝑥𝑥| − 1 36) 𝑦𝑦 = (𝑥𝑥)2 + 5

Page 8 of 19

37) 𝑦𝑦 = 𝑒𝑒𝑥𝑥 38) 𝑥𝑥 = −2

39) 𝑦𝑦 = (𝑥𝑥 + 3)2

Page 9 of 19

Section 6: Systems of Equations

Systems of equations occur when two or more equations need to work together. They can be solved in four ways: graphically, by substitution, by elimination, or with matrices.

Consider the following system: 2𝑥𝑥 + 𝑦𝑦 = 5 𝑥𝑥 − 4𝑦𝑦 = 7

Three of the above four ways of solving this system are illustrated below.

Graphically Substitution Elimination

Graph both lines. The point where the lines intersect is the solution.

If the lines are parallel, there

will be no solution. If the lines coincide, there will

be an infinite number of solutions.

Express one variable in terms of the other and substitute.

2𝑥𝑥 + 𝑦𝑦 = 5 𝑦𝑦 = 5 − 2𝑥𝑥

𝑥𝑥 − 4(5 − 2𝑥𝑥) = 7 𝑥𝑥 − 20 + 8𝑥𝑥 = 7 9𝑥𝑥 = 27 𝑥𝑥 = 3

Then use 𝑥𝑥 = 3 to find y:

2(3) + 𝑦𝑦 = 5 𝑦𝑦 = −1 Solution: (3,−1)

Eliminate one variable by adding equations. Use multiplication if needed.

4(2𝑥𝑥 + 𝑦𝑦 = 5) ⇒ 8𝑥𝑥 + 4𝑦𝑦 = 20

8𝑥𝑥 + 4𝑦𝑦 = 20 + 𝑥𝑥 − 4𝑦𝑦 = 7 9𝑥𝑥 = 27 𝑥𝑥 = 3

Then use 𝑥𝑥 = 3 to find y.

Solution: (3,−1)

Solve each system of equations. Use each method at least once.

40) � 2𝑥𝑥 + 𝑦𝑦 = 43𝑥𝑥 + 2𝑦𝑦 = 1

41) � 3𝑥𝑥 + 𝑦𝑦 = 62𝑥𝑥 − 2𝑦𝑦 = 4

42) �2𝑥𝑥 − 5𝑦𝑦 = 136𝑥𝑥 + 3𝑦𝑦 = 10

43) �5𝑥𝑥 − 3𝑦𝑦 = 1510𝑥𝑥 − 6𝑦𝑦 = 5

2

2

4

5

(3, -1)

Page 10 of 19

Section 7: Exponents

Product Rules

To multiply when two bases are the same, add the exponents.

𝑎𝑎𝑛𝑛𝑎𝑎𝑚𝑚 = 𝑎𝑎𝑛𝑛+𝑚𝑚 32 ∙ 33 = 32+3 = 35 = 243

To multiply when two bases are different and the exponents are the same, multiply the bases first.

𝑎𝑎𝑛𝑛𝑏𝑏𝑛𝑛 = (𝑎𝑎𝑏𝑏)𝑛𝑛 32 ∙ 42 = (3 ∙ 4)2 = 122 = 144

Quotient Rules

To divide when two bases are the same, subtract the denominator’s exponent from the numerator’s exponent.

𝑎𝑎𝑛𝑛

𝑎𝑎𝑚𝑚 = 𝑎𝑎𝑛𝑛−𝑚𝑚

34

32 = 34−2 = 32 = 9

To divide when two bases are different and the exponents are the same, divide the bases first.

𝑎𝑎𝑛𝑛

𝑏𝑏𝑛𝑛 = �

𝑎𝑎𝑏𝑏�𝑛𝑛

63

33 = �63�3

= 23 = 8

Power Rule

To raise a power to a power, multiply the exponents.

(𝑎𝑎𝑛𝑛)𝑚𝑚 = 𝑎𝑎𝑛𝑛𝑚𝑚 (22)3 = 22∙3 = 26 = 64

Zero Rule A base (other than zero) raised to the zero power is equal to 1.

𝑎𝑎0 = 1 40 = 1

One Rules

A base raised to the power of one is equal to itself.

𝑎𝑎1 = 𝑎𝑎 41 = 4

1 raised to any integer power is equal to 1.

1𝑛𝑛 = 1 13 = 1

Negative Exponents Rules

A base raised to a negative exponent is equal to 1 divided by the base raised to the corresponding positive exponent.

𝑎𝑎−𝑛𝑛 = 1𝑎𝑎𝑛𝑛

4−2 = 1

42=

116

Visit rapidtables.com for more exponent rules.

Page 11 of 19

Simplify. Use only positive exponents when needed.

44) 8𝑒𝑒0

45) 5𝑎𝑎𝑎𝑎−1

46)

2𝑒𝑒𝑓𝑓−1

𝑒𝑒 − 1

47) (𝑛𝑛3𝑝𝑝−2)4

(𝑛𝑛𝑝𝑝)−2

48) 3𝑥𝑥2 ∙ 2𝑥𝑥

49) (ℎ3)5

50) (−𝑏𝑏3𝑐𝑐4)5

51) 4𝑚𝑚(3𝑎𝑎2𝑚𝑚)

Page 12 of 19

Section 8: Polynomials

So far with polynomials you have learned to do the following: • Combine like terms • Distribute • Multiply (if two binomials, then FOIL) • Factor (including ONGF, swing method, and grouping) • Solve (set = 0 and use Multiplicative Property of Zero; may have to use quadratic equation) • Graph (parabola, including calculating and interpreting discriminant)

Combine like terms.

52) 2 2(6 5 7) (4 3 2)x x x x− + + − +

53) 3 2 3 2(14 2 5 ) (11 5 10)a a a a a− + − − +

Find each product. Simplify the answer.

54) 23 (2 7 1)x x x− +

55) 2 2( 5) ( 2)x x x x− + +

56) (4 1)(5 8)m m− +

57) 2(4 3)x +

58) ( 6 )( 6 )x y x y+ −

59) 2( 3)( 5 7)x x x+ − +

Write a polynomial expression to represent the area of the rectangle below.

60)

5𝑎𝑎 − 6

3𝑎𝑎 + 4

Factor completely.

61) 4 225 15a a+

62) 327 3p p−

63) 2 2 63x x− −

Page 13 of 19

64) 23 8 4a a+ +

65) 26 5x x− −

66) 9𝑐𝑐2 + 30𝑐𝑐 + 25

67) −16𝑛𝑛2 + 4

68) 2𝑚𝑚𝑛𝑛 − 2𝑚𝑚𝑚𝑚 + 2𝑠𝑠𝑛𝑛 − 2𝑠𝑠𝑚𝑚

Solve each equation.

69) 2 3 28 0x x+ − =

70) 22 9 4 0x x+ + =

71) 236 49x =

Page 14 of 19

Section 9: Polynomials, continued: Using Discriminants

The quadratic formula allows you to solve any quadratic for all its real and imaginary roots.

𝑥𝑥 = −𝑏𝑏±√𝑏𝑏2−4𝑎𝑎𝑎𝑎2𝑎𝑎

The number under the radical in the quadratic formula (b2 − 4ac) is called the discriminant (D). The value of the discriminant can tell you what kinds of roots you will have.

If b2 − 4ac > 0, you will have two real roots.

The graph touches the x-axis twice.

If b2 − 4ac = 0, you will have one real root.

The graph touches the x-axis once.

If b2 − 4ac < 0, you will have two imaginary roots.

The graph does not touch the x-axis.

For each quadratic equation on the next page, (a) find the value of the discrimant (b) describe the nature of the roots (c) solve the equation.

Use the following as an example.

ex. x2 + 2x + 3 = 0

(a) First, determine the values for a, b, and c: a = 1, b = 2, c = 3 Then, find the value of the discriminant (D).

𝐷𝐷 = 22 − 4 ∙ 1 ∙ 3 𝐷𝐷 = 4 − 8 𝑫𝑫 = −𝟖𝟖

(b) Since D < 0, there are two imaginary roots.

(c) Use the quadratic formula to solve the equation.

𝑥𝑥 =−2 ± �22 − 4 ∙ 1 ∙ 3

2 ∙ 1 =−2 ± 2𝑖𝑖√2

2 = −𝟏𝟏± 𝒊𝒊�𝟐𝟐

4

2

2

2

2

4

6

4

2

Page 15 of 19

72) 𝑥𝑥2 − 9𝑥𝑥 + 14 = 0

(a) value of the discriminant

(b) nature of the roots

(c) roots

73) 5𝑥𝑥2 − 2𝑥𝑥 + 4 = 0

(a) value of the discriminant

(b) nature of the roots

(c) roots

Page 16 of 19

Section 10: Using Function Notation

Evaluating a function at a particular value is done by plugging that value in.

example: For 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥2 − 5 , f(-3) = 2(-3)2 – 5 = 2(9)-5 = 13

Finding the inverse of a function involves switching x for y and solving again for y.

The symbol for the inverse of a function is 𝑓𝑓−1(𝑥𝑥) .

example: The inverse of 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥2 − 5 is found by solving 𝑥𝑥 = 2𝑦𝑦2 − 5 for y.

𝑥𝑥 + 5 = 2𝑦𝑦2

𝑥𝑥+52

= 𝑦𝑦2

±�𝑥𝑥+52

= 𝑦𝑦 or 𝑓𝑓−1(𝑥𝑥) = ±�𝑥𝑥+52

Composite functions are written as (𝑓𝑓𝑓𝑓𝑓𝑓)(𝑥𝑥) or 𝑓𝑓(𝑓𝑓(𝑥𝑥)) and are read “f of g of x.”

example: Given 𝑓𝑓(𝑥𝑥) = 2 − 𝑥𝑥 and 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 1,

𝑓𝑓�𝑓𝑓(𝑥𝑥)� = 2 − (𝑥𝑥2 − 1) = 3 − 𝑥𝑥2

The composite of a function and its inverse results in “x”.

𝑓𝑓�𝑓𝑓−1(𝑥𝑥)� = 𝑥𝑥

Use the following function definitions for problems 74-81.

𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝟐𝟐 − 𝟔𝟔𝒙𝒙 + 𝟏𝟏𝟐𝟐 𝒈𝒈(𝒙𝒙) = 𝟔𝟔𝒙𝒙 − 𝟕𝟕 𝒉𝒉(𝒙𝒙) = −𝟏𝟏𝟑𝟑𝒙𝒙

74) f(-4) 75) g(x + h)

76) f(x + 2) 77) f[g(x)]

78) h[f(3)] 79) h[g(x)]

80) g- 1(x) 81) h- 1 (x)

Page 17 of 19

Section 11: Algebraic Rational Expressions

Rational expressions involve fractions. To multiply or divide, factor the numerator and denominator completely. Cancel common factors if they appear in both the numerator and donominator. Remember to change division to multiplication by using the reciprocal of the second fraction. 𝑥𝑥2 + 10𝑥𝑥 + 21

5 − 4𝑥𝑥 − 𝑥𝑥2 ÷𝑥𝑥3 + 4𝑥𝑥 − 21𝑥𝑥𝑥𝑥2 + 2𝑥𝑥 − 15 =

𝑥𝑥2 + 10𝑥𝑥 + 215 − 4𝑥𝑥 − 𝑥𝑥2 ●

𝑥𝑥2 + 2𝑥𝑥 − 15𝑥𝑥3 + 4𝑥𝑥 − 21𝑥𝑥 =

(𝑥𝑥 + 7)(𝑥𝑥 + 3)−1(𝑥𝑥 + 5)(𝑥𝑥 − 1)●

(𝑥𝑥 + 5)(𝑥𝑥 − 3)𝑥𝑥(𝑥𝑥 − 3)(𝑥𝑥 + 7) =

𝑥𝑥 + 3−𝑥𝑥(𝑥𝑥 − 1)

To add or subtract you have to find a least common denominator.

3𝑥𝑥+1𝑥𝑥2+2𝑥𝑥

+ 5𝑥𝑥−42𝑥𝑥+4

= 3𝑥𝑥+1𝑥𝑥(𝑥𝑥+2)

+ 5𝑥𝑥−42(𝑥𝑥+2)

The common denominator is 2x(x+2). Multiply the first fraction by 2

2 and the second fraction by 𝑥𝑥

𝑥𝑥.

2(3𝑥𝑥+1)2𝑥𝑥(𝑥𝑥+2)

+ 𝑥𝑥(5𝑥𝑥−4)𝑥𝑥2(𝑥𝑥+2)

= 6𝑥𝑥+2+5𝑥𝑥2−4𝑥𝑥2𝑥𝑥(𝑥𝑥+2)

= 5𝑥𝑥2+2𝑥𝑥+22𝑥𝑥(𝑥𝑥+2)

Complex fractions can be simplified by rewriting the numerator as one fraction and the denominator as one fraction, then treating it as a division problem.

1+1𝑥𝑥2𝑥𝑥−1

=𝑥𝑥+1𝑥𝑥

2−𝑥𝑥𝑥𝑥

= 𝑥𝑥+1𝑥𝑥∙ 𝑥𝑥2−𝑥𝑥

= 𝑥𝑥+12−𝑥𝑥

Simplify.

82) 5𝑚𝑚3 + 𝑚𝑚2 −𝑚𝑚

3𝑚𝑚 83) 𝑥𝑥

2 − 25𝑥𝑥2 + 5𝑥𝑥

84) 10𝑎𝑎5

21𝑎𝑎2●

3𝑎𝑎5𝑎𝑎3

85) 𝑥𝑥2 − 5𝑥𝑥 + 6𝑥𝑥 + 4

●3𝑥𝑥 + 12𝑥𝑥 − 2

86) 6ℎ − 95ℎ + 1

÷6 − 13ℎ + 6ℎ2

15ℎ2 − 7ℎ − 2 87)

2𝑥𝑥5−𝑥𝑥3

88) 𝑏𝑏 − 𝑎𝑎𝑎𝑎2𝑏𝑏

+𝑎𝑎 + 𝑏𝑏𝑎𝑎𝑏𝑏2

89) 𝑥𝑥2 − 1𝑥𝑥2 + 𝑥𝑥

+4𝑥𝑥 + 43𝑥𝑥 − 3

90) 12 − 𝑥𝑥

𝑥𝑥 + 12

91) 𝑥𝑥 − 2

𝑥𝑥1𝑥𝑥2 − 𝑥𝑥

Page 18 of 19

Section 12: Logarithmic and Exponential Functions

Log Rules

1. log𝑏𝑏 1 = 0

2. log𝑏𝑏 𝑏𝑏 = 1

3. log𝑏𝑏 𝑏𝑏𝑥𝑥 = 𝑥𝑥

4. log𝑏𝑏 𝑥𝑥 = log𝑏𝑏 𝑦𝑦, if and only if 𝑥𝑥 = 𝑦𝑦

5. log𝑏𝑏(𝑢𝑢𝑢𝑢) = log𝑏𝑏 𝑢𝑢 + log𝑏𝑏 𝑢𝑢

6. log𝑏𝑏 �𝑢𝑢𝑣𝑣� = log𝑏𝑏 𝑢𝑢 − log𝑏𝑏 𝑢𝑢

7. log𝑏𝑏 𝑢𝑢𝑛𝑛 = 𝑛𝑛 log𝑏𝑏 𝑢𝑢

To solve an exponential or logarithmic equation, isolate the function and then change forms.

Example Solve 2 log3 𝑥𝑥 = 6.

• First, isolate the log: o log3 𝑥𝑥 = 3

• Then change to exponential form: o 33 = 𝑥𝑥

• Then evaluate: o 27 = 𝑥𝑥

Expand and simplify, if possible.

92) log4 16 93) log5 √25𝑥𝑥

94) ln3𝑥𝑥2

𝑦𝑦

Condense and simplify, if possible.

95) log7 𝑥𝑥 + 3 log7 𝑦𝑦 − 5 log7 49 96) ln𝑥𝑥 −12

ln 4

Solve, rounding decimals to three places as needed.

97) 3𝑥𝑥 = 10 98) 5 + log8 𝑥𝑥 = 12

99) 𝑒𝑒2𝑥𝑥 = 4 100) 200𝑒𝑒−.03𝑡𝑡 = 100

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