Dear Future Pre-Calculus Students,

Dear Future Pre-Calculus Students, Congratulations on your academic achievements thus far. You have proven your academic worth in Algebra II (CC), but the challenges are not over yet! Not to worry; this summer conditioning packet was created to get you ready for those challenges ahead. The following packet contains topics and definitions that you will be required to know in order to succeed in Pre-Calculus. You are advised to be familiar with each of the concepts and to complete the included problems. All of these topics were discussed in either Algebra I, II or Geometry and will be used frequently throughout the year. Ample writing space has been provided so that you may be able to show ALL WORK for ALL Problems. Credit will not be given to those answers lacking written work. This packet will be due the first day of class (Wednesday, September 5, 018). It will be discussed Thursday, September 6 th and you will have a test on Friday, September 7 th. For those of you who might possibly misplace this packet, you will find a fresh summer condition packet in the main office during the summer OR a link to it on my school website under the Pre-Calculus tab. If you find gaps in your knowledge I have provided some websites that you m ay find useful, listed below. You may also want to check other resources available online. Sincerely, Mrs. Baccaro

Pre-Calculus Summer Assignment 018-019 Academic Year Name: Assignment Due: Wednesday, September 5, 018 Assignment Test: Friday, September 7, 018

Summer Review Packet for Students Entering Pre-Calculus Radicals: To simplify means that 1) no radicand has a perfect square factor and ) there is no radical in the denominator (rationalize). Recall the Product Property ab = a! b and the Quotient Property Examples: Simplify 4 = 4! 6 find a perfect square factor = 6 simplify a = b a b Simplify 7 7 =! split apart, then multiply both the numerator and the denominator by 14 14 = = multiply straight across and simplify 4 If the denominator contains terms multiply the numerator and the denominator by the conjugate of the denominator The conjugate of 3 + is 3! (the sign changes between the terms). Simplify each of the following. 1. 3. 8 ( x ) 3. 3! 64 4. 8 49m n 5. 11 9 6. ( 5! 6)( 5 + ) Rationalize. 1 7. 8. 3! 5

Complex Numbers: Form of complex number - a + bi Where a is the real is part and bi is the imaginary part Always make these substitutions!1 = i and i =!1 To simplify: pull out the! 1 before performing any operation Example: " 5 = " 1! 5 Pull out! 1 Example: ( i 5) = i 5! i 5 List twice = i 5 Make substitution = i 5 Simplify = (! 1)(5) =! 5 Substitute Treat i like any other variable when +, -,!, or (but always simplify i =! 1) Example: i (3 + i) = (3i) + i( i) Distribute = 6i + i Simplify = 6 i + (! 1) Make substitution =! + 6i Simplify and rewrite in complex form Since i =! 1, no answer can have an i in the denominator RATIONALIZE!! Simplify. 9.! 49 10. 6! 1 11.! 6(! 8 i) + 3(5 + 7 i) 1. ( 3 4i)! 13. (6! 4 i)(6 + 4 i) Rationalize. 14. 1 + 6 i 5i 3

Geometry: Pythagorean Theorem (right triangles): a + b = c Find the value of x. 15. 16. 17. x 9 x x 5 8 x 1 5 1 18. A square has perimeter 1 cm. Find the length of the diagonal. * In 30! 60! 90 triangles, *In 45! 45! 90 triangles, sides are in proportion 1, 3,. sides are in proportion 1,1,. 60 45 1 1 30 45 3 1 Solve for x and y. 19. 0. 45 45 x y 4 x 45 45 y 4

1.. 60 4 60 y x 3 y 30 30 x Equations of Lines: Slope intercept form: y = mx + b Vertical line: x = c (slope is undefined) Point-slope form: y! y 1 = m(x! x 1 ) Horizontal line: y = c (slope is 0) Standard Form: Ax + By = C Slope: m = y! y 1 x! x 1 3. State the slope and y-intercept of the linear equation: 5x 4y = 8. 4. Find the x-intercept and y-intercept of the equation: x y = 5 5. Write the equation in standard form: y = 7x 5 Write the equation of the line in slope-intercept form with the following conditions: 6. slope = -5 and passes through the point (-3, -8) 7. passes through the points (4, 3) and (7, -) 8. x-intercept = 3 and y-intercept = 5

Graphing: Graph each function, inequality, and / or system. 9. 3x! 4y = 1 30. 6! x + y = 4 " $ x # y = 6 4 4-5 5-5 5 - - -4-4 -6-6 31. y <! 4x! 3. y + = x + 1 6 6 4 4-5 5-5 5 - - -4-4 -6-6 33. y > x! 1 34. y + 4 = ( x! 1) 6 6 4 4-5 5-5 5 - - -4-4 -6-6 Vertex: x-intercept(s): y-intercept(s): 6

Systems of Equations: 3x + y = 6 x! y = 4 Substitution: Elimination: Solve 1 equation for 1 variable. Find opposite coefficients for 1 variable. Rearrange. Multiply equation(s) by constant(s). Plug into nd equation. Add equations together (lose 1 variable). Solve for the other variable. Solve for variable. Then plug answer back into an original equation to solve for the nd variable. y = 6! 3x solve 1 st equation for y 6 x + y = 1 multiply 1 st equation by x! (6! 3x) = 4 plug into nd equation x! y = 4 coefficients of y are opposite x! 1 + 6x = 4 distribute 8 x = 16 add 8 x = 16 simplify x = simplify x = 3() + y = 6 Plug x = back into original 6 + y = 6 y = 0 Solve each system of equations. Use any method. 35.! x + y = 4 " # 3x + y = 1 36.! x + y = 4 " $ 3x # y = 14 37. " w! 5z = 13 # $ 6w + 3z = 10 7

Exponents: TWO RULES OF ONE 1. a 1 = a Any number raised to the power of one equals itself.. 1 a = 1 One to any power is one. ZERO RULE 3. a 0 = 1 Any nonzero number raised to the power of zero is one. PRODUCT RULE m n m+ n 4. a! a = a When multiplying two powers that have the same base, add the exponents. QUOTIENT RULE a m m n = a! n 5. a When dividing two powers with the same base, subtract the exponents. POWER RULE m n m! n 6. ( a ) = a When a power is raised to another power, multiply the exponents. NEGATIVE EXPONENTS! n 1 1 n 7. a = and = a n! n a a Any nonzero number raised to a negative power equals its reciprocal raised to the opposite positive power. Express each of the following in simplest form. Answers should not have any negative exponents. 38. 0 5a 39. 1 3c 40. c! ef e! 1 41.! 1 ( n p )! ( np) 3! 1 Simplify. 4. 3m m 43. 3 ( a ) 44. ( ) 3 4 5! b c 45. 4 m(3 a m ) 8

Polynomials: To add / subtract polynomials, combine like terms. EX: 8x! 3y + 6! (6y + 4x! 9) Distribute the negative through the parantheses. = 8x! 3y + 6! 6y! 4x + 9 Combine terms with similar variables. = 8x! 4x! 3y! 6y + 6 + 9 = 4x! 9y + 15 Simplify. 46. 3x 3 + 9 + 7x! x 3 47. 7m! 6! (m + 5) To multiplying two binomials, use FOIL. EX: (3x! )(x + 4) Multiply the first, outer, inner, then last terms. = 3x + 1x! x! 8 Combine like terms. = 3x + 10x! 8 Multiply. 48. (3a + 1)(a ) 49. (s + 3)(s 3) 50. (c 5) 51. (5x + 7y)(5x 7y) 9

Factoring. Follow these steps in order to factor polynomials. STEP 1: Look for a GCF in ALL of the terms. a.) If you have one (other than 1) factor it out front. b.) If you don t have one, move on to STEP. STEP : How many terms does the polynomial have? Terms a.) Is it difference of two squares? a! b = (a + b)(a! b) EX: x! 5 = (x + 5)(x! 5) b.) Is it sum or difference of two cubes? EX: m 3 + 64 = (m + 4)(m! 4m + 16) p 3! 15 = ( p! 5)( p + 5p + 5) a 3! b 3 = (a! b)(a + ab + b ) a 3 + b 3 = (a + b)(a! ab + b ) 3 Terms x + bx + c = (x + )(x + ) Ex: x + 7x + 1 = (x +3)(x + 4) x! bx + c = (x! )(x! ) x! 5x + 4 = (x! 1)(x! 4) x + bx! c = (x! )(x + ) x + 6x! 16 = (x! )(x + 8) x! bx! c = (x! )(x + ) x! x! 4 = (x! 6)(x + 4) 4 Terms -- Factor by Grouping a.) Pair up first two terms and last two terms b.) Factor out GCF of each pair of numbers. c.) Factor out front the parentheses that the terms have in common. d.) Put leftover terms in parentheses. Factor completely. Ex: x 3 + 3x + 9x + 7 = x 3 + 3x ( ) + 9x + 7 ( ) ( ) + 9( x + 3) ( ) x + 9 = x x + 3 = x + 3 ( ) 5. z + 4z! 1 53. 6! 5x! x 54. k + k! 60 10

55.!10b 4! 15b 56. 9c + 30c + 5 57. 9n! 4 58. 7z 3! 8 59. mn! mt + sn! st To solve quadratic equations, try to factor first and set each factor equal to zero. Solve for your variable. If the quadratic does NOT factor, use quadratic formula. EX: x! 4x = 1 Set equal to zero FIRST. x! 4x! 1 = 0 Now factor. (x + 3)(x! 7) = 0 Set each factor equal to zero. x + 3 = 0 x! 7 = 0 Solve each for x. x =!3 x = 7 Solve each equation. 60. x! 4x! 1 = 0 61. x + 5 = 10x 6. x! 14x + 40 = 0 DISCRIMINANT: The number under the radical in the quadratic formula (b! 4ac) can tell you what kinds of roots you will have. IF b! 4ac > 0 you will have TWO real roots. IF b! 4ac = 0 you will have ONE real root (touches x-axis twice) (touches the x-axis once) - - -4 IF b! 4ac < 0 you will have TWO imaginary roots. (Graph does not cross the x-axis) 4 11

QUADRATIC FORMULA allows you to solve any quadratic for all its real and imaginary roots. x = b ± b 4ac a EX: Solve the equation: x + x + 3 = 0 x + x + 3 = 0 Determi values for a,b,and c. a = 1 b = = 3 Find dicriminant. D = 4 1 3 D = 4 1 D = 8 There are two imaginary roots. Solve : x = ± 8 x = ± i x = 1 ± i Solve each quadratic. Use EXACT values. 63. x 9x + 14 = 0 64. 5x x + 4 = 0 Discriminant = Type of Roots: Roots = Discriminant = Type of Roots: Roots = 1

Long Division can be used when dividing any polynomials. Synthetic Division can ONLY be used when dividing a polynomial by a linear (degree one) polynomial. EX: x3 + 3x! 6x + 10 x + 3 Long Division x 3 + 3x! 6x + 10 x + 3 x! 3x + 3 + 1 x + 3 = x + 3 x 3 + 3x! 6x + 10 (!) (x 3 + 6x )! 3x! 6x (!) (!3x! 9x) 3x + 10 (!) (3x + 9) 1 Synthetic Division x 3 + 3x! 6x + 10 x + 3!3 3!6 10 "! 6 9! 9! 3 3 1 = x! 3x + 3 + 1 x + 3 Divide each polynomial using long division OR synthetic division. 65. c3! 3c + 18c! 16 c + 3c! 66. x 4! x! x + x + To evaluate a function for a given value, simply plug the value into the function for x. Evaluate each function for the given value. 67. f (x) = x! 6x + 68. g(x) = 6x! 7 69. f (x) = 3x! 4 f (3) = g(x + h) = 5!" f (x + ) # $ = 13

Composition and Inverses of Functions: Recall: ( f! g)(x) = f (g(x)) OR f [g(x)] read f of g of x Means to plug the inside function (in this case g(x) ) in for x in the outside function (in this case, f(x)). Example: Given f (x) = x + 1 and g(x) = x! 4 find f(g(x)). f (g(x)) = f (x! 4) = (x! 4) + 1 = (x! 8x + 16) + 1 = x! 16x + 3 + 1 f (g(x)) = x! 16x + 33 Suppose f (x) = x, g(x) = 3x!, and h(x) = x! 4. Find the following: 70. f!" g() # $ = 71. f!" g(x) # $ = 7. f!" h(3) # $ = 73. g!" f (x)# $ = To find the inverse of a function, simply switch the x and the y and solve for the new y value. Example: 3 f (x) = x + 1 Rewrite f(x) as y 3 y = x + 1 Switch x and y 3 x = y + 1 Solve for your new y ( x) 3 3 = y + 1 x 3 = y + 1 y = x 3! 1 ( ) 3 Cube both sides f!1 (x) = x 3! 1 Simplify Solve for y Rewrite in inverse notation Find the inverse, f -1 (x), if possible. 74. f (x) = 5x + 75. f (x) = 1 x! 1 3 14

Rational Algebraic Expressions: Multiplying and Dividing. Factor numerator and denominator completely. Cancel any common factors in the top and bottom. If dividing, change divide to multiply and flip the second fraction. EX: x + 10x + 1 5! 4x! x = x + x! 15 x 3 + 4x! 1x (x + 7)(x + 3) (x + 5)(x! 3) (5 + x)(1! x) x(x! 3)(x + 7) Factor everything completely. Cancel out common factors in the top and bottom. = (x + 3) x(1! x) Simplify. Simplify. 76. 5z3 + z! z 3z 77. m! 5 m + 5m 78. 10r5 1s 3s 5r 3 79. a! 5a + 6 a + 4 3a + 1 a! 80. 6d! 9 5d + 1 6! 13d + 6d 15d! 7d! 15

Addition and Subtraction. First, find the least common denominator. Write each fraction with the LCD. Add / subtract numerators as indicated and leave the denominators as they are. EX: 3x + 1 x + x + 5x! 4 x + 4 = 3x + 1 x(x + ) + 5x! 4 (x + ) Factor deno min ator completely. Find LCD (x)(x + ) = (3x + 1) x(5x! 4) + x(x + ) x(x + ) Rewrite each fraction with the LCD as the denomin ator. = 6x + + 5x! 4x x(x + ) Write as one fraction. = 5x + x + x(x + ) Combine like terms. 81. x 5! x 3 8. b! a a b + a + b ab 83.! a a + a + 3a + 4 3a + 3 16

Complex Fractions. Eliminate complex fractions by multiplying the numerator and denominator by the LCD of each of the small fractions. Then simplify as you did above EX: 1+ 1 a a! 1 Find LCD : a " 1+ 1 % # $ a& ' i a = " a! 1 % # $ & ' i a Multiply top and bottom by LCD. = a + a! a Factor and simplify if possible. = a(a + 1)! a 84. 1! 1 + 1 4 85. 1+ 1 z z + 1 86. 5 + 1 m! 6 m m! m 87. + 1 x! 1 x 1+ 4 x + 3 x 17

Solving Rational Equations: Multiply each term by the LCD of all the fractions. This should eliminate all of your fractions. Then solve the equation as usual. 5 x + + 1 x = 5 Find LCD first. x(x + ) x! 5 $! 1$! x(x + ) " # x + % & + x(x + ) " # x % & = 5 $ " # x % & x(x + ) Multiply each term by the LCD. 5x + 1(x + ) = 5(x + ) Simplify and solve. EX: 5x + x + = 5x + 10 6x + = 5x + 10 x = 8 Check : 5 8 + + 1 8 = 5 8 5 10 + 1 8 = 5 8 5 8 = 5 8 ' Check your answer. Sometimes they do not check! Solve each equation. Check your solutions. 88. 1 x + 3 4 = 3 89. x + 10 x! = 4 x 90. x 5! = x x! 5! 1 18

Logarithms y log a x is equivalent to x a y Product property: logbmn logbm logb n m Quotient property: lo gb logb m logbn n p Power property: logb m plogb m Property of equality: If logb m logb n, then m n logb n Change of base formula: loga n log a Solve each equation. Check your solutions. b x x 5 og( x ) 11 93. 1 10 7 91. (3) 5 9. 5l x 94. ln x ln( x ) 1 95. 3 l nx 8 96. 3e 4 9 19