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Download Algebra 3 Summer Assignment 2016
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Algebra III Summer Assignment 2016 The following packet contains topics and definitions that you will be required to know in order to succeed in Algebra III this year. You are advised to be familiar with each of the concepts and to complete the included problems by Thursday, September 1, 2016. All of these topics were discussed in either Algebra II or Geometry and will be used frequently throughout the year. All problems that you are to complete are marked in bold. All problems are expected to be completed. Section I: Lines Lines: π¦ βπ¦ Slope: π = π₯2 βπ₯1 Slope Intercept Form: y = mx + b Standard Form: ax + by = c Point-Slope Form: y - y1 = m(x β x1) 2 1 Intercepts: In order to find the x-intercept(s) of an equation, you have to set y equal to zero and solve the equation for x. In order to find the y-intercepts of an equation, you have to set x equal to zero and solve the equation for y. Parallel Lines: Two lines whose graphs have the same slope. Perpendicular Lines: Two lines whose graphs have opposite reciprocal slopes. Find the slope of the lines passing through each set of points: π (π, ππ) πππ (βπ, βπ) Find the slope of the line: π π (β π , π) πππ (β π , β π) Find the F Find the slope of the following Line: y ππ + ππ = βππ 4 2 β4 β2 O β2 β4 2 4 x Write the equation of the line in standard Find the point-slope form of the line form for a slope of -8 and through (-2, -2) through (-6, -4) and (2, -5). Write the equation of the line in slope-intercept Write the equation of the horizontal form through (βππ, βπ) and (βπ, π). line through (βπ, βπ). Graph the equation of 6x + 5y = 30 using the x- and y-intercepts. π π Graph π = β π β π Find the equation of the line that is parallel Find the equation of the line perpendicular to π = β ππ + π and goes through (βπ, π). to π = β π π + π and goes through (2, 6). π Give the slope-intercept form for the equation of the line that is perpendicular to ππ + ππ = ππ and contains (π, π). Which two lines are parallel? I. II. III. What must be true about the slopes of two perpendicular lines, neither of which is vertical? Section II: Polynomials Properties of Exponents: π₯ π = π₯ β π₯ β π₯ β β¦ β π₯ (n factors of x) 1. Whole number exponents: π₯ 0 = 1, π₯ β 0 2. Zero exponents: π₯ βπ = 3. Negative Exponents: 1 π₯π π βπ₯ = π β π₯ = π π 4. Radicals (principal nth root): 5. Rational exponents: π₯ 1β π 6. Rational exponents: π₯ πβ π π = βπ₯ π = βπ₯ π Operations with Exponents: π₯ π π₯ π = π₯ π+π 1. Multiplying like bases: π₯π 2. Dividing like bases: π₯π = π₯ πβπ π₯ π π₯π 3. Removing parentheses: (π₯π¦)π = π₯ π π¦ π ( ) = π (π₯ π )π = π₯ ππ π¦ π¦ For example: Simplify each of the following expressions: a. ο¨7a ο©ο¨ο 2a ο© 2 ο5 ο½ 7ο¨ο 2ο©a ( 2ο« ο5) ο½ ο14a ο3 ο 14 ο½ 3 a b. ο¨ο 2 x ο1 y2 ο© ο¨ ο© ο¨y ο© ο½ ο¨ο 2 ο© x ο1 3 ο3 ο½ ο8 x y ο½ ο 8y x3 6 c. 3 6 3 2ab 5 c 2 a 3bc 2 2 3 ο½ 2a (1ο3)b ( 5ο1) c ( 2 ο 2 ) ο½ 2a ο 2b 4 c 0 2b 4 ο½ 2 a Simplify the following expressions. Use only positive exponents. 1. ο¨3a ο©ο¨4a ο© 2. ο¨ο 4x ο©ο¨ο 2x ο© 2 6 13. x 4 x ο2 x ο5 ο2 2 14. ο¨12 x y ο© 15. ο¨4 p qο©ο¨ p q ο© 16. 4x3 2x 17. ο¨p ο© 18. ο 15 x 4 3x 19. r 2 s 3t 4 r 2 s 4 t ο4 20. xy 2 6 x ο 2 y2 21. ο¨s t ο© ο¨st ο© 2 3. ο¨4 x y ο© 4. ο¨2 x 5 2 3 ο5 y ο© 4 3 5 5. 6. 8a 2a 2 6x 7 y 5 3x ο1 ο¨4 x ο© 2 0 7. 8. 9. 2 xy5 ο¦ 3x 2 οΆ ο§ο§ ο·ο· ο¨ 2 οΈ 2 ο¨ο 6m n ο©ο¨3mnο© 2 6 2 8x 4 y 7 2 2 2 ο2 2 10. ο¨3x 4 y 5 ο© 2 3 ο3 22. ο¨3x ο3 y ο2 ο© ο2 11. ο¨2r ο© ο1 2 0 ο2 s t 2rs 23. 12. x 5 ο¨2x ο©3 24. ο¨h k ο© 4 5 0 s 2t 3 sr 3 ο r t 3 Section III: Radical Expressions and Complex Numbers Radical Expressions: π π π π π Multiplying Radical Expressions: If βπ πππ βπ are real numbers, then βπ β βπ = βππ π π π Dividing Radical Expressions: If βπ πππ βπ are real numbers and bβ 0, then βπ π π = βπ βπ π For example: Simplify each of the following rational expressions (be sure to rationalize the denominator if necessary): βπ₯ 3 β5π₯π¦ First use the Dividing Radical Expressions rule to rewrite with one radical. β Next cancel any common factors in the numerator and denominator. π₯3 5π₯π¦ π₯2 β 5π¦ βπ₯ 2 Now use the Dividing Radical Expressions rule to rewrite with two radicals. β5π¦ π₯ Simplify any possible square roots. β5π¦ π₯ β5π¦ β β5π¦ β5π¦ Rationalize the denominator to get you final answer. π₯β5π¦ 5π¦ Final Answer Simplify the following expressions. Use only positive exponents. 1) β2 5) β3 2 2) β3π₯ 6) β2π₯ 3 7) 3) 4) β10π₯π¦ 15β60π₯ 5 3β12π₯ 8) β3π₯π¦ 2 β5π₯π¦ 3 β5π₯ 4 π¦ β2π₯ 2 π¦ 3 10 β5π₯ 2 3β11π₯ 3 π¦ β2β12π₯ 4 π¦ Imaginary Numbers: For any positive real number a, βπ = π βπ π 2 = β1 ππ ββ1 = π For example: Simplify ββ8 by using the imaginary number i. ββ8 First factor out the -1 from -8 ββ1 β 8 πβ8 Next replace ββ1 with i πβ4 β 2 Then factor any perfect squares. 2πβ2 Take the square root of the perfect square. Simplify the following by using the imaginary number i. 1) ββ12 4) ββ68 2) ββ20 5) ββ1 3) ββ300 6) ββ121 Complex Numbers: A complex number can be written in the form π + ππ, where a & b are real numbers, including 0. For example: Simplify the expression (5 + 7π) + (β2 + 6π) (5 + 7π) + (β2 + 6π) Use the Commutative and Associative Properties of Addition to rewrite expression with like terms together (5 + β2) + (7π + 6π) (3 + 13π) Simplify For example: Simplify the expression (5π + 1)(β4π) (5π + 1)(β4π) (5π)(β4π) + 1(β4π) Use the Distributive Property to rewrite expression. β20π 2 β 4π Multiply Replace π 2 with -1 β20(β1) β 4π 20 β 4π Simplify Simplify the following by using the imaginary number i. 7) (2 + 4π) + (4 β π) 10) 6 β (8 + 3π) 8) (12 + 5π) β (2 β π) 11) (8 + π)(2 + 7π) 9) (β2π)(5π) 12) (9 + 4π)2 Section IV: Quadratic Equations Standard Form of a Quadratic Equation: ππ₯ 2 + ππ₯ + π = 0 Sum and Difference of Same Terms: (π’ + π£)(π’ β π£) = π’2 β π£ 2 Square of a Binomial: (π’ + π£)2 = π’2 + 2π’π£ + π£ 2 and (π’ β π£)2 = π’2 β 2π’π£ + π£ 2 For example: Write the following quadratic equations in standard form, then identify a, b, and c. 2 First move the β3π₯ to the other side of the equation by adding 3π₯ 2 to both sides of the equation Now move the 2 to the other side of the equation by subtracting 2 from both sides. So a=3 b=4 and c=-2 4π₯ = 2 β 3π₯ 2 +3π₯ 2 + 3π₯ 2 3π₯ 2 + 4π₯ = 2 -2 -2 3π₯ 2 + 4π₯ β 2 = 0 Write the following quadratic equations in standard form, then identify a, b, and c. 1) 2π₯ 2 β 11π₯ = β15 4) 2π₯ 2 β 1π₯ = β5 β 3π₯ 2) π₯ 2 + 7π₯ = 18 5) β7π₯ = 6π₯ 2 β 4 3) 16π₯ 2 = 8π₯ 6) β14π₯ β 2π₯ 2 = 6π₯ 2 + 3π₯ For example: Solve the following quadratic equations by factoring: π₯ 2 + 5π₯ + 4 = 0 To factor this quadratic equation, you first must split the π₯ 2 term. (π ) (π ) =π Now you must find two numbers that multiply to equal c (4) Either 1 * 4 or 2 * 2 will work. You must choose the set of numbers that and add to equal b (5) In this case 1 & 4 Set each set of parenthesis = 0 to and solve (π + π ) (π + π) =π (π + π ) = π ππ (π + π ) = π π = βπ ππ π = βπ Solve the following quadratic equations by factoring: Since the c=0 in the equation, you factor this quadratic by taking out the GCF β3π₯ 2 + 12π₯ = 0 The GCF of both terms is -3x, so factor this out of BOTH terms. (Note: you can check you factored correctly by re-distributing -3x, you should get the original equation.) (βππ)(π β π) = π Lastly, set each set of parenthesis = 0 to and solve βππ = π ππ π β π = π π = π ππ π = π Factor each completely: 7) 18 x3 ο 12 x 2 ο« 6 x ο 4 8) 64k 3 ο 56k 2 ο« 32k ο 28 9) 8k 3 ο 40k 2 ο 32k ο« 160 Factor the common factor out of each expression: 10) ο15 ο 30 x ο« 10 x 2 11) 20n 4 ο« 35n 2 ο« 35n Factor each completely: 12) v 2 ο 3v ο« 2 13) v 2 ο 9v ο 18 14) x 2 ο« 5 x ο 14 15) ο2n 2 ο 7v 16) 7 x 2 ο« 50 x ο« 7 17) 28n 2 ο« 240n ο 108 18) 8k 2 ο« 36k 19) 10b 2 ο« 12b 20) 10n 2 ο 14n 21) 27a 2 ο 36a ο« 12 22) 75r 2 ο 48 23) 9 x 2 ο 12 x ο« 4 24) n 2 ο 1 25) 48a 2 ο 27 26) 25n 2 ο 9 Solve each of the following by using the Quadratic Formula: π₯ = 27) ππ + ππ + π = π 29) βπ±βπ2 β4ππ 2π 28) πππ + ππ + π = π πππ β πππ β π = 0 30) ππ β ππ β π = π Expand each of the following (hint: FOIL): 31) (π + π)π 32) (π β π)π 33) (π β π)(π + π) 34) (π β π)(π β π)