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Intermediate Algebra Optional Pre-Final Exam Review  1 – Basic Algebra Review  2 – Graphs & Equations of Lines  3 – Solving Systems of Equations  4.

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Presentation on theme: "Intermediate Algebra Optional Pre-Final Exam Review  1 – Basic Algebra Review  2 – Graphs & Equations of Lines  3 – Solving Systems of Equations  4."— Presentation transcript:

1 Intermediate Algebra Optional Pre-Final Exam Review  1 – Basic Algebra Review  2 – Graphs & Equations of Lines  3 – Solving Systems of Equations  4 – Inequalities  5 – Polynomials & Factoring  6 – Rational Expressions & Functions  7 – Radical Expressions & Functions  8 – Quadratic Functions  9 – Exponents & Logarithms  10 – Conic Sections

2 1 – Basic Algebra Review

3 1 – Basic Algebra Review Rules for Order of Operations  To make sure an expression is always evaluated in the same way by different people, the Order of Operations convention was defined  Mnemonic: “Please Excuse My Dear Aunt Sally”  Parentheses  Exponents  Multiply/Divide  Add/Subtract  Always: Evaluate & Eliminate the innermost grouping first

4 1 – Basic Algebra Review

5 2 – Graphs & Equations of Lines Plotting Points aka Graphing Points (, )

6 2 – Graphs & Equations of Lines Solutions to Equations  Any point on a graphed equation is a Solution

7 2 – Graphs & Equations of Lines What is Slope & Why is it Important?  Using any 2 points on a straight line will compute to the same slope.

8 2 – Graphs & Equations of Lines The Slope-Intercept Form of a Line Remember the CoverUp Method?

9 Slopes of Parallel Lines m 1 = m 2 But wait! can you be sure that it’s not the same line ?

10 Slopes of Perpendicular Lines m 1 = -1 / m 2

11 3 – Solving Systems of Equations Using the Substitution Method

12 3 – Solving Systems of Equations Using the Elimination (Addition) Method

13 3 – Solving Systems of Equations Solution to 3 Equations  Adding (A) and (C) will eliminate y (A) 2x – y + 3z = 6 (C) 2x + y + z = -2 (D) 4x +4z = 4 first new equation in 2 variables  Adding (B) and 5·(C) will also eliminate y (B) 3x – 5y +4z = 7 5·(C) 10x + 5y + 5z = -10 (E) 13x + 9z = -3 second new equation in 2 variables  Solve (D) and (E) like a system of two equations (next page) Use Substitution or Addition

14  Well use substitution of x from (D) into (E) to find z (D) 4x +4z = 4 (D 1 ) x = 1 - z move 4z to the other side, divide by 4  Substitute x from (D 1 ) into (E) (E) 13x + 9z = -3 13(1 – z) + 9z = – 13z + 9z = -3 use distribution, then simplify -4z = -16  z = 4  Substitute z into (D) or (E) or (D 1 ) to find x (D) 4x + 4(4) = 4 4x + 16 = 4 4x = -12  x = -3  Substitute x and z into (A) or (B) or (C) to find y (C) 2(-3) + y + (4) = y + 4 = 4 y = 6Solution is (-3, 6, 4) Solution to (D) 4x + 4z = 4 Continued (E) 13x + 9z = -3

15 4 – Inequalities Intersections, Unions & Compound Inequalities  Set Diagrams  Intersections of Sets  Conjunctions of Sentences and  Unions of Sets  Disjunctions of Sentences or  Interval Notation  Domains

16 4 – Inequalities Expressing Domains With Interval Notation

17 4 – Inequalities Using the Absolute Value Principle  |x + 1| = 2x + 1 = 2 or x + 1 = -2  |2y – 6| = 02y – 6 = 0  |5x – 3| = -2 no solution

18 4 – Inequalities When an equation has 2 absolute values?

19 5 – Polynomials & Factoring Subtracting Polynomials  To subtract polynomials, add the opposite of the second polynomial.  (7x 3 + 2x + 4) – (5x 3 – 4) add the opposite! (7x + 2x + 4) + (-5x 3 + 4)  Use either horizontal or vertical addition.  Sometimes the problem is posed as subtraction: x 2 + 5x +6 make it addition x 2 + 5x +6 - (x 2 + 2x) _ of the opposite -x 2 – 2x__ 3x +6

20 5 – Polynomials & Factoring Multiplying Two Polynomials To multiply a polynomial by a polynomial, we use the distributive property repeatedly. Horizontal Method: (2a + b)(3a – 2b) = 2a(3a – 2b) + b(3a – 2b) = 6a 2 – 4ab + 3ab – 2b 2 = 6a 2 –ab – 2b 2 Vertical Method: 3x 2 + 2x – 5 4x + 2 6x 2 + 4x – 10 12x 3 + 8x 2 – 20x____ 12x x 2 – 16x – 10

21 5 – Polynomials & Factoring FOIL: Used to Multiply Two Binomials

22 5 – Polynomials & Factoring Find the Greatest Common Factor  7a – 21 = 7(? – ?) = 7(a – 3)  19x 3 + 3x = x(? + ?) = x(19x 2 + 3)  18y 3 – 12y 2 + 6y = 6y(? – ? + ?) = 6y(3y 2 – 2y + 1)

23 5 – Polynomials & Factoring Factor by Grouping  8t 3 + 2t 2 – 12t – 3  2t 2 (4t + 1) – 3(4t + 1)  (4t + 1)(2t 2 – 3)  4x 3 – 6x 2 – 6x + 9  2x 2 (2x – 3) – 3(2x – 3)  (2x – 3)(2x 2 – 3)  y 4 – 2y 3 – 12y – 3  y 3 (y – 2) – 3(4y – 1)  Oops – not factorable via grouping

24 5 – Polynomials & Factoring Using a Factor Table - Trial & Error  Let’s use x x + 36 as an example  Factors must both be sums: (x + ?)(x + ?)  Pairs=c=36 Sum=b=13  1, 3637  2, 1820  3, 1215  4, 913 ok quit!  x x + 36 = (x + 4)(x + 9)

25 The ac Grouping Method: ax 2 + bx + c Split bx into 2 Terms: Use a Table based on a · c  Let’s use 3x 2 – 10x – 8 as an example  ac = 3(-8) = -24 One factor is positive, the other negative and larger.  Pairs=ac=-24 Sum=b=-10  1,  2, quit!  3,  4,  3x 2 – 10x – 8 =  3x 2 + 2x – 12x – 8 = split the middle  x(3x + 2) – 4(3x + 2) = do grouping =  (3x + 2)(x – 4)

26 5 – Polynomials & Factoring Factoring Perfect Square Trinomials?  x 2 + 8x + 16 = (x + 4) 2  (x) 2 (4) 2 2(x)(4) = 8x yes, it matches   t 2 – 12t + 4 = not a PST  (t) 2 (-2) 2 2(t)(-2) = -4t no, it’s not -12t   25 + y y = (y + 5) 2  y y + 25 descending order  (y) 2 (5) 2 2(y)(5) = 10y yes, it matches   3x 2 – 15x + 27 = not a PST  3(x 2 – 5x + 9) remove common factor  (x) 2 (-3) 2 2(x)(-3) = -6x no, it’s not -5x PST Tests: 1. Descending Order 2. Common Factors 3. 1 st and 3 rd Terms (A) 2 and (B) 2 4. Middle Term 2AB or -2AB

27 5 – Polynomials & Factoring Difference of Squares Binomials  Remember that the middle term disappears? (A + B)(A – B) = A 2 - B 2  It’s easy factoring when you find binomials of this pattern A 2 – B 2 = (A + B)(A – B)  Examples: x 2 – 9 = (x) 2 – (3) 2 = (x + 3)(x – 3) 4t 2 – 49 = (2t) 2 – (7) 2 = (2t + 7)(2t – 7) a 2 – 25b 2 = two variables squared (a) 2 – (5b) 2 = (a + 5b)(a – 5b) 18 – 2y 4 = constant 1 st, variable square 2 nd 2 [ (3) 2 – (y 2 ) 2 ] = 2(3 + y 2 )(3 – y 2 )

28 5 - Factoring the Difference between 2 Cubes X 3 – Y 3 = (X – Y)(X 2 + XY + Y 2 )  F 3 – L 3 factors easily to (F – L)(F 2 + FL +L 2 )  Examine 27a 3 – 64b 3  (3a) 3 – (4b) 3  (3a – 4b)(9a ab + 16b 2 )  Remember to remove common factors and to factor completely  p 3 – 8 2x 6 – 128 = 2[x 6 – 64] (p) 3 – (2) 3 2[(x 2 ) 3 – 4 3 ] (p – 2)(p 2 + 2p + 4) 2(x 2 – 4)(x 4 + 4x ) 2(x + 2)(x – 2)(x 4 + 4x )

29 5 - Factoring the Sum of 2 Cubes X 3 + Y 3 = (X + Y)(X 2 – XY + Y 2 )  F 3 + L 3 factors easily to (F + L)(F 2 – FL +L 2 )  Examine 27a b 3  (3a) 3 + (4b) 3  (3a + 4b)(9a 2 – 12ab + 16b 2 )  Remember to remove common factors and to factor completely  p x = 2[x ] (p) 3 + (2) 3 2[(x 2 ) ] (p + 2)(p 2 – 2p + 4) 2(x 2 + 4)(x 4 – 4x )

30 5 - Definition Principle of Zero Factors

31 5 - Solving a Quadratic Equation by Factoring

32 6 – Rational Expressions & Functions Multiplying Fractions (Use parentheses for clarity) Factor expressions, then cancel like factors

33 6 – Rational Expressions & Functions Dividing Fractions Change Divide to Multiply by Reciprocal, follow multiply procedure

34 6 - Finding the LCD (must be done before adding or subtracting 2 or more RE’s) 1. Factor each denominator completely into primes. 2. List all factors of each denominator. (use powers when duplicate factors exist) 3. The LCD is the product of each factor to its highest power. 28z 3 = (2 2 ) (7)(z 3 ) 21z = (3)(7)(z) LCD=(2 2 )(3)(7)(z 3 ) a 2 – 25 = (a + 5)(a – 5) a + 7a + 10 = (a + 5) (a + 2) LCD = (a + 5)(a – 5)(a + 2)

35 1. Find the LCD. 2. Express each rational expression with a denominator that is the LCD. 3. Add (or subtract) the resulting rational expressions. 4. Simplify the result if possible. 6 - Adding or subtracting rational expressions with unlike denominators

36 6 - Simplifying Complex Rational Ex’s Method 1: Multiplying by 1

37 6 - Simplifying Complex Rational Ex’s Method 2: Multiplying by Reciprocal  Making the top and bottom into single expressions, then multiplying by reciprocal.

38 6 – Rational Expressions & Functions Rational Equations: False Solutions  Solve a Rational Equation by Multiplying BOTH SIDES by the LCD  Warning: Clearing an equation may add a False Solution  A False Solution is one that causes a divide by zero situation in the original equation  Before even starting to solve a rational equation, we need to identify values to be excluded  What values need to be excluded for these?  t ≠ 0 a ≠ ±5 x ≠ 0

39 6 – Rational Expressions & Functions Clearing & Solving a Rational Equation What gets excluded? x ≠ 0 What’s the LCD? 15x What’s the solution?

40 6 – Rational Expressions & Functions Dividing a Polynomial by a Polynomial Use the long- division process

41 7 – Radical Expressions & Functions Examples to Simplify

42 7 – Radical Expressions & Functions Practice Express using rational exponents: Simplify using rational exponents:

43 Rational Expressions Where the Numerator is greater than 1 Using Exponent Arithmetic, it’s a little easier

44 7 – Radical Expressions & Functions The Product Rule for Radicals

45 7 – Radical Expressions & Functions The Quotient Rule for Radicals

46 7 – Radical Expressions & Functions A Radical Expression is Simplified When: 1.Each factor in the radicand is to a power less than the index of the radical 2.The radicand contains no fractions or negative numbers 3.No radicals appear in the denominator of a fraction

47 7 – Radical Expressions & Functions Definitions  A Radical Equation must have at least one radicand containing a variable  The Power Rule:  If we raise two equal quantities to the same power, the results are also two equal quantities  If x = y then x n = y n  Warning: These are NOT equivalent Equations!

48 7 – Radical Expressions & Functions Equations Containing One Radical  To eliminate the radical, raise both sides to the index of the radical

49 Equations Containing Two Radicals  Make sure radicals are on opposite sides  Sometimes you need to repeat the process

50 7 – Radical Expressions & Functions i, The Basis of the Complex Number System

51 7 – Radical Expressions & Functions Powers of i

52 8 – Quadratic Functions The Square Root Principle Solve by factoring x 2 – 16 = 0 (x+4)(x-4)=0 x=4,-4 Then by the square root property x 2 – 16 = 0 x 2 = 16 x=4,-4

53 8 – Quadratic Functions Using Completing the Square to Solve an Equation

54 8 – Quadratic Functions Introducing … The Quadratic Formula!  The Quadratic Formula is used to find solutions to any quadratic equation  The formula was derived using completing the square and the square root property.

55 8 – Quadratic Functions Solving Quadratic Form Equations  The variable u is often used to replace squares of variables or expressions

56 8 – Quadratic Functions (parabolas) Graph f(x) = (x – 3) 2

57 Maximums & Minimums

58 9 – Exponents & Logarithms Graphing an Exponential Function

59 9 – Exponents & Logarithms Translating Right and Left

60 9 – Exponents & Logarithms General Definition of Logarithms

61 10 - Introduction to Conic Sections Parabola Circle Ellipse Hyperbola

62 10 – Conic Sections Remember Parabolas?  Two styles: Functions & Relations Find the Vertex: x = -b/(2a), (or y = -b/(2a)) solve for y or x

63 10 – Conic Sections A Circle has a Center and a Radius Find the center & radius

64 An Ellipse also has a Center and Foci

65 What Next? Getting a Good Grade on the Final! Thank you all for a good class. MrV


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