intermediate algebra optional pre-final exam review 1 – basic algebra review 2 – graphs &...
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Intermediate AlgebraOptional 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 appearwill
Exercises
1 – Basic Algebra Review
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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
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1 – Basic Algebra Review
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2 – Graphs & Equations of Lines Plotting Points aka Graphing Points
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2 – Graphs & Equations of Lines Solutions to Equations Any point on a graphed equation is a Solution
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.
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andif )4,5()5,2(
2 – Graphs & Equations of Lines The Slope-Intercept Form of a Line
int
int
2048
x
ym
xy
Remember the CoverUp Method?
Slopes of Parallel Lines m1 = m2
But wait! can you be sure that it’s not the same line ?
Slopes of Perpendicular Lines m1 = -1 / m2
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31 yxandxy
neitherorlarperpendicuParallel
3 – Solving Systems of EquationsUsing the Substitution Method
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5
yx
yx
B
A
3 – Solving Systems of Equations Using the Elimination (Addition) Method
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7
yx
yx
B
A
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 = 75·(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
22
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632
zyx
zyx
zyx
C
B
A
Well use substitution of x from (D) into (E) to find z (D) 4x +4z = 4
(D1) x = 1 - z move 4z to the other side, divide by 4
Substitute x from (D1) into (E)
(E) 13x + 9z = -3 13(1 – z) + 9z = -3
13 – 13z + 9z = -3 use distribution, then simplify -4z = -16 z = 4
Substitute z into (D) or (E) or (D1) 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) = 4
-6 + y + 4 = 4 y = 6 Solution is (-3, 6, 4)
Solution to (D) 4x + 4z = 4
Continued (E) 13x + 9z = -3
22
7453
632
zyx
zyx
zyx
C
B
A
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
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4 – Inequalities Expressing Domains With Interval Notation
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graphandsolve
4 – Inequalities Using the Absolute Value Principle
|x + 1| = 2 x + 1 = 2 or x + 1 = -2 |2y – 6| = 0 2y – 6 = 0 |5x – 3| = -2 no solution
9|53| x
4 – Inequalities When an equation has 2 absolute values?1|92||3| xx
5 – Polynomials & Factoring Subtracting Polynomials
To subtract polynomials, add the opposite of the second polynomial.
(7x3 + 2x + 4) – (5x3 – 4) add the opposite!(7x + 2x + 4) + (-5x3 + 4)
Use either horizontal or vertical addition. Sometimes the problem is posed as subtraction:
x2 + 5x +6 make it addition x2 + 5x +6 - (x2 + 2x) _ of the opposite -x2 – 2x__ 3x +6
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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) = 6a2 – 4ab + 3ab – 2b2 = 6a2 –ab – 2b2
Vertical Method: 3x2 + 2x – 5 4x + 2
6x2 + 4x – 10 12x3 + 8x2 – 20x____
12x3 + 14x2 – 16x – 10
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5 – Polynomials & Factoring FOIL: Used to Multiply Two Binomials )53)(25( yxyx
5 – Polynomials & Factoring Find the Greatest Common Factor 7a – 21 =
7(? – ?) = 7(a – 3)
19x3 + 3x = x(? + ?) = x(19x2 + 3)
18y3 – 12y2 + 6y = 6y(? – ? + ?) = 6y(3y2 – 2y + 1)
xyyxyx 493514 223
5 – Polynomials & Factoring Factor by Grouping 8t3 + 2t2 – 12t – 3 2t2(4t + 1) – 3(4t + 1) (4t + 1)(2t2 – 3)
4x3 – 6x2 – 6x + 9 2x2(2x – 3) – 3(2x – 3) (2x – 3)(2x2 – 3)
y4 – 2y3 – 12y – 3 y3(y – 2) – 3(4y – 1) Oops – not factorable via grouping
9664 23 xxx
5 – Polynomials & Factoring Using a Factor Table - Trial & Error
Let’s use x2 + 13x + 36 as an example Factors must both be sums: (x + ?)(x + ?) Pairs=c=36 Sum=b=13 1, 36 37 2, 18 20 3, 12 15 4, 9 13 ok quit!
x2 + 13x + 36 = (x + 4)(x + 9)
3652 xx
The ac Grouping Method: ax2 + bx + cSplit bx into 2 Terms: Use a Table based on a·c
Let’s use 3x2 – 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, -24 -23 2, -12 -10 quit! 3, -8 -5 4, -6 -2
3x2 – 10x – 8 = 3x2 + 2x – 12x – 8 = split the middle x(3x + 2) – 4(3x + 2) = do grouping = (3x + 2)(x – 4)
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5 – Polynomials & Factoring Factoring Perfect Square Trinomials?
x2 + 8x + 16 = (x + 4)2
(x)2 (4)2 2(x)(4) = 8x yes, it matches
t2 – 12t + 4 = not a PST (t)2 (-2)2 2(t)(-2) = -4t no, it’s not -12t
25 + y2 + 10y = (y + 5)2
y2 + 10y + 25 descending order (y)2 (5)2 2(y)(5) = 10y yes, it matches
3x2 – 15x + 27 = not a PST 3(x2 – 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. 1st and 3rd Terms (A)2 and (B)2
4. Middle Term 2AB or -2AB
5 – Polynomials & Factoring Difference of Squares Binomials Remember that the middle term disappears? (A + B)(A – B) = A2 - B2
It’s easy factoring when you find binomials of this pattern A2 – B2 = (A + B)(A – B)
Examples: x2 – 9 = (x)2 – (3)2 = (x + 3)(x – 3) 4t2 – 49 = (2t)2 – (7)2 = (2t + 7)(2t – 7) a2 – 25b2 = two variables squared (a)2 – (5b)2 = (a + 5b)(a – 5b) 18 – 2y4 = constant 1st, variable square 2nd 2 [ (3)2 – (y2)2 ] = 2(3 + y2)(3 – y2)
5 - Factoring the Difference between 2 CubesX3 – Y3 = (X – Y)(X2 + XY + Y2)
F3 – L3 factors easily to (F – L)(F2 + FL +L2) Examine 27a3 – 64b3
(3a)3 – (4b)3 (3a – 4b)(9a2 + 12ab + 16b2) Remember to remove common factors and to factor completely
p3 – 8 2x6 – 128 = 2[x6 – 64] (p)3 – (2)3 2[(x2)3 – 43] (p – 2)(p2 + 2p + 4) 2(x2 – 4)(x4 + 4x2 + 16) 2(x + 2)(x – 2)(x4 + 4x2 + 16)
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5 - Factoring the Sum of 2 Cubes X3 + Y3 = (X + Y)(X2 – XY + Y2)
F3 + L3 factors easily to (F + L)(F2 – FL +L2) Examine 27a3 + 64b3
(3a)3 + (4b)3 (3a + 4b)(9a2 – 12ab + 16b2) Remember to remove common factors and to factor completely
p3 + 8 2x6 + 128 = 2[x6 + 64] (p)3 + (2)3 2[(x2)3 + 43] (p + 2)(p2 – 2p + 4) 2(x2 + 4)(x4 – 4x2 + 16)
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5 - Definition
Principle of Zero Factors
5 - Solving a Quadratic Equation by Factoring 028102 23 xxx
solve
6 – Rational Expressions & Functions Multiplying Fractions
(Use parentheses for clarity)Factor expressions,
then cancel like factors
6 – Rational Expressions & Functions Dividing Fractions
Change Divide to Multiply by Reciprocal,follow multiply procedure
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.
28z3 = (22) (7)(z3)
21z = (3)(7)(z)
LCD=(22)(3)(7)(z3)
a2 – 25 = (a + 5)(a – 5)
a + 7a + 10 = (a + 5) (a + 2)
LCD = (a + 5)(a – 5)(a + 2)
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
6 - Simplifying Complex Rational Ex’s Method 1: Multiplying by 1
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6 - Simplifying Complex Rational Ex’s Method 2: Multiplying by Reciprocal Making the top and bottom into single
expressions, then multiplying by reciprocal.
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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
6 – Rational Expressions & Functions Clearing & Solving a Rational Equation
What gets excluded? x ≠ 0What’s the LCD? 15xWhat’s the solution?
6 – Rational Expressions & Functions Dividing a Polynomial by a Polynomial
Use the long-division process
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7 – Radical Expressions & Functions Examples to Simplify
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7 – Radical Expressions & Functions Practice
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Express using rational exponents:
Simplify using rational exponents:
2
3
2
3
8
273
13
1 3
Rational Expressions Where the Numerator is greater than 1
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2 42 323
2221616
62342
342
3
22216
Using Exponent Arithmetic, it’s a little easier
7 – Radical Expressions & Functions The Product Rule for Radicals
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333 33 43 3025322532240
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7 – Radical Expressions & Functions The Quotient Rule for Radicals
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n
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b
a
b
a
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8
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x
xy
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3 33
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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
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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 xn = yn
Warning: These are NOT equivalent Equations!
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7 – Radical Expressions & Functions Equations Containing One Radical To eliminate the radical,
raise both sides to the index of the radical
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Check
Equations Containing Two Radicals Make sure radicals are on opposite sides Sometimes you need to repeat the process
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7 – Radical Expressions & Functions i, The Basis of the Complex Number System
7 – Radical Expressions & Functions Powers of i
1
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4
3
2
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1
11
8
7
6
5
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8 – Quadratic Functions The Square Root Principle
Solve by factoring
x2 – 16 = 0 (x+4)(x-4)=0 x=4,-4
Then by the square root property
x2 – 16 = 0 x2 = 16 x=4,-4
08811 2 x
8 – Quadratic Functions Using Completing the Square to Solve an Equation
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sidesbothofrootSqxandx
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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.
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8 – Quadratic Functions Solving Quadratic Form Equations The variable u is often used to replace squares
of variables or expressions
ixandxxandx
uxSolveNowuandu
uusolvenowuu
rewriteandxuletxx
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8 – Quadratic Functions (parabolas) Graph f(x) = (x – 3)2
Maximums & Minimums382 2 xxykhxaxf 2)()(
9 – Exponents & Logarithms Graphing an Exponential Function
9 – Exponents & Logarithms Translating Right and Left
9 – Exponents & Logarithms General Definition of Logarithms
3125log2100log
32log
02:0log
510
2
2
impossiblealso
solutionnoimpossible y
10 - Introduction to Conic Sections Parabola Circle Ellipse Hyperbola
10 – Conic SectionsRemember Parabolas? Two styles: Functions & Relations
Find the Vertex:
x = -b/(2a),
(or y = -b/(2a))
solve for y or x
10 – Conic SectionsA Circle has a Center and a Radius
Find the center & radius
An Ellipse also has a Center and Foci
What Next? Getting a Good Grade on the Final!Thank you all for
a good class.
MrV