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Final Exam Review Slides Example: Absolute Value Simplify the following expressions using your knowledge of absolute values: a. 21 7 14 b. c. 3 3 The Order of Mathematical Operations Order of Operations 1. If the expression is a fraction, simplify the numerator and denominator individually, according to the guidelines in the following steps. 2. Parentheses, braces and brackets are all used as grouping symbols. Simplify expressions within each set of grouping symbols, if any are present, working from the innermost outward. 3. Simplify all powers (exponents) and roots. 4. Perform all multiplications and divisions in the expression in the order they occur, working from left to right. 5. Perform all additions and subtractions in the expression in the order they occur, working from left to right. Properties of Exponents and Their Use Properties of Exponents Throughout this table, a and b may be taken to represent constants, variables or more complicated algebraic expressions. The letters n and m represent integers. Property 1. Product Rule a n a m a n m 2. Quotient Rule 3. Zero Exponent Rule an n m a am a 0 1. 4. Negative Exponent Rule a n 1 n a Properties of Exponents and Their Use Power Rules n m 5. a a nm 6. ab n n a nb n n a a 7. n b b Example: Scientific Notation The distance from Earth to the sun is approximately 93,000,000 miles. Scientific notation takes advantage of the observation that multiplication of a number by 10 moves the decimal point one place to the right, and we can repeat this process as many times as necessary. 7 Thus, in scientific notation, 93,000,000 9.3 10 . . . Example 2: Scientific Notation a. The mass of an electron, in kilograms, is approximately 0.000000000000000000000000000000911. Scientific notation takes advantage of the observation that multiplication of a number by 101 moves the decimal point one place to the left, and we can repeat this process as many times as necessary. Thus, in scientific notation, 0.000000000000000000000000000000911=9.11 1031. Simplifying Radical Expressions In the following properties, a and b may be taken to represent constant variable, or more complicated algebraic expressions. The letters n and m represent natural numbers. Property 1. Product Rule 2. Quotient Rule n ab n a n b n a b n a n b Simplifying Radical Expressions Rationalizing Denominators Denominator consists of two terms, one or both of which are square roots. Once again, remember that we cannot multiply the denominator by A – B unless we multiply the numerator by this same factor. Thus, multiply the fraction by A– B . A– B The factor A – B is called the conjugate radical expression of A + B. Common Factoring Methods • Method 1: Greatest common factor. • Method 2: Factoring by grouping. • Method 3: Factoring special binomials. • Method 4: Factoring trinomials. – Case 1: Leading coefficient is 1. – Case 2: Leading coefficient is not 1. Example: The Algebra of Complex Numbers Simplify the following complex number expressions. 2 i 1 2i 4 3i The product of two complex numbers leads to four products via the distributive property. Roots and Complex Numbers We earlier said that if a and b are real numbers, then a a a b ab and . If a and b are not real b b numbers, then these properties do not necessarily hold. For instance: 9 4 9 4 3i 2i 36 6 6i 2 6 In order to apply either of these two properties, first simplify any square roots of negative numbers by rewriting them as pure imaginary numbers. Types of Equations There are three types of equations: 1. A conditional equation has a countable number of solutions. For example, x + 7 = 12 has one solution, 5. The solution set is {5}. 2. An identity is true for all real numbers and has an infinite number of solutions. For example, x( x 1) x 2 x is true for all real number values of x . The solution set is R. 3. A contradiction is never true and has no solution. For example, x 6 x is false for any value of x. The solution set is Ø. Solving Linear Equations To solve a linear equation (in x): 1. Simplify each side of the equation separately by removing any grouping symbols and combining like terms. 2. Add or subtract the same expression(s) on both sides of the equation in order to get the variable term(s) on one side and the constant term(s) on the other side of the equation and simplify. 3. Multiply or divide by the same nonzero quantity on both sides of the equation in order to get the numerical coefficient of the variable term to be one. 4. Check your answer by substitution in the original equation. Example: Solving Absolute Value Equations 3x 2 5 Solve: Step 1: Rewrite the absolute value equation without absolute values. Step 2: Solve the two equations 3x 2 5 or 3x – 2 = -5 3x 7 or 3x = -3 7 3 or x 1 x Example: Absolute Value Equations Solve: 4x 3 2 0 |4x + 3| = -2 False, absolute value is never negative. No solution; the solution set is Ø. Solve: |6x – 2| = 0 6x – 2 = 0 6x = 2 x=⅓ If |ax + b| = 0, then ax + b = 0. Example: Linear Inequalities Solve the following linear inequality. 10 4( x 2) (2 x) Step 1: Distribute. Step 2: Combine like terms. Step 3: Divide by 5 . Note the reversal of the inequality sign. 10 4 x 8 2 x 4 x 18 2 x 5 x 20 x4 Solution is 4, Example: Solving Compound Inequalities Solve the compound inequality. 12 5 2 x 15 12 10 5 x 15 Note: each inequality is reversed since we are dividing by a negative number! 2 5 x 25 2 x 5 5 2 5 x 5 2 Solution is 5, 5 Example: Solving Absolute Value Inequalities Solve the following absolute value inequality. 5y 3 2 9 Step 1: Subtract 2. Step 2: Rewrite the inequality without absolute values. Step 3: Solve as compound inequality. 5y 3 7 7 5 y 3 7 4 5 y 10 4 y2 5 4 Solution is ,2 5 Example: Solving Quadratic Equations by Factoring Solve the quadratic equation by factoring. 3x 2 x 5 5 2 Step 1: Multiply both sides by 5 . Step 2: Subtract 2 from both sides so 0 is on one side. Step 3: Factor and solve the two linear equations. 5 x 2 3x 2 5 x 2 3x 2 0 (5 x 2)( x 1) 0 5x 2 0 2 x 5 or x 1 0 or x 1 Example: The Quadratic Formula Solve using the quadratic formula. 4 x 2 7 x 15 c a2 b 4x 7 x 15 0 b b 2 4ac x 2a x 7 7 289 x 8 7 17 x 8 5 x 3, 4 7 4 4 15 2 4 2 Operations with Rational Expressions • To add or subtract two rational expressions, a common denominator must first be found. • To multiply two rational expressions, the two numerators are multiplied and the two denominators are multiplied. • To divide one rational expression by another, the first is multiplied by the reciprocal of the second. • No matter which operation is being considered, it is generally best to factor all the numerators and denominators before combining rational expressions. Example: Add/Subtract Rational Expressions Subtract the rational expression. 5x 2 2x x2 5x 6 x2 4 5x 2 2x x 2 x 3 x 2 x 2 x 2 x 3 5x 2 2x x 2 x 2 x 3 x 3 x 2 x 2 5x2 8x 4 2 x2 6 x x 2 x 2 x 3 x 2 x 2 x 3 3x 2 14 x 4 , x 2 x 2 x 3 x 2, 2, 3 Step 1: Factor both denominators. Step 2: Multiply to obtain the least common denominator (LCD) and simplify. Example: Divide Rational Expressions Divide the rational expression. x 2 x 20 x 2 10 x 25 2x 6 x3 x 4 x 5 2x 3 6 x3 x 5 2 6 x x 4 x 5 3 2 x x 5 3x 2 x 4 x 5 2 , x 0,5 2 Example: Complex Rational Expressions Simplify the complex rational expression. Step 1: Multiply the numerator and the denominator by the LCD x z x . Step 2: Cancel the common factor of z. 1 1 xz x z 1 1 x z x x z x z x z x 1 x x z z x z x z z x z x 1 , x 0, z 0, x z 0 xx z Example: Work-Rate Problem One hose can fill a swimming pool in 10 hours. The owner buys a second hose that can fill the pool in half the time of the first one. If both hoses are used together, how long does it take to fill the pool? 1 The work rate of the first hose is 10 1 The work rate of the second hose is 5 1 1 1 Step 1: Set up the problem. 10 5 x Step 2: Multiply both sides by the x 2 x 10 LCD 10x , and solve. 3 x 10 1 x 3 hours 3 Example: Solving Radical Equations Solve the radical equation. 1 x 1 x 1 x x 1 1 x 2 x 1 2 1 x x2 2x 1 0 x 2 3x 0 x x 3 x 0, 3 x0 Note that 1 ( 3) ( 3) 1, so -3 is an extraneous solution. Recognizing Linear Equations in Two Variables Linear Equations in Two Variables A linear equation in two variables, say the variables x and y, is an equation that can be written in the form ax by c where a , b, and c are constants and a and b are not both zero. This form of such an equation is called the standard form. Example: Linear Equations Determine if the equation is a linear equation. 2 x 3 y 5x x 2 y 1 2 x 6 y 5x x 2 y 1 3x 6 y x 2 y 1 3x 6 y x 2 y 1 4 x 4 y 1 The equation is linear. Example: Intercepts Find the x- and y -intercepts of the equation and graph. 3 x 4 y 12 3 0 4 y 12 y 3 y-intercept: 0, 3 3x 4 0 12 x4 x-intercept: 4,0 Example: Finding Slope Using Two Points Determine the slopes of the line passing through the following points. y2 y1 m x2 x1 8,1 and 2,3 3 1 m 2 8 2 m 10 1 m 5 Slope-Intercept Form of a Line If the equation of a non-vertical line in x and y is solved for y, the result is an equation of the form y mx b. The constant m is the slope of the line, and the line crosses the y-axis at b ; that is, the y -intercept of the line is 0,b . If the variable x does not appear in the equation, the slope is 0 and the equation is simply of the form y b . Point-Slope Form of a Line Given an ordered pair x1 , y1 and a real number m, an equation for the line passing through the point x1 , y1 with slope m is y y1 m x x1 . Note that m, x1 , and y1 are all constants, and that x and y are variables. Note also that since the line, by definition, has slope m, vertical lines cannot be described in this form. The Slopes of Parallel and Perpendicular lines Important Parallel lines have the same slope. For example: 1 1 m1 and m2 . 2 2 Perpendicular lines have slopes that are negative reciprocals of each other. For example: 2 3 m1 and m2 . 3 2 Example: Slopes of Parallel Lines Find the equation, in slope-intercept form, for the line which is parallel to the line 8 x 2 y 10 and which passes through the point 1,5 . 8 x 2 y 10 Step 1: Write equation in 2 y 10 8 x slope-intercept form. y 4 x 5 Use slope m 4 to write a new equation that passes through the point 1,5 . Step 2: Use point-slope form. y 5 4 x 1 Step 3: Solve for y to obtain y 5 4 x 4 slope-intercept form. y 4 x 1 Functions and the Vertical Line Test Functions A function is a relation in which every element of the domain is paired with exactly one element of the range. Equivalently, a function is a relation in which no two distinct ordered pairs have the same first coordinate. Implied Domain of a Function The domain of the function is implied by the formula used in defining the function. It is assumed that the domain of the function consists of all real numbers at which the function can be evaluated to obtain a real number: any values for the argument that result in division by zero or an even root of a negative number must be excluded from the domain. Quadratic Functions and Their Graphs Vertex Form of a Quadratic Function 2 The graph of the function g x a x h k where a, h and k are real numbers and a 0 is a parabola whose vertex is (h,k). The parabola is narrower than f x x 2 if a 1 and is broader than f x x 2 if 0 a 1 . The parabola opens upward if a is positive and downward if a is negative. Commonly Occurring Functions Piecewise-Defined Function A piecewise-defined function is a function defined in terms of two or more formulas, each valid for its own unique portion of the real number line. In evaluating a piecewise-defined function f at a certain value for x, it is important to correctly identify which formula is valid for that particular value. Example: Commonly Occurring Functions Sketch the graph of the function 2 x 2 if x 1 f x 2 . x if x 1 The function f is a linear function on the interval , 1 and a quadratic function on the interval 1, . To graph f, we graph each portion separately, making sure that each formula is applied only on the appropriate interval. The complete graph appears to the right, with the points f(–4) = 6 and f(2) = 4 noted in particular. Also note the use of a closed circle at (–1,0) to emphasize that this point is part of the graph, and the use of an open circle at (–1,1) to indicate that this point is not part of the graph. Example: Variation Problems For the following phrases, write the general formula that applies. a. “y varies inversely as the n ͭ ͪ power of x” y k xn b. “y is directly proportional to the n ͭ ͪ power of x” y kx n c. “y is inversely proportional to the n ͭ ͪ power of x” k y n x d. “y varies directly as the n ͭ ͪ power of x” y kx n Shifting, Stretching and Reflecting Graphs Horizontal Shifting Let f(x) be a function whose graph is known, and let h be a fixed real number. If we replace x in the definition of f by x – h, we obtain a new function g x f x h . The graph of g is the same shape as the graph of f, but shifted to the right by h units if h > 0 and shifted to the left by h units if h < 0. Shifting, Stretching and Reflecting Graphs Vertical Shifting Let f(x) be a function whose graph is known and let k be a fixed real number. The graph of the function g x f x k is the same shape as the graph of f, but shifted upward if k > 0 and downward if k < 0. Shifting, Stretching and Reflecting Graphs Reflecting With Respect to the Axes Let f(x) be a function whose graph is known. 1. The graph of the function g(x)= –f(x) is the reflection of the graph f with respect to the x-axis. 2. The graph of the function g(x) = f(–x) is the reflection of the graph of f with respect to the y-axis. Shifting, Stretching and Reflecting Graphs Stretching and Compressing Let f(x) be a function whose graph is known, and let a be a positive real number. 1. The graph of the function g(x) = a f(x) is stretched vertically compared to the graph of f if a > 1. 2. The graph of the function g(x) = a f(x) is compressed vertically compared to the graph of f if 0 < a < 1. Combining Functions Arithmetically Addition, Subtraction, Multiplication and Division of Functions 1. f g x f x g x 2. f g x f x g x 3. f g x f x g x f x f x , provided that g x 0 4. g x g The domain of each of these new functions consists of the common elements (or the intersection of elements) of the domains of f and g individually. Composing Functions Composing Functions Let f and g be two functions. The composition of f and g, denoted f g , is the function defined by f g x f g x . The domain of f g consists of all x in the domain of g for which g(x) is in turn in the domain of f. The function f g is read “f composed with g,” or “f of g.” Example: Composing Functions Given f(x) = x2 + 2 and g(x) = x + 5 , find: Again, we know by definition f g x f g x that f g x f g x . f x 5 = (x + 5)2 + 2 = x2 +10x + 25 + 2 = x2 +10x + 27 Finding Inverse Function Formulas To Find a Formula for f 1 Let f be a one-to-one function, and assume that f is defined by a formula. To find a formula for f 1, perform the following steps: 1. Replace f x in the definition of f with the variable y. The result is an equation in x and y that is solved for y. 2. Interchange x and y in the equation. 3. Solve the new equation for y. 4. Replace the y in the resulting equation with f 1 x . Long and Synthetic Division Compare the division of 2 x3 8 x 2 9 x 7 by x 2 below, using long division on the left and synthetic division on the right. 2 x 2 4 x 1 x 2 2 x 3 8 x 2 9 x 7 2x 3 4 x 2 4 x2 9 x 7 4x 2 8 x x 7 x 2 5 2 2 8 9 7 4 8 2 2 4 1 5 Note: the numbers in blue are the coefficients of the dividend and the numbers in pink are the coefficients of the quotient and remainder. Synthetic Division Step 1: Write k and the coefficients of the dividend. Copy the leading coefficient of the dividend in the first slot below the horizontal line. Step 2: Multiply this number by k and write the result directly below the second coefficient of the dividend. Step 3: Add the two numbers in that column and write the result in the second slot below the horizontal line. Step 4: Repeat the process until the last column is completed and the last number written down is the remainder. Example: Solving Systems by Elimination Solve the system by the method of elimination. 5 x 2 y 6 2 x 3 y 10 To eliminate a variable, you’ll need to multiply each equation by a unique constant. Let’s eliminate y. To do so, notice that we’ll have to multiply the first equation by 3 and the second equation by 2. 3 5 x 2 y 6 2 2 x 3 y 10 5 2 2 y 6 y 2 15 x 6 y 18 4 x 6 y 20 19 x 38 x2 Thus, the solution is 2, 2 .