First Semester Final - Algebra 2 w/Trig Algebra 2 w/Trig...

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Unformatted text preview: Algebra 2 w/Trig Algebra 2 w/Trig First Semester PowerPoint Index of Sections Index of Sections P1 P2 P3 P4 P5 P6 P Quiz 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1 Quiz 2.1 2.2 2.3 2.4 2.5 2.6 2 Quiz 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3 Quiz P.1 : Real Numbers P.1 : Real Numbers By: Juliana Pajdo and Lindsay Smith Real Numbers Real Numbers • Real Numbers – Are used in everyday life to • describe quantities. They are represented by numbers like… 4 π , , 0.333, 2, 3 27, 4 3 Real Numbers Real Numbers • A real number is rational only if it can be written as a fraction… • The decimal representation must either repeat or terminate. • A number that does not fulfill these requirements are… 1 = .333 3 125 = 1.126 111 1 = .50 2 Irrational! Irrational! • Irrational numbers have infinite non­repeating decimal representations like… π = 3.141592... 2 = 1.4142135 The Real Number Tree The Real Number Tree Real Numbers Irrational Rational Integers Non-integers Fractions Negative Numbers Whole numbers Natural Numbers Zero Graphing Graphing • Real Numbers are represented graphically by a real number line. • The point of zero is the origin; numbers right of zero are positive, and those left are negative. |­­­|­­­|­­­|­­­|­­­|­­­|­­­|­­­|­­­|­­­|­­­|­­­| ­5 ­4 ­3 ­2 ­1 0 1 2 3 4 5 6 7 P.2 P.2 Exponents and Radicals Pages 12­23 Integer Exponents Integer Exponents Exponential Form a • a • a • a • a = a5 a = a • a • a • ...a n a is the base, n is the exponent. an is read “a to the nth power” Properties of Exponents Properties of Exponents x 2 + x3 = x5 1. Multiplication of like bases is done by adding exponents. x5 = x3 x2 2. Division of like bases is done by subtracting exponents. 3. a −n 1 1 = n = ,n ≠ 0 a a n 4. a = 1, a ≠ 0 0 5. ab ) = a mb m ( m **Be careful, 6. (a mn ) = a mn m ( −2 ) 4 ≠ −24 7. a a =m b b 2 2 m 8. a = a =a 2 Scientific Notation Scientific Notation 650, 000, 000, 000, 000 = 6.5 ×1014 0.000004 = 4 ×10−6 Find the location of the digits in the number after which a decimal should be placed to form a number between 1 and 10. Place a decimal point there. Count decimal places to find the exponent to which 10 should be multiplied. Moving left indicates a positive exponent, and moving to the right indicates a negative exponent. Radicals and Their Properties Radicals and Their Properties Squared root of a number is one of its two equal factors. Example: Cube root of a number is one of its three equal factors. Example: 5g = 25 5 53 = 125 Example 36 = 6 − 36 = −36 5 −32 = −2 125 5 = 64 4 4 3 −81 Properties of Radicals Properties of Radicals n a m = ( n a )m n n n a g b = n ab a na = ,b ≠ 0 n b b a = mn a n n mn ( a) = a For n even, n an = a a =a n For n odd, n Simplifying Radicals Simplifying Radicals A radical is in simplest form when… • All possible factors have been removed from the radical. • All fractions have radical –free denominators. • The index of the radical is reduced n a (index is the n in .) Example with even roots Example with even roots 4 48 = 4 16 • 3 = 4 24 • 3 = 2 4 3 Example with odd roots 3 24 = 3 8 • 3 = 3 23 • 3 = 2 3 3 Rationalizing Denominators And Rationalizing Denominators And Numerators • To rationalize a denominator or numerators of a −b m a form multiply both numerator and a+b m denominator by a conjugate: Example: Rationalize the expression . 2 2 3− 7 3+ 7 = = = 3+ 7 • 3− 7 3(3 − 7 ) (3) 2 −( 7 ) 2 2 3+ 7 2(3 − 7 ) 2 =3 − 7 P.3. P.3. Polynomials and Factoring Definitions! Definitions! • An algebraic expression is a An collection of variables and real numbers also known as real ax k a is the coefficient and k is the degree When a polynomial in x is written with descending powers of x it is written in: POLYNOMIALS!!! Examples: 2x + 7 4x − 6x + x + 4 4 2 Standard Form! −5 x + 4 x + 3 x − 2 7 2 4 x y + xy + 3 3 2 FOIL Method (Not Al) FOIL Outer First (2 x − 3)(7 x + 5) Inner Last Answer :14 x 2 −11x −15 Special Products Special Sum and Difference of Same Terms: (u + v) 2 = u 2 + 2uv + v 2 (u − v) 2 = u 2 − 2uv + v 2 Square of a Binomial: (u + v)(u − v ) = u 2 − v 2 Cube of a Binomial: (u + v) = u = 3u v + 3uv + v 3 3 2 2 3 (u − v)3 = u 3 − 3u 2 v + 3uv 2 − v3 More Definitions! More The process of writing a polynomial as a product is called factoring. If a polynomial cannot be factored using integer coefficients, it is prime or irreducible over the integer. A polynomial is completely factored when each of its factors is prime so: x − x + 4 x − 4 = ( x −1)( x − 4) 3 2 2 FACTORED! Factoring Special Polynomial Forms Factoring Special Polynomial Forms Difference of Two Squares: u 2 − v 2 = (u + v)(u − v) Perfect Square Trinomial: 2 2 u + 2uv + v = (u + v) 2 2 2 u − 2uv + v = (u − v) 2 Sum or Difference of Two Cubes: u + v = (u + v)(u − uv + v ) 3 3 2 2 u 3 − v 3 = (u − v)(u 2 + uv + v 2 ) Factoring By Grouping Factoring x − 2 x − 3x + 6 3 2 Then group them the terms ( x 3 − 2 x 2 ) − (3 x − 6) Then factor out the common factor. (x-2) x 2 ( x − 2) − 3( x − 2) FINAL ANSWER……… ( x − 3)( x − 2) 2 AMAZING FACTOR METHOD! AMAZING • • • • • • ax + bx + c 1. Multiply 1. ag c 2 1x − 5 x + 6 2. List all factors of 2. ag c 6g = 6 1 3. Identify any pair of 3. 1, 6 1x 1x number that’ll add to 2,3 −2 −3 equal b −2, −3 4. Write each factor as a 4. fraction. fraction. ( x − 2)( x − 3) 5. Simplify the fraction 6. Write as (ax + f1 )(ax + f 2 ) 6. 2 Now you try! Now FOIL: (1 − x )(4 x ) 3 Factor out GCF: 4x − 4x4 FOIL: 4 x 3 − 6 x 2 + 12 x 2 x(2 x − 3 x + 6) 2 ( x +10)( x −10) x −100 2 More Practice! More Completely Factor the Expressions: 2( x +1)( x − 3) − 3( x +1) ( x − 3) 2 2 −( x +1)( x − 3)( x + 9) 9 x +10 x +1 2 (9 x + 1)( x + 1) P.4 P.4 Rational Expressions Domain of an Algebraic Expression Domain of an Algebraic Expression Two algebraic expressions are equivalent if they have the same domain and yield the same values for all numbers in their domain. Examples: 2 x + 5x − 4 2 The domain is the set of all real numbers. The domain is all real numbers greater than or equal to 2. The domain is all real numbers except for 3 because the denominator cannot equal zero x −2 x+2 x −3 Simplifying Rational Expressions Simplifying Rational Expressions First, simplify fractions normally and identify any restrictions of the domain. Second, factor the polynomial. Next, simplify again. Then list restrictions of the domain. ac a = ,c ≠ 0 bc b x + 4 x −12 ( x + 6)( x − 2) x + 6 = = ,x ≠2 3x − 6 3( x − 2) 3 2 Simplifying Rational Expressions Simplifying Rational Expressions • Examples: Solution: 12 x + x − x 2 Write 2 in simplest form. 2x − 9x + 4 12 x + x − x 2 (4 − x)(3 + x) = 2 2 x − 9 x + 4 (2 x −1)( x − 4) −( x − 4)(3 + x) = (2 x −1)( x − 4) 3+ x =− ,x ≠4 2 x −1 Operations with Rational Operations with Rational Expressions: Division/Multiplication • Remember that to divide fractions, you flip the second fraction and multiply. x3 − 8 x 2 + 2 x + 4 Example: x 2 − 4 ÷ x3 + 8 x3 − 8 x 2 + 2 x + 4 ÷ 2 x −4 x3 + 8 x3 − 8 x3 + 8 =2 •2 x − 4 x + 2x + 4 ( x − 2)( x 2 + 2 x + 4) ( x + 2)( x 2 − 2 x + 4) = • ( x + 2)( x − 2) ( x 2 + 2 x + 4) = x 2 − 2 x + 4, x ≠ ±2 Operations with Rational Operations with Rational Expressions: Addition/Subtraction • Use the LCD method to add and subtract. a c ad ± bc ±= , b ≠ 0 and d ≠ 0 bd bd Operations with Rational Operations with Rational Expressions: Addition/Subtraction 3 2 x+3 Example: 3 2 x +3 −+2 x −1 x x − 1 −+ x − 1 x ( x + 1)( x − 1) 3( x)( x + 1) 2( x + 1)( x − 1) ( x + 3) x = − + x( x + 1)( x − 1) x( x + 1)( x − 1) x( x + 1)( x − 1) 3( x)( x + 1) − 2( x + 1)( x − 1) + ( x + 3) x = x( x + 1)( x − 1) 3x 2 + 3x − 2 x 2 + 2 + x 2 + 3x = x( x + 1)( x − 1) (3 x 2 − 2 x 2 + x 2 ) + (3 x + 3 x) + 2 = x( x + 1)( x − 1) 2x2 + 6x + 2 = x( x + 1)( x − 1) 2( x 2 + 3 x + 1) = x( x + 1)( x − 1) Complex Fractions Complex Fractions Fractional expressions with separate fractions in the numerator, denominator, or both are called complex fractions. 1 Examples: ( ) x x 2 +1 To simplify: First simplify the top and bottom separately. Then multiply the numerator by the reciprocal of the denominator. Simplifying a Complex Fraction SExample: implifying a 2 −3 2 − 3 x 1 1 − x −1 1 1− x −1 2 − 3( x) x = 1( x −1) −1 x −1 2 − 3x x = x −2 x −1 2 − 3 x x −1 = • x x −2 (2 − 3 x )( x −1) = , x ≠1 x( x − 2) x Section P.5 Section The Cartesian Plane What is a Cartesian Plane? What • It allows you to represent ordered pairs of real numbers by It points in a plane. points • Formed by using two real number lines intersecting at right Formed angles; the horizontal line is the x-axis and the vertical line is the y-axis. the Plotting Points on a Cartesian Plane Plotting (x ,y) Coordinate (x The “x” value indicates the horizontal movement on the x-axis. The “y” value indicates the vertical movement on the y-axis. - A positive number moves the point right that many places on the x-axis positive and up that many places on the y-axis and A negative number moves the point left that many places on the x-axis and negative down that many places on the y-axis. down Coordinate Plotted: (5,2) Sketching a Scatter Plot Sketching Components of a Scatter Plot: - Must represent each pair of data values with an ordered pair data - X & Y axis represent a axis different variable different - Points are not connected - Represents a change over Represents time, etc. time, - Unit labels - Consecutive intervals The Distance Formula The • d is the distance between points two pairs of (x,y) coordinates d = ( x2 − x1 ) + ( y2 − y1 ) 2 2 Finding a Distance Finding • Use the distance formula and substitute the Use • coordinates in for x and y coordinates EX: Find the distance between points (-2,1) and EX: (3,4) (3,4) d = [3 − (−2)]2 + (4 −1) 2 = (5) 2 + (3) 2 = 34 ≈ 5.83 Now you try! Now • Find the distance between Find points (2,3) and (4,1) points d = (4 − 2) 2 + (1 − 3) 2 = (2) + (−2) 2 2 = 8 ≈ 2.83 Verifying a Right Triangle Verifying Using the distance formula& the triangles vertices, you can find the lengths of the 3 sides of that said triangle. Example: Right triangle with the vertices (2,1) (4,0) & (5,7) Use substitution to put coordinates into the formula (5 − 2) 2 + (7 − 1) 2 = 9 + 36 = 45 First side ^ (4 − 2) 2 + (0 − 1) 2 = 4 + 1 = 5 Second side ^ (5 − 4) 2 + (7 − 0) 2 = 1 + 49 = 50 Third side ^ Finding the Midpoint Finding the Midpoint x1 + x2 y1 + y2 2,2 Use substitution to solve (4, 2) & (4,4) 4+4 2+4 2,2 midpoint= (4, 3) Writing the Equation of a Circle Writing the Equation of a Circle The standard form of the equation of a Circle: ( x − h) + ( y − k ) = r 2 2 2 Substitute: (x, y) for a given point on the radius (h, k) for the center point & “r” for the radius Then………… SOLVE!! Translating Points on a Plane Translating Points on aPlane Ways to translate: 1. Adding # to the x­coordinate moves the point vertically 2. Adding # to the y­coordinate moves the point horizontally postive # will move it up& to the right and negative # will move it down& to the left Fin Fin P.6 P.6 Exploring Data: Representing Data Graphically By Mihret Moges & Unmool Akram Definitions Definitions • Line Plot­ Uses a portion of real number line to • • order numbers. Frequency­ Measures the number of times of value occurs in a data set. Bar Graphs and Histograms: The differences between these two are bar graphs can be horizontal or vertical and the labels of the bars are not necessary numbers. Bars in a bar graph are usually separated spaces. Examples Examples • Constructing a Line Plot Numbers given: 70 76 67 64 76 66 74 69 64 74 76 76 74 70 x x x x x x x x x x x x x ­*­­­­*­­­*­­­­*­­­*­­­­*­­­­­­*­­­­­­*­­­­­*­­­­*­­­*­­­­­*­­­*­­­­­­*­­­­­­ 65 70 75 Range is the difference between the greatest and least data values. Range: 12 or (64,76) • Construct a Histogram 100, 101, 102,103,104,105,106,107,108,109,110,1 07,102, 109,103, 101 Tallies 100 ­102 103­105 106 ­108 109 ­111 lllll llll llll lll Unit Sales 6 Number of sales representative 5 4 3 2 1 0 100- 103- 106- 109102 105 108 111 units sold Series1 Example Example Graph: Example Example • Construct Bar Graph January February March April May June July August September October 3.7 3.0 3.4 3.6 5.2 5.4 3.2 3.8 4.3 4.5 November December 4.2 3.7 Amount of rain(inches) Monthly percipatation( in inches) Graph : 6 5 4 3 2 1 0 ar ch Ju ly M Se pt M No v Ja n ay Series1 Month Example Example Decades 1901­1910 1911­1920 1921­1930 1931­1940 1941­1950 1951­1960 1961­1970 1971­1980 1981­1990 1991­ 2000 • Construct a Line Graph 8795 5736 4107 528 1035 2515 3322 4493 7338 9095 Numbers U.S. Immigrations 10000 # of Immigrants (in thousands) 8000 6000 4000 2000 0 19 01 -1 19 910 21 -1 19 930 41 -1 19 950 61 -1 19 970 81 -1 99 0 Series1 Decades In Thousands How to put it in your Calculator How to put it in your Calculator • Line Graph­ Chart of Scattered data Calc – Stat – Enter – 2nd – Y= ­ Enter ­ Enter Extra help for u Extra help for u • See page 63­65 For line plot # 2 & 3 For Histograms # 5 For Bar Graphs # 7 & 8 Line Graphs # 16 & 17 Chapter P Quiz ! Chapter P Quiz ! Chapter P Quiz ! Chapter P Quiz ! Chapter P Quiz ! Chapter P Quiz ! Chapter P Quiz ! Chapter P Quiz ! Chapter P Quiz Chapter P Quiz ! ! Chapter Chapter P Quiz ! Chapter P Quiz ! Chapter P Quiz ! Chapter P Quiz ! Chapter P Quiz ! L E S S O N 1! #1. Determine which numbers are (a) natural numbers, (b) whole numbers, (c) integers, (d) rational numbers, and (e) irrational numbers. 85 11, − 14, − , , 6, 0.4 92 SOLUTION 85 ( a)11 (b)11(c)11, − 14 (d )11, − 14, − , , 0.4 (e) 6 92 #2 Find the distance between a & b . a = -74 b = 48 SOLUTION | -74 – 48 | = 122 # 3 Use absolute value notation to describe the situation The distance between x and 8 is at least 3 SOLUTION x −8 ≥ 3 The distance between y and 30 is less than 5 SOLUTION - y + 30 < 5 #4 Evaluate the expression for each value of x. Expression 10x - 3 -2x2 – x +3 Values x= -1 , x= 3 x = 3 , x= -3 SOLUTIONS -13 , 27 - 18 , -12 #5 Perform the operation in simplest form. 28 + 39 39 ÷ 16 2 SOLUTION 14 9 SOLUTION 1 24 L E S S O N 2 ! #6 Simplify each expression (a) (-2z)3 (b) (a2b4)(3ab-2) (c) 62u3v-3 12u-2v (d) 3-4m-1n-3 9-2mn-3 SOLUTION SOLUTION SOLUTION -8z3 3a3b2 3u5 v4 m-2 SOLUTION #7 Simplify by removing all possible factors from the radical (a) 4 x 81 (b) 144 2 x3 (c ) 3 27 4 SOLUTION 2x 3 4 2 SOLUTION SOLUTION x3 2 3 #8 Simplify the expression 50 − 18 8x + 2 x 3 SOLUTION 22 SOLUTION 3 3x 2 x (2 x + 1) 8 3x − 5 3x SOLUTION #9 Simplify the expression 81 2 3 2 SOLUTION 5 25 (−3 x 5 )(−2 x 2 ) 1 SOLUTION 729 L E S S O N 3 ! #10 Perform the operations and write the results in standard form. PROBLEM SOLUTION −(3 x + 2 x) + (1 − 5 x) 2 −3x 2 − 7 x + 1 x5 − 2 x 4 + x3 − x 2 + 2 x − 1 ( x 2 − 2 x + 1)( x3 − 1) (2 x3 − 5 x 2 + 10 x − 7) + (4 x 2 − 7 x − 2) 2 x3 − x 2 + 3x − 9 GRAPHING REAL NUMBERS GRAPHING REAL NUMBERS a=2 |---|---|---|---|---|---|---|---|---|---|---|---| -5 -4 -3 -2 -1 0 1 2 3 4 5 6 |---|---|---|---|---|---|---• ---|---|---|---|---| -5 -4 -3 -2 -1 0 1 2 3 4 5 6 GRAPHING REAL NUMBERS GRAPHING REAL NUMBERS b =− 5 -|---|---|---|---|---|---|---|---|---|---| -5 -4 -3 -2 -1 0 1 2 3 4 5 6 Interpreting Inequalities Interpreting Inequalities Interval Type Notation Inequality Graph [--------] A B (--------) A B [--------) A B (--------] A B Closed! Open! [a,b] (a,b) ----- a< x<b a ≤ x <b a ≤ x ≤b [a,b) ------ (a,b] a< x≤b Unbounded Intervals on the Real Unbounded Intervals on the Real Number Line Notation Interval Type Inequality Graph ­­ Open ­­ x≥ a x> a x≤ b x< b −∞ < x < ∞ ­­­­[­­­­­ x ­­­­(­­­­­ x ­­­­]­­ x ­­­­)­­ x ­­­­­­ x (−∞, b) (−∞, ∞) Open Entire real line Absolute Value Absolute Value • Definition: the distance between the origin and the point representing the real number on the number line. – So, if a is a real number, the absolute value of a is: | a |= a, if a ≤ 0 | a |= −a, if a > 0 Algebraic Expressions Algebraic Expressions • Variables – letters that stand for unknown quantities, plus any coefficients (the number(s) that come before the letters). • Terms – parts of an algebraic expression that are separated by addition. Chapter 1 Chapter Section 1.1 Vocabulary Vocabulary • Equation- the relationship between to Equationquantities • Solution point- if x=a and y=b satisfies the Solution equation then (a, b) is a solution point equation • Graph of the equation- the set of all solution Graph points in an equation points Example of determining solution points points Determine whether (a) (2,13) and (b) (-1,-3) lie on the graph of y=10x-7 A) y = 10 x − 7 13 = 10(2) − 7 13 = 13 B.) y = 10 x − 7 −3 = 10( −1) − 7 −3 ≠ 17 • Sketching the graph of an equation by point Sketching plotting plotting 1. If possible, rewrite the equation so that one of the If variables is isolated on one side of the equation. variables 2. Make a table of values showing several solution Make points points 3. Plot these points on a rectangular coordinate Plot system system 4. Connect points with a smooth curve or line Connect Example of Point Plotting Example -1 9 (-1,9) 0 6 (0,6) 1 3 (1,3) 2 0 (2,0) 3 -3 (3,-3) See next slide for Graph 10 8 6 4 2 -10 -5 5 10 -2 -4 -6 Using a Graphing Utility to Graph and Equation Equation 1. Rewrite the equation so that y is isolated on Rewrite 2. 3. 4. the left side. the Enter the equation into the graphing utility. Determine a viewing Window that shows all Determine viewing important features of the graph. important Graph the equation Graphing a Circle Graphing x 2 + y 2 = 9 is a circle whose center is the origin and whose radius is 3. Solve for y. Solution x + y =9 2 2 y =9− x 2 2 2 y = ± 9− x Example from homework Example from homework • Sketch the graph of the equation y = −4 x +1 See next slide for graph f( -4 ⋅ x) x+1 = 4 2 5 5 2 4 Example from homework Example from homework • Complete the table x y Solution point ­1 0 1 3/2 x y ­1 5 0 3 (0,3) 1 1 (1,1) 3/2 0 (3/2,0) Solution (­1,5) points 1.2 1.2 Lines in the Plane The Slope of a Line The Slope of a Line • Slope – The number of units the line rises or falls vertically for each unit of horizontal change from left to right. • The Variable for slope is m • Equation: y2 − y1 ∆y m= = x2 − x1 ∆x Tips for Finding Slope Tips for Finding Slope • Order of subtraction is very important!!! • Correct Examples: y2 − y1 m= x2 − x1 y1 − y2 m= x1 − x2 • Wrong! y2 − y1 m= x1 − x2 Finding The Slope of a Line Finding The Slope of a Line Example 1: (­2,0) and (3,1) y 2 − y2 1−0 1 1 m= = = = x2 − x1 3 − (−2) 3 + 2 5 Example 2: (0,4) and (1,­1) y2 − y1 −1 − 4 −5 m= = = = −5 x2 − x1 1−0 1 Undefined Slope Undefined Slope • The definition of slope does not apply to vertical lines because the slope of a vertical line is undefined. • Example: y2 − y1 2 −2 0 m= = = =0 x2 − x1 2 − (−1) 3 The Slope of a Line The Slope of a Line 1. A line with positive slope (m>0) rises from left to right. 1. A line with negative slope (m<0) falls from left to right. 1. A line with zero slope (m=0) is horizontal. 1. A line with undefined slope is vertical. The Point­Slope Form of the Equation of a Line The Point­Slope Form of the Equation of a Line 1.3 1.3 Functions By: Kaci Casper and Yesenia De Leon Vocabulary Vocabulary • Relation­ pairs of quantities that are related to each other by some rule of correspondence A =πr x y x y 1 9 1 9 2 • Function­ A relationship between two variables such that to each value of the independent variable there corresponds exactly one value of the dependent variable. 23456 13 15 15 12 10 12345 13 15 15 12 10 Vocabulary Vocabulary • Domain: set of inputs. • Input: x­value. • Range: Set of outputs. • Output: y­value Domain/Input Range/Output x y 1 9 2 3 4 5 6 13 15 15 12 10 Function Rules and Testing Functions Function Rules and Testing Functions 1. 2. 3. 4. Each element of A must be matched with an element of B. Some elements of B may not be matched with any element of A. Two or more elements of A may be matched with the same element of B. An element of A (the domain) cannot be matched with two different elements of B. Vocabulary Vocabulary f • Function Notion: y = (x) • is the name of the function. • Y is the Dependent Variable, or the output value. • X is the Independent variable, or the input value. • (x) is the value of the function at x. f f Vocabulary Vocabulary • Implied Domain­ Consists of all real numbers for which the expression is defined • Difference Quotient­ The ratio of the equation. – Ex: f ( x + h) − f ( x ) , h ≠0 h Section 1.4 Section 1.4 Graphs of a Function Graphs of a Function Graphs of a Function • The graph of a function f is the collection of ordered pairs (x, f(x)) where x is the domain of f. x= the directed distance from the y­axis f(x)= the distance from the x­axis • • f ( x) = x + 2 x − 8 2 5 5 2 (− , ) ∞∞ Domain= − ∞ ( 9, ) Range= 2 • • 4 6 8 Finding the Domain and Range of a Function Finding the Domain and Range of a Function Range­the vertical distance of the function Domain­ the horizontal distance of the function • In the graph to the left, the domain and the range are all real numbers. 4 2 5 5 Finding the Domain and Range of a Function Finding the Domain and Range of a Function f ( x) = x − 4 x −4 ≥ 0 x≥4 • This is because there cannot 2 be a negative number under the radical, so the domain is the set of all real numbers greater than or equal to 4. 5 10 -2 Find the domain and range yourself… Find the domain and range yourself… f ( x) = x + 8 x +15 2 0 = x +8 x + 15 2 0 = ( x + 3)( x + 5) x = −3 x = −5 Vertical Line Test for Functions Vertical Line Test for Functions • If there are any points in the graph that cross vertically, it is not a function. Increasing and Decreasing Functions Increasing and Decreasing Functions • If increasing on any interval, for any x1 < x2 implying f ( x1 ) < f ( x2 ) • If decreasing on any interval, for any x1 < x2 implying f ( x1 ) > f ( x2 ) • If constant on any interval, f ( x1 ) = f ( x2 ) Increasing and Decreasing Graph Increasing and Decreasing Graph 2 <decreasing -5 5 <increasing increasing> 2 Relative Minimums and Maximums Relative Minimums and Maximums • Relative minimum of f if there is an interval • x < x < x implies f (a ) ≤ f ( x) ( x1 , x2 ) 1 2 Relative maximum of f if there is an interval ( x1 , x2 ) x1 < x < x2 implies f (a ) ≥ f ( x) minimum 2 4 maximum 1 8 2 2 1 6 2 0 1 4 1 8 Even Function Even Function A function with a graph that is symmetric with respect to the y-axis. 2 f ( −x ) = f ( x) 5 5 2 Odd Function Odd Function A function with a graph that is symmetric with respect to the origin. 2 f (− x) = − f ( x) 5 5 2 Checking If Equations Are Even or Odd Checking If Equations Are Even or Odd f ( x) = x − 7 x 3 f ( x) = x − x 4 2 f (−x) = −x + 7 x 3 This is an odd equation!!! This is an even equation!!! f (−x) = x 4 − x 2 Section 1.5 Section 1.5 Shifting, Reflecting and Stretching Graphs Graphs of Common Functions Graphs of Common Functions • Constant Function, f(x)= 5 f(x) = 5 4 2 -5 5 -2 -4 Graphs of Common Functions Graphs of Common Functions • Identity (Linear) Function, f(x)=x g(x) = x 4 2 5 5 2 4 Graphs of Common Functions Graphs of Common Functions • Absolute Value Function, f(x)= x 8 6 h(x) = x 4 2 -5 5 -2 Graphs of Common Functions Graphs of Common Functions • Square Root Function, f(x)= r(x) = x x 4 2 -5 5 -2 -4 Graphs of Common Functions Graphs of Common Functions • x2 Quadratic Function, f(x)= 8 6 s( x) = x2 4 2 5 5 2 Graphs of Common Functions Graphs of Common Functions • Cubic Function, f(x)= x 3 t( x) = x3 4 2 5 5 2 4 Transformations Transformations • Shifts, stretches, shrinks and reflections are all Transformations, that form complicated graphs which are derived from more simple graphs (look at common graphs). Vertical and Horizontal Shifts Vertical and Horizontal Shifts • To perform a vertical shift you must add or • subtract a constant to the equation. x2 x2 + 2 Ex: f(x)= (solid) f(x)+2= (Dashed) 8 6 u( x) = v(2+2 x) =x 4 x2 2 5 5 2 Vertical and Horizontal Graphs Vertical and Horizontal Graphs • To perform a horizontal shift you must add or • 8 subtract a constant directly to the variable. 2 2 x − 2) ( x Ex: f(x)= (solid) f(x­2)= (dashed) 6 w( x) = x2 f1( ) x)2 =( x-2 4 2 5 5 1 0 2 Vertical and Horizontal Shifts Vertical and Horizontal Shifts • Vertical shift c units upward: f(x)=h(x) + c • Vertical shift c units downward: f(x)= h(x)+ c • Horizontal shift c units to the right: f(x)= h(x­c) • Horizontal shift c units to the left: f(x)= h(x + c) C is a positive real number. Exercises Exercises • x3 + 3 f(x)= What graph is this equation shifted to the left 2 units? A CORRECT ANSWER Equation: g 1( x) = x3+3 6 4 B 4 2 2 5 5 5 5 2 f ( x) = ( x + 2)3 + 3 4 2 4 4 1 0 To shift 2 units left, you must put (x+2) for C D x. 2 5 5 8 6 4 2 2 5 5 4 Exercises Exercises • A x3 + 3 f(x)= What is the graph for this equation shifted down 3 units? 4 6 B 2 4 2 CORRECT ANSWER! 5 5 2 5 5 2 4 Equation: f ( x) = x 3 4 4 C 2 To shift down 3 units, you must subtract 3 from the right side of the equation. D 3 8 6 5 5 5 5 2 f ( x) = ( x + 3) − 3 f ( x) = x 3 4 2 2 4 Reflecting Graphs Reflecting Graphs • To reflect over the x­axis you must negate the • entire equation: f ( x) = x 2 +1 − f ( x ) = − x 2 −1 (solid) (dashed) 4 2 5 5 2 4 Reflections Reflections • To reflect over the y­axis you must negate only the variable: • (solid) (dashed) f ( x) = x3 + x − 2 f ( − x) = (− x)3 − x − 2 4 2 5 5 2 4 Examples Examples • Reflect this graph over the x­axis. f ( x) = x − 3x + 2 3 2 A CORRECT ANSWER! 4 2 5 5 B 2 1 0 8 Equation: f ( x) = − x + 3x − 2 3 2 4 6 4 2 C You must negate the entire equation to reflect over the x-axis. D 4 2 5 5 4 2 5 5 5 5 2 2 4 4 Examples Examples A 4 f ( x) = x + 9 Reflect over the y-axis. B 2 CORRECT ANSWER! 2 -5 4 -5 5 Equation: f ( x ) = −x + 9 5 -2 -4 -2 -4 C D 4 2 You must negate the variable alone to reflect over the y-axis. 4 2 -5 5 -5 5 -2 -2 -4 -4 Nonrigid Transformations Nonrigid Transformations • Nonrigid Transformations cause distortions, or change • in the shape of the original graph. f ( x) = x f ( x) = 4 x (solid) (dashed) 1 0 8 6 4 2 5 5 Transformation represented by cf(x)=cx, in this case c=4. Combinations of Functions Combinations of Functions By Aharon Brown and Robby Browder Arithmetic Combinations of Functions Arithmetic Just as two real numbers can be Just combined by the operations of addition, subtraction, multiplication, and division to form other real numbers, two functions can be combined to create )new(functions. 3) +( x − create +g x) =(2 x − f (x 1) 2 =x 2 +2 x − 4 f ( x ) −g ( x ) = x − − x 2 − (2 3) ( 1) =− 2 +2 x −2 x f ( x )g ( x ) =(2 x − g 3)( x 2 − 1) =2 x 3 − x 2 −2 x + 3 3 f ( x) 2x − 3 =2 , g ( x) x− 1 x ≠± 1 Sum Difference Product Quotient Sum, Difference, Product, and Quotient of Functions Sum, Difference, Product, and Quotient of Functions f • Let and be two functions g with overlappping domains. Then, for all common to both domians, thesum, the g f difference the product, and the quotient of and are defined as follows. 1. Sum : f + g x = f x + g x )( ) ( ) ( ) 2.Difference : ( f − g ) ( x ) = f ( x ) − g ( x ) 3.Product : ( fg ) ( x ) = f ( x ) ⋅ g ( x ) f ( x) f 4.Quotient : ( x ) = , g ( x) ≠ 0 g ( x) g ( Definition of Composition of Two Functions Definition f • The composition of the function (f) The with the function (g) is denoted. with ( f o g )( x ) = f ( g ( x )). • The domain of The f og • Is the set of all x in the domain of g Is such that g(x) is in the domain of f. such Example Example 36. f ( x ) = 3 x − 1 g ( x ) = x3 + 1 find f o g , g o f f og 3 go f ( x + 1) − 1 3 ( 3 x −1 + 1 x −1 +1 x ) 3 Examples Examples 71. find the domains of , f , g , and f o g find domain of f . All real numbers find the domain of g All real numbers except x = ±2 find the domain of f o g All real numbers except x = ±2 Example Example 5( 4 − x) + 4 20 − 5 x + 4 24 − 5 x 45. a ) find ( f o g ) ( x ) b) find ( g o f 4 − ( 5x + 4) −5 x ) ( x) c) 24 − 5 x ≠ −5 x 1.7 I NVERSE FUNCTI ONS 1.7 By: Claire and Salia Properties of Inverse Functions Properties of Inverse Functions f – The domain of is equal to the range of (inverse of ) – When you form the inverse of a function or vice versa, you obtain the identity function. f −1 f Ex. f ( f ( x)) = f ( x − 4) = ( x − 4) + 4 = x f ( f ( x)) = f ( x + 4) = ( x + 4) − 4 = x −1 −1 Definition of an Inverse Function Definition of an Inverse Function Let f and g be two functions such that for every x in the domain of g f ( g ( x)) = x and for every x in the domain of f g ( f ( x)) = x Under these conditions, the function g is the inverse function of f. f −1 The function g is denoted by . SO: f ( f −1 ( x)) = x and f −1 ( f ( x)) = x f The domain of must be equal to the range of , and the range of f −1 f −1 f must be equal to the domain of . The Graph of an Inverse Function The Graph of an Inverse Function • The graphs of a function and its inverse f function are related to each other in the f −1 following way. If the point (a,b) lies on the graph of , then the point (b,a) must lie on the graph f −1 of ,and vice versa. This means that the f −1 f graph of is a reflection of the graph of in f y=x the line . Existence of an Inverse Function Existence of an Inverse Function • Definition of a one to one function 1. A function is one­to­one if, for a and b in its f a =b domain, implies that . f (a ) = f (b) Existence of an inverse function Existence of an inverse function (cont.) • Existence of an Inverse Function 1. A function has an inverse function if −1 f f f and only if is one­to­one. Finding Inverse Functions Finding Inverse Functions Algebraically • Finding an Inverse Function 1. Use the horizontal line test to decide if has an inverse f function. 2. In the equation for , replace by y. f ( x) f ( x) 3. Interchange the roles of x and y, and solve for y. f −1 ( x) 4. Replace y by in the new equation. −1 f 5. Verify that and are inverse functions of each f other by showing that the domain of is equal to the f −1 range of , the range of is equal to the domain of f f −1 −1 f , and and equal x. f ( f −1 ( x)) f ( f ( x)) 2.1 2.1 Pages 162­171 Vocabulary Vocabulary • Identity­ An equation that is • true for every real number in the domain of the variable. Conditional Equation­ A equation that is true for some (or even none) of the real numbers in the domain of the variable. Linear in one Variable x­ An equation that can be written in the standard form ax+b=0 (a, b, are real. A not 0.) Extraneous solution­ One that does not satisfy the original equation. Formulas­ Ready made equations. • • • Solving an Equation with Fractions Solving an Equation with Fractions x x −= 3 5 2 Original Equation xx 10( − ) =10(3) 52 Multiply Each side by the LCD Put x together Simplify Divide by -3 on both sides Simplify your answer 2 x − 5 x = 30 −x = 3 30 − x( 3 1 1 ) =30( ) − 3 − 3 x =− 10 An Equation with an Extraneous Solution An Equation with an Extraneous Solution x 6 x 19 + = 2 7 14 −2 6(−2) 19 + = 2 7 14 −1 + −12 19 =− 7 7 Does x equal -2? Solve to find if it is an extraneous solution. Plug in -2 for x Simplify 19 19 − ≠ 7 14 Obliviously not equal A Distance Problem A Distance Problem Equation! D= rt Distance=Rate x Time You are driving the Canadian Freeway on your way to see a moose, which is 300 kilometers from your house. After thirty minutes, you are fifty kilometers away from your house. Assume you are going the same speed for the rest of the epic adventure, how long will it take you to get to the moose (if it is still there)? Plug in numbers to the distance equation. 50 300 = x 30 50 (300)30 =( x )30 30 Multiply both sides by 30 9000 = x 50 Simplify 9000 50 x = 50 50 Divide each side by 50 x = 180 min or 3 hours Height Height To obtain the height of a barn silo, you measure the silo’s shadow, and find that it is 80 feet long. You also measure the shadow of a 4 foot stake, and find that it is 3 ½ feet long. Height of building Height of stake = Length of building ' s shadow Length of stake ' s shadow X 4 ft = 1 80 ft 3 ft 2 Insert known variables into the equation. 3 1 X = 320 ft 2 Cross multiply Get X by its self 320 X= 1 3 2 X = 91.4 ft Simplify Formulas Formulas A = s2 A =lw A = r2 π P = 4s p = 2l + 2w C = 2π r 1 A = bh 2 P = a+b+c V =s 3 V = lwh V =πr h 2 43 V = πr 3 Using A Formula Using A Formula The triangular sail of a 125 foot long pirate ship has an area of 182.25 square feet. The sail has a base of 13.5 feet. What is the height of the sail? 1 A = bh 2 182.25 = 1 (13.5) h 2 Triangle Area Equation Insert known variables Multiply by 2 to get rid of the 1/2 Simplify Divide each side by 13.5 to get h by its self Simplify for the answer 1 2(182.25) = ( (13.5)h)2 2 364.5 = 13.5h 364.5 13.5h = 13.5 13.5 h = 27 ft Practice Problems Practice Problems an identity or a conditional Determine whether the equation below is equation. 2( x −1) = 2 x − 2 It is an identity because every real number is a domain of the variable. Practice Problems Practice Problems Solve for x. Solve for h 7 8x − = −4 2x +1 2x −1 x=11/6 1 A = bh 2 2 A = bh 2A h= b Section 2.2 Section 2.2 Baylor Dowdy and Ryan Murphy Intercepts, Zeros, and Solutions Intercepts, Zeros, and Solutions • X­intercept – The point (a,0) if it is a solution point of the equation. • Y­intercept – The point (b,0) if it a solution point of the equation. To find the intercepts To find the intercepts • X­intercept – To find the X­intercept, set the y equal to 0 and solve for x • Y­intercept – To find the y­intercept, set x equal to 0 and solve for y • Note! There may be more than one x or y intercept! There may also be no intercept! Your turn! Your turn! 1. y = 5 x − 10 3. x = 5 y 2. 2 x − 3 y = 4 4. x = 10 − y 1. y = 5 x − 10 • X­intercept – Set y=0 • New equation: 0=5x­10 – Add 10 to each side • New equation: 10=5x – Divide by 5 • New equation – Final solution: x=2 10/5=x • Y­intercept – Set x=0 – New equation: y=5(0)­10 – y= ­10 2. 2 x − 3 y = 4 • X­intercept – Set y=0 • New equation: – Divide both sides by 2 • New equation: – Final solution: x=2 x=4/2 2x­3(0)=4 • Y­intercept – Set x=0 • New equation: – Divide both sides by ­3 4 • New equation: – Final solution: 4 2(0)­3y=4 y =− 3 y =− 3 3. x = 5 y • X­intercept • Y­intercept – Set y=0 • New equation: x=5(0) – Final solution: x=0 – Set x=0 • New equation: 0=5y – Divide both sides by 5 0 • New equation: y= 5 – Final solution: y=0 4. x = 10 − y • X­intercept – Set y=0 • Y­intercept • New equation: x=10­0 – Set x=0 • New equation: 0=10­y – Subtract 10 from both sides • New equation: ­y = ­10 – Divide both sides by ­1 • New equation: − y −10 = −1 −1 – Final Solution: x=10 – Final solution: y=10 Zeros of Functions Zeros of Functions • A zero of a function • y=f(x) is a number a such that f(a)=0. To find the zeros of a function, you must solve the equation f(x)=0. 1. The point (a,0) is an 2. 3. x­intercept of the graph of y=f(x) The number a is a zero of the function f. The number a is a solution of the equation f(x)=0. Try one! Try one! Verify that the real numbers -2 and 3 are zeros of the function f ( x) = x − x − 6 2 f ( x) = x − x − 6 2 • Verify that ­2 is a solution – Substitute ­2 for x – New equation: • Verify that 3 is a solution – Substitute 3 for x – New equation: • Simplify • ­2 is a zero of 2 • Simplify • 3 is a zero of 2 f ( x) = x − x − 6 f ( x) = x − x − 6 Finding Solutions Graphically Finding Solutions Graphically Graphical Approximations of Solutions of an Equation 1. Write the equation in general form, f(x)=0, with the nonzero terms on one side of the equation and zero on the other side. 2. Use a graphing utility to graph the function y=f(x). Be sure the viewing window shows all the relevant features of the graph. 3. Use the zero or root feature or the zoom and trace features of the graphing utility to approximate the x-intercepts of the graph of f. Technology Tip! Technology Tip! 1. With each successive zoom­in, adjust the x­ 2. 3. scale (if necessary) so that the resulting viewing window shows at least the two scale marks between which the solution lies. The accuracy of the approximation will always be such that the error is less than the distance between two scale marks. If you have a trace feature on your graphing utility, you can generally add one more decimal place of accuracy without changing the viewing window. Your turn! Your turn! Use a graphing utility to approximate the solutions of 2 x − 3x + 2 = 0 2 Your graph should look something Your graph should look something like this… Points of Intersection of Two Points of Intersection of Two Graphs Point of Intersection- ordered pair ( things like (a,b)) on the graph that is a solution of two different equations. Use a graphing utility to find the points of intersection. y = x+ 2 2 y = x − 2x − 2 Graph Graph Check! Check! • • • • • • Check that (­1, 1) is a solution Equation 1:y=­1 +2=1 Equation 2: y=(­1)^2­2(­1)­ 2=1 Check that (4,6) is a solution Equation 1: y=4+2=6 Equation 2: y=(4)^2­2(4)­2=6 Finding Points of Intersection Finding Points of Intersection Find the points of intersection of the graphs of: 2 x − 3 y = −2 4x − y = 6 Solve Algebraically! Solve Algebraically! • Solve for y! • 2 2 y = x+ 3 3 Set them equal to y =4 x −6 each other! 2 2 x + = 4x − 6 3 3 2 x + 2 = 12 x −18 −10 x = −20 x =2 When x=2, then y=2. So, the point of intersection is (2,2). Try some ON YOUR OWN! Try some ON YOUR OWN! 1.Solve. M = 20.5t + 557, 0 ≤ t ≤ 11 E = −34.4t +1070, 0 ≤ t ≤ 11 2.Solve. ( x + 2) 2 = x 2 − 6 x +1 3.Verify the zero. x + 2 x −1 − −1 3 5 x =1 Problem 1 Problem 1 • 20.5t+557 = ­34.4t + • • 1070 54.9t +557=1070 54.9t= 513 513 t= 54.9 t ≈ 9.34 Problem 2 Problem 2 x2 + 4x + 4 = x2 − 6x + 1 10 x = −3 x =− 3 10 Problem 3 Problem 3 1+ 2 1−1 − −1 3 5 3 − 0 −1 3 1 −1 = 0 0=0 2.3 2.3 COMPLEX NUMBERS DIRECTOR: KANE STRATMAN EXECUTIVE PRODUCER: JACOB TODD STARRING: i THE IMAGINARY UNIT i • Some equations have Some • no solution no You can use i tto solve o them them x +1 = 0 2 x = −1 2 • x = −1 i = −1 COMPLEX NUMBERS COMPLEX • You can use a • multiple of i to express these numbers these The standard form is The a+bi a+b −9 − 5 = 3 ( −1) − 5 = 3 −1 − 5 2 3i − 5 = −5 + 3i DEFINITION OF COMPLEX # DEFINITION OF COMPLEX # • If a and b are real numbers, a=bi is a complex number, and it is said to be written in standard form. If b=0, the number a+bi is called an imaginary number. A number of the form bi, where b is not equal to 0, is called a pure imaginary number. Equality of Complex Numbers Equality of Complex Numbers c+di, written • Two complex numbers a+bi c+di, written in standard form, are equal to each other a+bi=c+di If and only if a=c and b=d Addition and Subtraction Addition and Subtraction • SUM: (a+bi)+(c+di) = (a+c) + (b+d)i • DIFFERENCE: (a+bi) – (c+di) = (a­c) + (b­d)i Example Example (3­i)+(2+3i)=3­i+2+3i =3+2­i+3i =(3+2)+(­1+3)i =5+2i Multiplying Multiplying • (a+bi)(c+di)=a(c+di)+bi(c+di) ac=(ad)i+(bd)i2 ac+(ad)i + (bc)i + (bd)(­1) ac­ bd+ (ad)i +(bc)i = (ac­bd) + (ad+bc)i Multiplying Conjugates Multiplying Conjugates • The product of two complex numbers can be a real number (3­5i)(3+5i)=32­(5i)2 9 ­ 25i2 = 9 ­ 25 (­1) = 34 Fractals Fractals • To help give an idea of how complex numbers can be applied, fractal geometry is used Imaginary Axis 2+3i 2 4 -1+2i -5 4 5 Real Axis -2 -3i -4 Mandelbrot Set Mandelbrot Set • A. The complex number ­2 is in the Mandelbrot Set because for c=­2, the corresponding Mandelbrot sequence is ­2,2,2,2… which is bounded • B. i is also in the Set because for c=i, the corresponding sequence is i, ­1+i, i, ­1+i,…… • C. The complex Number 1+i is not in the Set because for c=1+I, the corresponding sequence is i, 1+3i, ­7+7i, 1­97i, ­9407­193i,….. Fin Fin Section 2.4 Section 2.4 Quadratic Equations Quadratic Equation in x Quadratic Equation in x An equation that can be written in the general form ax + bx + c = 0 2 Also known as a second­degree polynomial equation in x Solving Them Solving Them • • • • Factoring Extracting Square Roots Completing the Square Quadratic Formula Factoring Factoring If ab = 0, then a = 0 or b = 0. x −−= x 6 0 (x − 3)( x + = 2) 0 x− = 3 0 x+ = 2 0 x= − 3, 2 2 x − x − =0 2 8 TRY IT! TRY IT! 2 x =4, − 2 ( x −4)( x +2) Extracting Square Roots Extracting Square Roots c >0 u =c If , where , then . 2 u=± c (x + 3) = 16 x + =4 3 ± x =3 −± 4 x=− 1 ,7 2 (x − 12) = 16 2 TRY IT! TRY IT! x= 16, 8 x −12 = ±4 x = 12 ± 4 Completing the Square Completing the Square If , then x + bx = c 2 x + bx + ( ) = c + ( ) 2 b2 2 b2 2 (x + ) = c + b2 2 b 4 2 Example Example 2 2 x +6 x =5 x +6 x + =5 + 3 3 2 2 2 (x + 3) = 14 x + =± 14 3 x =− ± 14 3 TRY IT! TRY IT! 2 x −2 x −3 =0 x − 2 x +1 = 3 +1 2 x − 2x = 3 2 x = 3, −1 x −1 = ± 4 x = 1± 2 ( x −1) = 4 2 Quadratic Formula Quadratic Formula −b ± b − 4ac 2a 2 Example Example 2 x +3 x −1 =0 2 − ± 3 −4(2)( − 3 1) 2(2) 2 − ± 17 3 4 TRY IT! TRY IT! x 2 +x − = 8 4 0 −8 ± 8 − 4(1)(−4) 2(1) 2 −8 ± 80 2 −4 ± 2 5 Completing the Square when the Completing the Square when the Leading Coefficient is Not 1= 2x + x + 8 3 0 2 • Subtract 3 from • • • • • both sides Divide each side by 2 3 2 2 x +x+ 4 2= − +2 2 2 2 Add 2 to each 5 2 side (x + 2) = 2 Simplify 5 Take square x+ = 2 ± 2 root of each 10 side x+ = 2 ± 2 Rationalize 10 denominator x = 2± − 2 2x2 + x = 3 8 − 3 x2 + x = 4 − 2 Polynomial Equations of a higher Polynomial Equations of a higher Degree x 4 − 3x 2 + 2 = 0 • Write in quadratic • • • • • form Partially factor Factor completely Set first factor equal to 0 Set 2nd factor equal to 0 Set 3rd factor equal to 0 ( x ) − 3( x ) + 2 = 0 22 2 2 ( x − 1)( x − 2) = 0 2 ( x + 1)( x − 1)( x 2 − 2) = 0 x +1 = 0 x −1 = 0 x2 − 2 = 0 x = −1,1, ± 2 Solving an Equation Involving Two Solving an Equation Involving Two Radicals 2x + 6 − x + 4 = 1 • • • • • • • • Isolate 2x + 6 Square each side Isolate 2 x +4 Square each side Write in general form Factor Set 1st factor equal to 0 Set 2nd factor equal to 0 2x + 6 = 1 + x + 4 2 x + 6 = 1 + 2 x + 4 = ( x + 4) x +1 = 2 x + 4 x 2 + 2 x + 1 = 4( x + 4) x 2 − 2 x − 15 = 0 ( x − 5)( x + 3) = 0 x −5 = 0 x+3= 0 x = 5, −3 Solving Equations Involving Solving Equations Involving Fractions 2 3 • Multiply each term by • • • • • the LCD Simplify Write in general form Factor Set 1st factor equal to 0 Set 2nd factor equal to 0 x− 2 2 3 x( x − 2) = x( x − 2) − x( x − 2)(1) x x− 2 2( x − 2) = 3x − x( x − 2) x ≠ 0, 2 x x 2 − 3x − 4 = 0 ( x − 4)( x + 1) = 0 x− 4= 0 x +1= 0 x = 4, − 1 = −1 Solving an Equation Involving Solving an Equation Involving Absolute Value x 2 − 3 x = −4 x + 6 First Equation… • Use positive expression • Write in general form • Factor • Set 1st factor equal to 0 • Set 2nd factor equal to 0 x − 3 x = −4 x + 6 2 2 x + x−6 = 0 ( x + 3)( x − 2) = 0 x+3= 0 x−2=0 x = −3, 2 CONTINUED2 CONTINUED x − 3 x = −4 x + 6 Second Equation • Use negative expression • Write in general form • Factor • Set 1st factor equal to 0 • Set 2nd factor equal to 0 −( x − 3 x ) = −4 x + 6 2 x −7x + 6 = 0 2 ( x −1)( x − 6) = 0 x −1 = 0 x −6 = 0 x = 1, 6 Check Check • • • • • • • • Substitute ­3 for x ­3 checks Substitute 2 for x 2 does not check Substitute 1 for x 1 checks Substitute 6 for x 6 does not check (−3) 2 − 3(−3) = −4(−3) + 6 18 = 18 22 − 3(2 ) = −4(2) + 6 2 ≠ −2 12 − 3(1) = −4(1) + 6 2 =2 62 − 3(6 ) = −4(6) + 6 18 ≠ −18 x = −3,1 Other Helpful Equations Other Helpful Equations 2 s = −16t + v0t + s0 • Position Equation: • For finding falling time • Pythagorean • Theorem: For finding the hypotenuse of a right triangle a +b = c 2 2 2 2.5 Solving Inequalities Algebraically and Graphically Algebraically By Ashley Davis and Rachel Berger Words to Remember Words • Solve inequalities by finding all values of x that satisfy the Solve • inequality. Graph of the inequality: set of all points on the real number Graph line that represent the solution set line Properties of Inequalities Properties Let a, b, c, and d be real numbers. 1. Transitive Property a < b and b < c a < c and 2. Addition of Inequalities a < b and c < d a + c < b + d and 3. Addition of a Constant a<ba+c<b+c 4. Multiplying by a Constant For c > 0, a < b ac < bc For For c < 0, a < b ac > bc For Solving a Linear Inequality Solving Example 1 Solve Solution Solution 5 x − 7 > 3x + 9 5 x − 7 > 3x + 9 2x − 7 > 9 2 x > 16 x >8 Solving an Inequality Solving Example 2 3 1− x ≥ x − 4 Solve 2 Solution 1− 3 x ≥ x −4 2 2 − 3x ≥ 2 x − 8 2 − 5 x ≥ −8 −5 x ≥ −10 x≤2 Solving a Double Inequality Solving Example 3 Solve −3 ≤ 6 x − 1 and and Solution 6 x −1 < 3 −3 ≤ 6 x − 1 < 3 −2 ≤ 6 x < 4 1 2 − ≤x< 3 3 Solving an Absolute Value Inequality Solving Let x be a variable or an algebraic expression and let a be a real number such that Let a ≥0 1. The solutions of x <a are all values of x that lie between –a and a. that x <a if and only if − a < x < a x−5 > 2 2. The solutions of x > a are all values of x that are less than –a or greater 2. than a. a. x > a iif and only if x < a or x > a f These rules are also valid if < is replaced by ≤ and > is replaced by ≥ . Solving Absolute Value Inequalities Solving Example 4 Solve Each of the following x − 5 < 2 and x − 5 > 2 Solve x −5 < 2 Solution −2 < x − 5 < 2 3< x < 7 Polynomial Inequalities Polynomial Critical values: points of intersection or undefined values To determine the intervals on which is entirely negative and To 2 x − x which it + 2)( x − 3) those on− 6 = ( xis entirely positive, factor the quadratic as (−∞, −2), (−2,3), and . The critical values occur at x = -2and (3, ∞) x = 3. So, test the intervals for the quadratic are x2 − x − 6 Interval X-Value Value of Polynomial Sign of Polynomial Positive Negative Positive (−∞, −2) x = −3 (−3) 2 − (−3) − 6 = 6 x=0 (−2,3) (0) 2 − (0) − 6 = − 6 (5) 2 − (5) − 6 = 14 (3, ∞) x=5 Solving a Polynomial Inequality Solving Example 5 Solve 2 x 2 + 5 x > 12 Solution 2 x 2 + 5 x − 12 > 0 ( x + 4)(2 x − 3) > 0 Critical Values: Critical Test Intervals: 3 x = −4, x = 2 33 ( −∞, −4), (−4, ), ( , ∞) 22 3 (−∞, −4) ∪ ( , ∞) 2 Solving a Rational Inequality Solving Example 6 Solve 2 x − 7 Solution x −5 ≤3 2x − 7 −3≤ 0 x−5 2 x − 7 − 3 x + 15 ≤0 x −5 2x − 7 ≤3 x −5 −x + 8 ≤0 x−5 Section 2.6 Section 2.6 Exploring Data: Linear Models and Scatter Plots Vocabulary Vocabulary • Positive Correlation: If • • y increases as x increases. Negative Correlation: If y decreases as x increases. No Correlation: Ordered pairs scattered randomly with no trend. y (x,y) = (1,2) 7 6 5 4 3 2 1 x 1 2 3 4 5 6 7 8 9 Creating a Scatter Plot Creating a Scatter Plot 1. Represent the data in ordered pairs from the table of data. Plot each point in a coordinate plane. It has a positive correlation! 8 7 6 5 4 3 2 1 y at (1,2), (3,4), (5,6) x 1 3 5 y 2 4 6 2. x 1 2 3 4 5 6 7 8 9 Plotting on a Calculator Plotting on a Calculator 1. To activate the stat plots, hit 2nd and Y=. 2. 3. Then highlight 1: Plot 1 and press Enter to get to the options. Once here, select On to activate. Press clear to go back to the main screen. Then, press stat and select edit…. Fill in your x values in for L1 and your y values in for L2. To view the scatter plot, press graph. Practice Practice • Graph the data in the x table on a scatter plot 1.8 on your calculator. 3.7 7.5 2.8 y 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 x 9 y 2.3 6.4 4.6 1.7 Answer: Interpreting Correlation Interpreting Correlation 8 y 1. Construct scatter 7 6 5 4 3 2 1 plot from data. 2. Determine if the points have a positive, negative, or no correlation. x 1 2 3 4 5 6 7 8 9 This has a weak positive correlation. Finding Linear Regression Using a Finding Linear Regression Using a Calculator 1. Press stat and hit the right arrow. Press 4 and then press vars, right arrow, and enter twice. 2. Once you see the main screen showing LinReg(ax+b) Y1, press enter. want to see the line in the graph, press graph. 3. Your values will then be displayed. If you Chapter 2 Quiz With Answers Problem 1 Problem 1 Determine whether each value of x solves the equation 1 x= − and 4 2 5 4 − =3 2x x Problem 2 Problem 2 x = −2 and 1 x 6 x 19 + = 2 7 14 Problem 3 Problem 3 x = −1 and − 2 1 3+ =4 x +2 Problem 4 Problem 4 Solve for X xx − =3 52 Problems 5,6,7 Problems 5,6,7 find the x and y intercepts y = x−5 y = x + x −2 2 y = x x +2 Problem 8 Problem 8 write in standard form 4+ − 25 Problems 9,10 Problems 9,10 Perform the indicated operation (4 + i ) + (7 − 2i ) (− 1 + − 8) + (8 − − 50) Problems 11,12,13 Problems 11,12,13 solve by your favorite method 3 + 5x − 2 x = 0 2 x − 2x −8 = 0 2 x + 4 = 12 2 Problem 14,15,16 Problem 14,15,16 Solve the inequalities −8 ≤ 1 − 3( x − 2) < 13 4( x +1) < 2 x + 3 0 ≤ 2 − 3( x +1) < 20 Problems 17,18,19,20 Problems 17,18,19,20 use a graphing utility to approximate the solution 5 − 2x ≥1 20 < 6 x −1 4( x − 3) ≤ 8 − x 3( x + 1) < x + 7 Solutions Solutions • • • • • • • • • • • 1. ½,yes, 4,no 2. ­2,no 1,yes 3. ­1 yes, ­2,no 4. x=­1 multiply reciprocals to eliminate fraction For problems 5,6,7 use a graphing utility to verify your answer. 5. y int:0,­5 X int:5,0 6. y int:0,­2 X int: 1,0 and ­2,0 7. y and x int:0,0 8. 4+25i 9. 11­i 10. 7+49i • 12 x2 − x − = 2 8 0 (x + 2)( x − = 4) 0 x= − 4, 2 • 11 − ± 52 − − 5 4( 2)(3) =x 2( − 2) 1 x= − 3, 2 • 13 x 2 + x− 4 12 = 0 (x − 2)( x + = 6) 0 x =2, − 6 Remember that when the inequalities are divided the Remember that when the inequalities are divided the direction of the arrows changes • • • • • • • 14. 5 ≥ x > −2 15. x<1 16. x<­1 17. x<3 18. x>7/2 19. x<2 4 x≤ 20. Section 3.1 Section Quadratic Functions Graph of Quadratic Equation Cont’d Definition of a Quadratic Function Let a, b, and c be real numbers with a ≠ 0. The function given by f ( x ) = ax + bx + c 2 is called a quadratic function. Quadratic Function The graph of a quadratic function is a U-shaped curve called a parabola 8 6 4 2 Parabola 5 1 0 1 0 5 2 4 6 8 All parabolas are symmetric with respect to a line called the axis of symmetry, which intersects the parabola at the vertex Graph of a Quadratic Function Definition of a Polynomial Function Let n be a nonnegative integer and let numbers with an ≠ 0. The function given by n−1 an , an −1 ,..., a2 , a1 , a0 2 be real f ( x) = an x + an −1 x n + ... + a2 x + a1 x + a0 Is called a polynomial function of x with degree n. Polynomial functions are classified by degree: • f(x) = a, • f(x) = mx + b, a≠0 constant function m ≠ 0 linear function Graph of Quadratic Equation Cont’d Graphing Simple Quadratic Equations g ( x) = 2 x 2 Describe how the graph of each function is related to the graph of y=x^2 A. Each output of f “shrinks” by a factor of 1/3. the result is a parabola that opens upward and is broader. B. Each output of g “stretches” by a factor of 2, creating a narrower parabola. C. H is obtained by a reflection in the x-axis and a vertical shift one upward. −x +1 2 k ( x) = ( x + 2) 2 − 3 D. The graph of k is obtained by a horizontal shift to the left 2 units and a vertical shift 3 units down The Standard Form of a Quadratic Function Standard Form of a Quadratic Equation The standard for of the quadratic function given by is in standard form. The graph of f is a parabola whose axis is the vertical line x = h and whose vertex is the point (h,k). If a > 0, the parabola opens upward, and if a < 0, the parabola opens downward. 8 f ( x ) = a ( x − h) + k 2 a≠0 6 4 2 Vertex (h,k) 5 1 0 1 0 5 2 4 Axis of symmetry (line x = h) 6 8 Example: Identifying the Vertex of a Quadratic Function Problem: Describe the graph f ( x) = 2 x 2 + 8 x + 7 and identify the vertex Solution: Write the quadratic function in standard form by completing the square. Recall 2 that the first step is to factor out any coefficient of x that is not 1. f ( x) = 2 x 2 +8 x +7 = 2( x 2 + 4 x ) +7 = 2( x 2 + 4 x + 4 −4) +7 4 2 2 Write original function Factor 2 out of x terms Because b = 4, add and subtract (4/2)squared within parentheses = 2( x 2 + 4 x + 4) −2(4) +7 Regroup terms = 2( x + 2) 2 −1 Write in standard form From the standard form, you can see that the graph is a parabola opening upward with vertex (-2,-1). Finding x -Intercepts of a Quadratic Function Function Example: Solution: Identifying x-Intercepts of a Quadratic Function Problem: Describe the graph f ( x) = − x 2 + 6 x − 8 and identify the x-intercepts f ( x) =−x 2 +6 x −8 =−( x 2 −6 x ) −8 =−( x 2 −6 x +9 −9) −8 6 2 2 Write original function Factor -1 out of x terms Because b = 6, add and subtract (6/2)squared within parentheses = 2( x 2 + 4 x + 4) −2(4) +7 Regroup terms = 2( x + 2) 2 −1 Write in standard form To find x-Intercepts − ( x 2 − 6 x + 8) = 0 − ( x − 2)( x − 4) = 0 x intercepts = 2 or 4 Factor out -1 Factor ANSWER Section 3.2 Section Polynomial Functions of Higher Degree Kareem El-Khodary, Jacob Mumford, & Shaan Erickson Graphs of Polynomial Functions Graphs • All graphs of polynomial functions have smooth rounded curves, unlike the one below. 8 6 4 2 -10 -5 5 10 -2 -4 -6 -8 -0 1 5 5 • All graphs of polynomial functions All 1 0 are continuous, meaning they have no breaks, holes or gaps. breaks, Definition of a Polynomial Function Definition • The polynomial function of degree n has the form: f ( x) = an n in a+ an −1 x n −1 + ... + a2 x 2 + a1 x + a0 x where s positive integer and a ≠ 0. n 8 6 • f ( x) = x 2 • • If n is even, the graph is similar to the graph of f(x)=x2 and touches the axis at the x-intercept If n is odd, the graph is similar to the graph of f(x)=x3 and crosses the axis at the x-intercept The greater the value of n, the flatter the graph near the origin 4 2 1 0 5 5 1 0 2 8 4 6 6 4 f ( x) = x3 5 10 2 8 -10 -5 -2 -4 -6 -8 Basic Characteristics of the Cubic Function Basic f ( x) = x 3 8 6 4 2 -10 -5 5 10 Graph of f(x)=x3 Domain: (-∞, ∞) Domain: Range: (-∞, ∞) Range: (Intercept: (0, 0) Increasing on: (-∞, ∞) Increasing (Odd Function Origin symmetry -2 -4 -6 -8 Leading Coefficient Test Leading • Whether the graph of a polynomial eventually rises or falls can be determined by Whether the function’s degree (even or odd) and by its leading coefficient. the • • • When n is odd: If the leading coefficient is positive, the graph falls to the left and rises to the right. If the leading coefficient is negative, the graph rises to the left and falls to the right. When n is even: If the leading coefficient is positive, the graph rises to the left and right. If the leading coefficient is negative, the graph falls to the left and right. • • • Zeros of Polynomial Functions Zeros It can be shown that for a polynomial function f of the degree n, the following It the statements are true: statements • The function f has at most n real zeros. The • The graph of f has at most n – 1 relative extrema (relative minima or maxima) The extrema minima maxima • Recall that a zero of a function f is a number x for which f(x)=0. Recall zero Finding the zeros of polynomial functions is one of the most important problems in Finding algebra algebra If f is a polynomial function and a is a real number, the following statements are If equivalent: equivalent: 1. x=a is a zero of the function f. x=a zero f. 2. x=a is a solution of the polynomial equation f(x)=0 x=a solution f(x)= 3. (x-a) is a factor of the polynomial f(x). is factor f(x) 4. (a, 0) is an x-intercept of the graph of f. 4. is Repeated Zeros Repeated For a polynomial function, a factor of (x-a)k, k > 1, yields a repeated zero x=a of repeated x=a multiplicity k multiplicity 1. If k is odd, the graph crosses the x-axis at x=a 1. crosses x-axis x=a 2. If k is even, the graph touches the x-axis (but does not cross the x-axis) at x=a touches -axis x-axis) x=a • Multiplying the functions by any real number does not change the zeros of the function Intermediate Value Theorem The Intermediate Value Theorem concerns the existence of real zeros of polynomial functions. The The theorem states that if (a, f(a)) and (b, f(b)) are two points on the graph of a polynomial b, function such that f(a) ≠ f(b) then for any number d between f(a) and f(b) there must be a f(a) f(b) f(b) number c between a and b and my brain is slowly frying such that f(c) = d. f(c) This theorem helps you to locate the real zeros of a polynomial function in the following way: This if you can find a value x = a at which a polynomial function is positive and another value x = b at which it is negative, you can conclude thatf the)function 2has at least one real zero between ( x = x 3 + x +1 these two values. For example the function is negative when x = -2 and positive when x = -1. Therefore it follows from the Intermediate Value Theorem that f must have a real zero somewhere between -2 and -1 must EXAMPLES EXAMPLES • Finding Zeros of a Polynomial Function – Find all real zeros of Find f ( x) = x3 − x 2 − 2 x f ( x) = x3 − x 2 − 2 x 0 = x3 − x 2 −2 x 0 = x( x 2 − x − 2) 0 = x( x − 2)( x + 1) x = 2, x = −1, x = 0 More EXAMPLES Leading Coefficient Test Sketch the graph of each function f ( x ) = −x 2 5 f ( x) = ( x + 1) 2 5 -5 4 -5 5 -2 -2 Since the leading coefficient is negative and the exponent is odd The graph shifts one unit to the left of the graph of x4 . Since the exponent is even and the coefficient is positive, it goes from positive to positive Example for “Finding a Polynomial Function with Given Zeros” Example for “Finding a Polynomial Function with Given Zeros” Find polynomial functions with the following zero. (There are many correct solutions.) 1 − ,3,3 2 (x + 1 ) 2 or (2 x + 1) Find possible factors for 1 − 2 or (2 x + 1) Choose either (x + 1 ) 2 (2 x + 1) ( x − 3) 2 f ( x) = (2 x + 1)( x − 3) 2 Find the factors for 3 Use Foil Method Final Answer 2 x 3 − 11x 2 + 12 x + 9 Example for “Sketching the Graph of a Polynomial Function” Example for “Sketching the Graph of a Polynomial Function” Sketch the graph of f ( x ) = 3 x 4 − 4 x 3 by hand q ( x ) = 3⋅ x 4 ­4 ⋅ x 3 4 2 -5 5 -2 -4 Example for “Approximating the Zeros of a Function” Example for “Approximating the Zeros of a Function” Find three intervals of length 1 in which the polynomial is guaranteed to have a zero. f ( x) = 12 x 3 − 32 x 2 + 3 x + 5 ( h ( x ) = 12⋅ x 3 ­32 ⋅ x 2 )+3⋅ x+5 4 2 -5 5 -2 -4 ...
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