2 Quadratic Function A function that can be written in the form: f(x) = ax2 + bx + c (standard form)where “a”, “b”, and “c” are real numbers with a ≠ 0A second degree polynomial functionExamples:f(x) = x2 – 8x + 15f(x) = -4x2 – 16x + 33
3 Graphs of Quadratic Functions Every quadratic function has a graph that is a “U-shaped” curve called a “parabola” that opens upward when “a” is positive and downward when “a” is negativeThe highest point on a parabola that opens downward and the lowest point on a parabola that opens upward is called the “vertex”The vertical line through the vertex is called the “axis”The parabola is always “symmetric” with respect to its axis (if the graph were folded along the axis it would lay on top of the other half
4 “Vertex Form” of a Quadratic Function Every quadratic function can be written in “vertex form”:f(x) = a(x – h)2 + kwhere “a”, “h”, and “k” are real numbers with a ≠ 0, and where (h, k) is the vertex of the parabolaAs before, parabola opens upward when “a” is positive and downward when “a” is negativeExamples:f(x) = 2(x – 3)2 – 4parabola that opens upward with vertex (3, -4)f(x) = -3(x + 5)2 + 1parabola that opens downward with vertex (-5, 1)
5 Converting a Quadratic Function to Vertex Form Given f(x) = ax2 + bx + c, factor “a” from variable terms leaving variables in parenthesesComplete the square on the expression inside the parentheses and balance on the same side of the equationFactor the expression inside parentheses and combine like terms outside parentheses to get vertex form:
6 Example One: Converting to Vertex Form f(x) = x2 – 8x + 15Factor “a” from variable terms:f(x) = 1(x2 – 8x ) + 15Complete the square on the expression inside parentheses and balance on the same side of the equation:f(x) = 1(x2 – 8x + 16) + 15 – 16Factor the expression inside parentheses and combine like terms on outside:f(x) = 1(x – 4)2 – 1Note: parabola opens upward with vertex (4, -1)
7 Example Two: Converting to Vertex Form f(x) = -4x2 – 16x + 33f(x) = -4(x2 + 4x ) + 33f(x) = -4(x2 + 4x + 4)f(x) = -4(x + 2)2 + 49Note: parabola opens downward with vertex (-2, 49)
8 Graphing a Quadratic Function Convert to “vertex form” to find vertex, (h, k)Note the “x” value of vertex, h, and pick two other x values that are bigger than h, or two values that are smaller than h, and calculate corresponding y valuesPlot these three points on a graph and use symmetric properties of parabola to plot two more points on the other side of the axisConnect points with smooth curve with arrows at both ends
9 Example of Graphing Quadratic Function f(x) = x2 – 8x + 15f(x) = 1(x – 4)2 – 1Parabola opens upward with vertex (4, -1)Note: h = 4 (x coordinate of vertex)If we pick x = 3, we get y = 32 – 8(3) + 15(3, 0) is a point on parabolaIf we pick x = 2, we get y = 22 – 8(2) + 15(2, 3) is a point on parabola
11 Example of Graphing Quadratic Function f(x) = -2x2 + 4x + 3f(x) = -2(x - 1)2 + 5Parabola opens downward with vertex:(1, 5)If x = 0, then y = -2(0)2 + 4(0) + 3 = 3(0, 3) is a point on the graphIf x = -1, then y = -2(-1)2 + 4(-1) + 3 = -3(-1, -3) is a point on the graph
13 Equation of Axis, Domain and Range of Quadratic Functions Given vertex (h, k), the equation of the axis is:x = hUnless otherwise specified, the domain of a quadratic function is “all real numbers”The range depends on whether the parabola opens upward or downward:If upward the range is from “k” to positive infinityIf downward, the range is from negative infinity to “k”Example:If vertex is (-2, 3) and “a” is positive:Equation of axis is:Domain is:Range is:
14 Homework Problems Section: 3.1 Page: 303 Problems: All: 1 – 4, Odd: 13 – 25MyMathLab Assignment 3.1 for practiceMyMathLab Quiz 3.1 is due for a grade on the date of our next class meeting
15 Polynomial FunctionsA polynomial function in “x” is defined as f(x) = a finite sum of terms, each of which has the form anxn where “n” represents a whole number and an is the coefficient of the variable xnThe largest exponent in the finite sum of terms is called the degree of the function (the only exception being where the polynomial function is of the form: f(x) = 0. This is called the “zero polynomial” and has no degree.)
17 Dividing Polynomial Function by “x – k” Using Long Division This method has been discussed in previous classes:First write each polynomial in descending powersIf a term of some power is missing, write that term with a zero coefficientComplete the problem exactly like a long division problem in basic math
19 Dividing Polynomial Function by “x – k” Using Synthetic Division First write each polynomial in descending powersIf a term of some power is missing, write that term with a zero coefficientSet the problem up as a long division problem, but write only the coefficients of the dividend inside division symbol and “k” outsidePut a blank line under dividend coefficients with an underlineDrop the first coefficient below this lineMultiply this coefficient by “k” and write the answer above the line below the second coefficientAdd the two numbers and put answer beneath the lineRepeat this process until you reach the endThe numbers beneath the line represent coefficients of the answer and the remainder where the first term of the answer is of degree one less than the dividend and the last number is the remainder
21 Homework Problems Section: 3.2 Page: 319 Problems: Odd: 1 – 17 MyMathLab Assignment 3.2(a) for practice
22 Remainder Theorem When a polynomial, ,is divided by the remainder is Consider the last example where was divided by using synthetic division:The Theorem says thatAnd it is:
23 Evaluating a Polynomial It is usually easiest to evaluate a polynomial for a specific value of “x” by using synthetic division and the remainder theoremExample: GivenFind:
24 Homework Problems Section: 3.2 Page: 319 Problems: Odd: 27 – 37 MyMathLab Assignment 3.2(b) for practice
25 Zeros of Polynomial Functions Zeros of polynomial functions are the values of “x” that make “y = 0”“k” is a zero of f(x) if f(k) = 0If a zero, k, of a polynomial function is a real number, then k is also an x-intercept of the graph of f(x)Polynomial functions may also have zeros that are non-real complex numbers (these are not x-intercepts of the graph)
26 Relationship Between Zeros and Factors If k is a zero of f(x) and f(x) has degree “n”, then x – k is one factor of f(x), and the other factor is the polynomial of degree “n – 1” whose coefficients are shown in the bottom row of synthetic divisionExample: Given:Find:
27 Example ContinuedSynthetic Division shows that is a zero of , so is one factor, and the other factor is:
28 Homework Problems Section: 3.2 Page: 320 Problems: Odd: 39 – 55 MyMathLab Assignment 3.2(c) for practiceMyMathLab Quiz 3.2 is due for a grade on the date of our next class meetingSection: 3.3Page: 329Problems: Odd: 5 – 21MyMathLab Assignment 3.3(a) for practice
29 Rational Zeros Theorem Given a polynomial function , with integer coefficients, where is the coefficient of the highest degree term and is the coefficient of the constant term, rational zeros must be the ratio of some factor of to some factor ofExample: Find all possible rational zeros:
31 Using Rational Zeros Theorem on Polynomials with Non-Integer Coefficients “k” is a zero of f(x), if and only if “k” is also a zero of the polynomial formed by multiplying a constant by f(x)To find zeros of f(x) when f(x) does not have integer coefficients, find zeros of af(x) where “a” is a factor that forms a polynomial with integer coefficients
32 Relationship Between Degree of a Polynomial Function and the Number of Distinct Zeros A polynomial function of degree “n” has at most “n” distinct zerosExamples: A polynomial of:degree 2, has at most ___ distinct zerosdegree 5, has at most ___ distinct zeros
33 Other ExamplesGivenWe can find zeros by solving
34 Non-Real Complex Zeros of Polynomial Functions If a non-real complex number is a zero of f(x), then its conjugate is also a zero of f(x)Examples:If 3 - 2i is a zero, then _____ is also a zeroIf 1+ 5i is a zero, then _____ is also a zero
35 Exact Number of Zeros in a Polynomial of Degree “n” A polynomial function, f(x), of degree “n” has exactly “n” zeros, but they may not all be distinctA number “k” is said to be a zero of multiplicity “m” if “m” is greater than 1 and is the largest exponent for which (x – k)m is a factor of f(x)In this case “k” counts as “m” of the “n” zeros of f(x)
36 Example Given: Degree: Exact Number of Zeros: Number of Distinct Zeros:Distinct Zeros:What is the multiplicity of each zero?
37 Finding Zeros of a Polynomial Function Use Rational Zeros Theorem to find all possible rational zerosUse Synthetic Division and Remainder Theorem to try to find one rational zero (remainder will be zero)If “n” is a rational zero, factor original polynomial as (x – n)q(x)Test remaining possible rational zeros in q(x). If one is found, factor again as in previous stepContinue in this way until all rational zeros have been foundSee if additional irrational or non-real complex zeros can be found by solving a quadratic equation
38 Example Find all zeros of: Find all solutions to: Rational Zeros Theorem says the only possible rational zeros are:See if -1 is a zero:Conclusion:
39 Example ContinuedThis new factor has the same possible rational zeros:Check to see if -1 is also a zero of this:Conclusion:
40 Example Continued This new factor has as possible rational zeros: Check to see if -1 is also a zero of this:Conclusion:
41 Example ContinuedCheck to see if 1 is a zero:Conclusion:
42 Example ContinuedCheck to see if 2 is a zero:Conclusion:
44 Homework Problems Section: 3.3 Page: 330 Problems: Odd: 29 – 47 MyMathLab Assignment 3.3(b) for practiceMyMathLab Quiz 3.3 (Shortened Version) is due for a grade on the date of our next class meeting
45 Further Hints on Finding All Zeros of a Polynomial Function In trying to find all zeros of a polynomial function, it would be useful to know the number of positive and negative real zerosDescartes’ Rule of Signs - If f(x) is a polynomial function with real coefficients, written in descending powers, with a non-zero constant termThe number of positive real zeros is equal to the number of sign changes in the terms of f(x), or is less than the number of sign changes by an even positive integerThe number of negative real zeros is equal to the number of sign changes in the terms of f(-x), or is less than the number of sign changes by an even positive integerIn both considerations, missing terms do not count as a sign change
46 Application of Descartes’ Rule of Signs Find the number of positive and negative real zeros of:How many sign changes?Number of positive real zeros?Find f(-x):Number of negative real zeros?
47 Further Consideration of Previous Example Consider all possible zeros:Total Zeros:Positive Negative Non-Real** Since non-real zeros occur only as conjugate pairs there must always be an even number
48 Homework Problems Section: 3.3 Page: 331 Problems: All: 73 – 78 MyMathLab Assignment 3.3(c) for practice
49 Example of Finding Polynomial Functions with Specific Zeros Given the fact that k is a zero of f(x) if and only if (x – k ) is a factor:Find a polynomial with zeros: 3, 1, and -2Note: Other polynomial functions with the same zeros will be non-zero multiples of this oneOther examples:
50 Additional Considerations Relative to Last Example For the previous example, if we are also told that f(2) = 8, find the polynomial function with the specified zeros
51 Homework Problems Section: 3.3 Page: 330 Problems: Odd: 49 – 71 MyMathLab Assignment 3.3(d) for practiceMyMathLab Quiz 3.3 is due for a grade on the date of our next class meeting
52 Turning Points of Polynomial Functions The point at which the graph of a polynomial function changes from increasing to decreasing or vice versa is called a turning point of the function
53 Turning Points of Polynomial Functions A polynomial function of degree n has at most n – 1 turning points with at least one turning point between successive zerosGiven that the following function has zeros at 1 and -1, what do you know about its turning points?It has at most how many?Where is at least one of those located?
54 Graphs of Polynomial Functions The domain of every polynomial function is:Polynomial functions are continuous over their domain (entire graph can be drawn without lifting pencil)Even polynomial functions, , are symmetric with respect to y-axisOdd polynomial functions, , are symmetric with respect to originNote: Many polynomial functions are neither even nor odd
55 Graphs of Polynomial Functions End behavior of graphs of polynomial functions is determined by characteristics of highest degree term:If degree of polynomial is odd and coefficient of highest degree is positive, ends look like:If degree of polynomial is odd and coefficient of highest degree is negative, ends look like:If degree of polynomial is even and coefficient of highest degree is positive, ends look like:If degree of polynomial is even and coefficient of highest degree is negative, ends look like:
56 Graphs of Polynomial Functions x-intercepts of polynomial functions are real zeros of the functionThe y-intercept of a polynomial function, f(x), is f(0)
57 Sketching Graphs of Polynomial Functions Find the real zeros of the function and plot them as x-interceptsFind and plot the y-interceptUse synthetic division to find values of the polynomial function between zerosShow end behavior of graph based on characteristics of highest degree term
58 Sketch Graph: Descartes’ Rule of Signs: Possible rational roots: Test for rational roots:
59 Example Continued Solving equation by zero factor: Find y-intercept: Find values of function between zeros:
60 Example ContinuedFind f(2) by synthetic division:
63 Intermediate Value Theorem If “a” and “b” are in the domain of a polynomial function, f(x), and f(a) and f(b) have opposite signs, then f(x) has a least one real zero between “a” and “b”We can use this theorem to approximate the value of irrational zeros to any desired degree of accuracy as shown by the following example
64 Find a Zero to the Nearest Tenth: Find f(1)Find f(2)There must be a zero between 1 and 2Find f(1.5)Now we know there is a zero between:
65 Find a Zero to the Nearest Tenth: Find f(1.8)There must be a zero between:Find f(1.7)Find f(1.75)
66 Find a Zero to the Nearest Tenth: Since andWe have now found that, to the nearest tenth, a zero is:
67 Homework Problems Section: 3.4 Page: 343 Problems: Odd: 43 – 51 Also, without using a graphing calculator, approximate the zero discussed in problems 45 and 47 to the nearest tenthMyMathLab Assignment 3.4(b) for practice
68 Boundedness TheoremGiven a polynomial function, f(x), with real coefficients, of degree 1 or more, and a real number “c” in the domain of f(x), and finding f(c) by synthetic division:If c > 0 and all numbers on bottom row of synthetic division are non-negative, then f(x) has no zero greater than cIf c < 0 and all numbers on bottom row of synthetic division alternate in sign (with 0 considered positive or negative as needed), then f(x) has no zero less than c
69 Example of Boundedness Theorem In a previous example:we found f(2) by synthetic division:Notice bottom row. What does this tell us about zeros?
70 Example of Boundedness Theorem Using the same function:find f(-3) by synthetic division:Notice bottom row. What does this tell us about zeros?
71 Homework Problems Section: 3.4 Page: 343 Problems: Odd: 53 – 59 MyMathLab Assignment 3.4(c) for practiceMyMathLab Quiz 3.4 is due for a grade on the date of our next class meeting
72 Rational FunctionsA “rational function” is a function that can be written as:Any “real “ zeros of will make the rational function undefined and will establish “vertical asymptotes” for the graph of the functionWhen the zeros of are non-real complex numbers, there will be no vertical asymptotes
73 Vertical AsymptoteA “vertical asymptote” is a vertical line located at each real zero of the denominator polynomial,On either side a vertical asymptote the value of the rational function, ,approaches as as x gets closer to the real zero ofThe possible behaviors of the graph of a rational function near a vertical asymptote are illustrated on the next slide
74 Behavior of Rational Function Graphs Near Vertical Asymptotes
75 Finding Vertical Asymptotes of Rational Functions Given a rational function:Find zeros ofFor each number, , that is a real zero of, is a vertical asymptote
76 Example of Finding Vertical Asymptotes Find the vertical asymptotes:Find zeros of denominator:
79 Linear Systems of Equations When two or more linear equations are considered simultaneously they are referred to as a system of equationsExample of System of Linear Equations in Two Variables:2x – y = 73x + 7y = 2The solution to this system is the set of all points, (x, y) pairs, that make both equations true at the same timeIf these represent different non-parallel lines, the solution is a single point. If they represent parallel lines, there is no solution. If they are different forms of the same line, then any point on the line is a solution.
80 Solving a System of Linear Equations in Two Variables by Substitution Solve either equation for one variable (choose easiest)Substitute this value into the second equationSolve the second equationSubstitute the solution to the second equation back into the first equation and solve itThe solution to the system is shown as an (x, y) pair
81 Example2x – y = 73x + 7y = 2It is easiest to solve first equation for y:2x – 7 = ySubstitute into the second equation:3x + 7(2x – 7) = 23x + 14x – 49 = 217x = 51x = 3Substitute into first equation:2(3) – y = 7- y = 1y = -1Solution: (3, -1)
82 Homework Problems Section: 5.1 Page: 484 Problems: Odd: 7 – 17 MyMathLab Assignment 5.1(a) for practice
83 Solving a System of Linear Equations in Two Variables by Elimination Multiply one or both equations by a constant so that when the two equations are added, the result is a single equation in one variableSolve the resulting equation for that variableSubstitute that value back into either original equation to find value of other variableShow solution as an ordered pair
84 Example of Solving a System of Linear Equations by Elimination 2x – y = 73x + 7y = 2Multiply the first equation by 7 and add the result to the second equation to eliminate “y”:14x – 7y = 493x + 7y = 217x = 51x = 3Substitute into either equation (first):2(3) – y = 7y = -1Solution: (3, -1) (Same as before)
85 Homework Problems Section: 5.1 Page: 484 Problems: Odd: 19 – 29 MyMathLab Assignment 5.1(b) for practiceMyMathLab Quiz 5.1(Shortened Version) is due for a grade on the date of our next class meetingThis is the last assignment for a grade!!!!
86 Linear Equations in Three Variables Equations of the form:Ax + By + Cz = DSolutions to these equations are “ordered triples”Example:4x + 3y – z = -5There are an infinite number of ordered triples that are solutions.One solution is (3, -5, 2)
87 Solving a System of Three Linear Equations by Elimination Multiply one or two equations by a constant and add to eliminate one variableUse the same process on two different equations to eliminate the same variableNow find the solution to this system of two equations in two variables as previously learnedFind the number solution for the third variable by substituting into any of the three original equations
88 Example 3x + 9y + 6z = 3 2x + y – z = 2 x + y + z = 2 Multiply second by 6 and add to first to eliminate “z”:3x + 9y + 6z = 312x + 6y – 6z = 1215x + 15y = 15Also add second and third to eliminate “z”:3x + 2y = 4
89 Example Continued 3x + 2y = 4 Multiply second by -5 and add to first: Solve the system:15x + 15y = 153x + 2y = 4Multiply second by -5 and add to first:15x + 15y = 15-15x – 10y = -205y = -5y = -1Substitute into second equation at top to get:3x + 2(-1) = 43x = 6x = 2
90 Example ContinuedSubstitute both x = 2 and y = -1 into any of the original equations (last):x y + z = 2(2) + (-1) + z = 21 + z = 2z = 1Solution is:(2, -1, 1)
91 Homework Problems Section: 5.1 Page: 485 Problems: Odd: 47 – 57 MyMathLab Assignment 5.1(c) for practiceMyMathLab Quiz 5.1 is due for a grade on the date of our next class meeting
92 Determinant Solution of Linear Systems A “matrix” is a rectangular arrangement of numbersA matrix is designated as being an “m x n matrix” with “m” representing the number of rows, and “n” representing the number of columnsExamples:A 2 x 2 matrix has two rows and two columnsA 3 x 4 matrix has three rows and four columns
93 Determinant of n x n Matrix Every matrix, A, that has an equal number of rows and columns has a real number associated with it called its “determinant”The determinant of a matrix, A, is indicated by the symbol: |A|The way in which the determinant is found depends on the number of rows and columns
94 Determinant of a 2 x 2 Matrix The determinant of a 2 x 2 matrix:a bc d is defined as: ad – bcExample:If matrix B is:4 -23 5 then |B| is:(4)(5) – (3)(-2) = = 26
95 Solving a System of Two Linear Equations in Two Variables using Cramer’s Rule Write each equation in standard formMake a 2 x 2 “coefficient” matrix by taking the coefficients of “x” as the first column and the coefficients of “y” as the second columnCall the determinant of this “coefficient matrix” by the name “D”Make an “X” 2 x 2 matrix by substituting in the “coefficient” matrix the constants on the right side of the equal sign for the “x” coefficients and find the determinant of this matrix, DxMake a “Y” 2 x 2 matrix by substituting in the “coefficient” matrix the constants on the right side of the equal sign for the “y” coefficients and find the determinant of this matrix, DyThe solution to the equation will be:x = Dx/D and y = Dy/Dexcept in the case where D = 0 when this method fails
96 Example of Solving a System by Cramer’s Rule 2x – y = 73x + 7y = 2Coefficient Matrix:2 -13 7D = (2)(7) – (3)(-1) = 17“X” Matrix:7 -12 7Dx = (7)(7) – (2)(-1) = = 51“Y” Matrix:2 73 2Dy = (2)(2) – (3)(7) = 4 – 21 = -17
97 Example Continued From previous page: D = 17, Dx = 51, and Dy = -17 x = Dx / Dx = 51 / 17 = 3y = Dy / Dy = -17 / 17 = -1Solution for System: (3, -1)
98 Cramer’s Rule Applied to “Larger Systems” Cramer’s Rule can be applied to systems containing “n” equations and “n” variables, but will not be discussed at this time.
100 Solving Systems of Equations by Gauss-Jordan Method Must have same number of equations as variablesPut each equation in standard form (variable terms in alphabetical order on left side of = sign and constants on right side)Make an “augmented matrix” consisting of a rectangular arrangement of variable coefficients, a vertical line, and constants as appear on the right side of the = sign
101 Gauss-Jordan Method Continued With a goal of achieving an arrangement of “ones” down the diagonal and “zeros” in all other positions on the left side of the vertical line in the augmented matrix:Interchange any two rowsMultiply numbers in any row by any nonzero numberReplace any row by adding to its numbers the multiples of the numbers of another row
102 Gauss-Jordan Method Continued When “ones” have been obtained down the diagonal from upper left to lower right, and “zeros” are in all other positions, the solutions to the system will be seen in alphabetical order on the right side of the vertical bar arranged from top to bottom
103 Example of Gauss-Jordan Method Applied to System with 2 Variables 2x + 3y = -15x – 2y = 26Augmented Matrix:
111 Direct VariationWe say that y varies directly as x (or y is directly proportional to x) if there is a constant, k, such that:y = kxTo solve a problem involving direct variation, the key is to find the value of k, from knowing one pair of (x, y) valuesOnce the k value is determined, it establishes a formula, that can then find a value of y for any value of x
112 Direct Variation Example Assuming that the pressure on an object underground varies directly as its depth, and the pressure is 5 psi when the depth is 10 feet, what would the pressure be at a depth of 30 feet?
113 Inverse VariationWe say that y varies inversely as x (or y is inversely proportional to x) if there is a constant, k, such that:To solve a problem involving inverse variation, the key is to find the value of k, from knowing one pair of (x, y) valuesOnce the k value is determined, it establishes a formula, that can then find a value of y for any value of x
114 Inverse Variation Example Assuming that the illumination of light on an object varies inversely as the square of its distance from the object, and the illumination is 50 candela at 5 meters, what would the illumination be at 20 meters?
115 Joint VariationJoint variation occurs when y varies with multiple variables, directly with some and inversely with othersIn such cases, we write y as being equal to a constant, k, times the product of the other variables or inverses of the other variables, as appropriateTo solve a problem involving joint variation, the key is to find the value of k, from knowing the value of one set of variablesOnce the k value is determined, it establishes a formula, that can then find a value of y given the values of the other variables
116 Joint Variation Example Assuming that the gravitational attraction of an object varies directly as its mass and inversely as the square of the distance from the center of the mass, and the gravitational attraction is 33 newtons at 21 meters from the center when the mass is 2079 kilograms. What would the gravitational attraction be for the same mass at a distance of 3 meters from the center?.
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