There are four irrational roots with approximate values of

Similar documents
Basic Equation Solving Strategies

Roots are: Solving Quadratics. Graph: y = 2x 2 2 y = x 2 x 12 y = x 2 + 6x + 9 y = x 2 + 6x + 3. real, rational. real, rational. real, rational, equal

Chapter 2. Polynomial and Rational Functions. 2.5 Zeros of Polynomial Functions

P.3: Linear Equations and Inequalities p28: #3, 4, 15, 17, 20, 22, 23, 26, 27, 32, 33, 38, 39, 40, 45, 47, 48, 52, 53, 54, 70 73

3 Polynomial and Rational Functions

Final Exam C Name i D) 2. Solve the equation by factoring. 4) x2 = x + 72 A) {1, 72} B) {-8, 9} C) {-8, -9} D) {8, 9} 9 ± i

Section 3.2 Quadratic Functions & Their Graphs

Final Exam A Name. 20 i C) Solve the equation by factoring. 4) x2 = x + 30 A) {-5, 6} B) {5, 6} C) {1, 30} D) {-5, -6} -9 ± i 3 14

Chapter 1: Precalculus Review

Solving Polynomial and Rational Inequalities Algebraically. Approximating Solutions to Inequalities Graphically

6.1 Polynomial Functions

M30-1: Polynomial, Radical and Rational Functions, Graphs and Equations Exam /20

( ) = 1 x. g( x) = x3 +2

( ) - 4(x -3) ( ) 3 (2x -3) - (2x +12) ( x -1) 2 x -1) 2 (3x -1) - 2(x -1) Section 1: Algebra Review. Welcome to AP Calculus!

ALGEBRA 2/TRIGONMETRY TOPIC REVIEW QUARTER 2 POWERS OF I

CHAPTER 4: Polynomial and Rational Functions

Rational Functions 4.5

To get horizontal and slant asymptotes algebraically we need to know about end behaviour for rational functions.

SUMMER MATH PACKET College Algebra and Trigonometry A COURSE 235 and Pre-Calculus A COURSE 241

MPM2D - Practice Mastery Test #5

Chapter 4: Radicals and Complex Numbers

MAT116 Final Review Session Chapter 3: Polynomial and Rational Functions

Algebra II Midterm Exam Review Packet

a factors The exponential 0 is a special case. If b is any nonzero real number, then

Algebra Review. Finding Zeros (Roots) of Quadratics, Cubics, and Quartics. Kasten, Algebra 2. Algebra Review

Key Features of a Graph. Warm Up What do you think the key features are of a graph? Write them down.

What students need to know for... ALGEBRA II

GUIDED NOTES 5.6 RATIONAL FUNCTIONS

30 Wyner Math Academy I Fall 2015

PreCalculus: Semester 1 Final Exam Review

Algebra II Chapter 5: Polynomials and Polynomial Functions Part 1

Lesson 5b Solving Quadratic Equations

Use a graphing utility to approximate the real solutions, if any, of the equation rounded to two decimal places. 4) x3-6x + 3 = 0 (-5,5) 4)

APPLICATIONS OF DIFFERENTIATION

11 /2 12 /2 13 /6 14 /14 15 /8 16 /8 17 /25 18 /2 19 /4 20 /8

SOLUTIONS FOR PROBLEMS 1-30

Polynomial Functions and Models

Exponential Functions Dr. Laura J. Pyzdrowski

Solving Linear and Rational Inequalities Algebraically. Definition 22.1 Two inequalities are equivalent if they have the same solution set.

. As x gets really large, the last terms drops off and f(x) ½x

1. The graph of a quadratic function is shown. Each square is one unit.

9.5. Polynomial and Rational Inequalities. Objectives. Solve quadratic inequalities. Solve polynomial inequalities of degree 3 or greater.

Unit 3: HW3.5 Sum and Product

TEST 150 points Write neatly. Show all work. Write all responses on separate paper. Clearly label the exercises.

3.3 Real Zeros of Polynomial Functions

Mock Final Exam Name. Solve and check the linear equation. 1) (-8x + 8) + 1 = -7(x + 3) A) {- 30} B) {- 6} C) {30} D) {- 28}

of multiplicity two. The sign of the polynomial is shown in the table below

BARUCH COLLEGE MATH 1030 Practice Final Part 1, NO CALCULATORS. (E) All real numbers. (C) y = 1 2 x 5 2

October 28, S4.4 Theorems about Zeros of Polynomial Functions

Determine whether the formula determines y as a function of x. If not, explain. Is there a way to look at a graph and determine if it's a function?

3.4. ZEROS OF POLYNOMIAL FUNCTIONS

Lesson 2.1: Quadratic Functions

Chapter 2 Polynomial and Rational Functions

MAT 129 Precalculus Chapter 5 Notes

Graphing Rational Functions

Chapter 2 Polynomial and Rational Functions

UNIT 4: RATIONAL AND RADICAL EXPRESSIONS. 4.1 Product Rule. Objective. Vocabulary. o Scientific Notation. o Base

Rational Functions. A rational function is a function that is a ratio of 2 polynomials (in reduced form), e.g.

Horizontal and Vertical Asymptotes from section 2.6

Day 6: 6.4 Solving Polynomial Equations Warm Up: Factor. 1. x 2-2x x 2-9x x 2 + 6x + 5

f ( x ) = x Determine the implied domain of the given function. Express your answer in interval notation.

King Fahd University of Petroleum and Minerals Prep-Year Math Program Math Term 161 Recitation (R1, R2)

Herndon High School Geometry Honors Summer Assignment

G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s ( 3 0 S ) Final Practice Exam

PreCalculus Notes. MAT 129 Chapter 5: Polynomial and Rational Functions. David J. Gisch. Department of Mathematics Des Moines Area Community College

2. If the values for f(x) can be made as close as we like to L by choosing arbitrarily large. lim

Procedure for Graphing Polynomial Functions

Instructor Quick Check: Question Block 12

Midterm Review. Name: Class: Date: ID: A. Short Answer. 1. For each graph, write the equation of a radical function of the form y = a b(x h) + k.

NAME DATE PERIOD. Power and Radical Functions. New Vocabulary Fill in the blank with the correct term. positive integer.

Math 119 Main Points of Discussion

Section 3.6 Complex Zeros

Solving a Linear-Quadratic System

Section 3.1 Quadratic Functions

Carnegie Learning High School Math Series: Algebra I Indiana Standards Worktext Correlations

Skill 6 Exponential and Logarithmic Functions

Pre-Calculus Summer Packet

1.2 Functions and Their Properties Name:

Summer Work for students entering PreCalculus

4.1 Exponential Functions

This is the summer math packet if you are entering Algebra 3 Trig next year. I

NOTES: Chapter 11. Radicals & Radical Equations. Algebra 1B COLYER Fall Student Name:

Unit 8 - Polynomial and Rational Functions Classwork

Algebra I Unit 1: Data & Functions

UP AND UP DOWN AND DOWN DOWN AND UP UP AND DOWN

MCPS Algebra 2 and Precalculus Standards, Categories, and Indicators*

Chapter 4: Radicals and Complex Numbers

Basic Fraction and Integer Operations (No calculators please!)

Algebra 2 Honors: Final Exam Review

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. C) x 8. C) y = x + 3 2

Intermediate Algebra Sample Final Exam Fall 2016

Chapter 8. Exploring Polynomial Functions. Jennifer Huss

Skill 6 Exponential and Logarithmic Functions

Chapter 3: Polynomial and Rational Functions

2014 Summer Review for Students Entering Algebra 2. TI-84 Plus Graphing Calculator is required for this course.

Polynomials. Exponents. End Behavior. Writing. Solving Factoring. Graphing. End Behavior. Polynomial Notes. Synthetic Division.

ALGEBRA 2 Summer Review Assignments Graphing

Solving Exponential Equations

Complex Numbers: Definition: A complex number is a number of the form: z = a + bi where a, b are real numbers and i is a symbol with the property: i

Subtract 6 to both sides Divide by 2 on both sides. Cross Multiply. Answer: x = -9

Transcription:

Power of the Quadratic Formula 1 y = (x ) - 8(x ) + 4 a = 1, b = -8, c = 4 Key 1. Consider the equation y = x 4 8x + 4. It may be a surprise, but we can use the quadratic formula to find the x-intercepts of graph because the ratio of the two exponents is : 1. Use the quadratic formula to first solve for x. Once we know the value of x, we can find The fundamental theorem of algebra tells us that the polynomial equation x 4 8x + 4 = 0 has four roots where some of the roots may be multiple roots. Roots of polynomial equations come in two flavors, real or complex. Real roots also come in two flavors, rational and irrational. From Descartes' rule of signs we can deduce that the polynomial has exactly or 0 positive real roots and exactly or 0 negative real roots. We will see that the polynomial x 4 8x + 4 has exactly positive irrational roots and negative irrational roots. 8 ± 64-4(4) 8 ± 48 8 ± 3g16 8 ± 4 3 4 ± 3 7.464101615 or 0.5358983849 ±.73050808 or ± 0.730508076 There are four irrational roots with approximate values of +.73050808 or + 0.730508076 Tip to check answers for TI graphing calculator users: The <ANS> key is a wonderful convenience. Follow the steps below to plug.73050808 into the equation x 4-8x + 4. 1. Enter (4 + (3)). Make sure to press the <ENTER> key.. Enter the expression ANS^4-8ANS^ + 4 3. The output from step () above should equal * 10-1 = 0.000 000 000 00 0. Note: If.73050808^4 8*.73050808^ + 4 is entered, the output = 1.63 * 10-8 0. Keep in mind that many numerical results have a rounding error! Only humans know the exact answer.

. Consider the equation y = x 4 8x - 4. We can use the quadratic formula to find the x-intercepts of graph because the ratio of the two exponents is : 1. Use the quadratic formula to first solve for x. Once we know the value of x, we can find The fundamental theorem of algebra tells us that the polynomial equation x 4 8x - 4 = 0 has four roots where some of the roots may be multiple roots. Roots of polynomial equations come in two flavors, real or complex. Real roots also come in two flavors, rational and irrational. From Descartes' rule of signs we can deduce that the polynomial has exactly 1 positive real root and exactly 1 negative real root. Therefore the other two roots of x 4 8x 4 must be complex roots. How can one tell by inspection of the equation y = x 4 8x 4 that the function y = f(x) is an even function or the graph has symmetry with respect to the y-axis? 1 y = (x ) - 8(x ) - 4 a = 1, b = -8, c = - 4 8 ± 64-4( -4) 8 ± 80 8 ± 5g16 8 ± 4 5 4 ± 5 8.47135955 or -0.47135955 ± 4 + 5 or ± i 4-5 ±.91069338 or ± 0.687114994i There are two irrational and two complex roots with approximate values of +.91069338 or + 0.68711499i Follow the steps below to plug in 0.68711499i into the equation x 4-8x - 4. 1. Press the <MODE> button and set the number mode to <a + bi>. Enter 0.68711499i. ( i is the button above the decimal point key) Make sure to press the <ENTER> key. 3. Enter the expression ANS^4-8ANS^ - 4 4. The output from step (3) above should equal -5.47 * 10-9 = -0.000 000 005 47 0. Keep in mind that many numerical results will have a rounding error! Only humans know the exact answer.

3. Consider the equation y = x /3 x 1/3-3. It may be a surprise, but we can use the quadratic formula to find the x-intercepts of graph because the ratio of the two exponents is : 1. Use the quadratic formula to first solve for x 1/3. Once we know the value of x 1/3, we can find (Descartes' rule of signs only applies to polynomials with whole number exponents.) /3 1/3 x - x - 3 = 0 1/3 1/3 ( x ) ( x ) 1 - - 3 = 0 a = 1, b = -, c = - 3 ± 4-4( - 3) ± 16 = 3 or -1 1/3 x 1/3 1/3 3 3 3 = 7 or (-1) = -1 There are two rational roots with values of -1 or 7. 4. Consider the equation y = x 3/ x - 3. We can use the quadratic formula to find the x-intercepts of graph because the ratio of the two exponents is : 1. Use the quadratic formula to first solve for x. Once we know the value of x 4/3, we can find 3/ x - x - 9 = 0 ( ) ( ) 1 x - x - 9 = 0 a = 1, b = -, and c = -9 ± 4-4( - 9) ± 40 ± 4g10 ± 10 1 + 10 or 1-10 4.167766 or -.167766 -.1677 ( ) 4/3 66 is not possible. Why? 4/3 4.167766 = 6.6953793 There is one irrational root with an approximate value of 6.6953793.

5. Consider the equation y = x 9 (x) + 4. It may be a surprise, but we can use the quadratic formula to find the x-intercepts of graph because the ratio of the two exponents is : 1. Use the quadratic formula to first solve for x 1/. Once we know the value of x 1/, we can find It would be faster to solve the equation by factoring since x 9 (x) + 4 = ( (x) - 1)( (x) 4). Students should always first try the factoring method to solve the equation. If you don't see how to factor the expression after a brief amount of time, use the quadratic formula. It always works. y = x - 9 x + 4 1/ 1/ ( ) ( ) 1 y = x - 9 x + 4 a =, b = -9, c = 4 9 ± 49 9 ± 7 4 4 1 1/ 1/ 1/ 4 or 1/ 1/ ( ) ( ) 4 or 1 16 or 4 æ 1 ö ç è ø There are two rational roots with values of 16 or 1/4. Some things to think about. What is it about the equation y = x 9 (x) + 4 that causes the graph to have no points to the left of the y-axis? Consider the equation y = x 9 ( x ) + 4. Can you predict what the graph of this equation looks like? Can you find all roots of the equation without any additional work?

6. Consider the implicitly defined relation x 3xy = 4y -. Even though b and c are y-variable expressions, we can use the quadratic formula to solve the equation for x. Of course, we can also solve the equation for y. All three equations express the relationship between the x-coordinate and y- coordinate for every point (x,y) on the graph, but in a different way. It is truly amazing to see that the output graphs of all three equations are identical. Solve x 3xy = 4y - for y. Observations: x + = 3xy + 4y x + = y(3x + 4) x + y = 3x + 4 Since the numerator can never equal zero, y = f(x) has no real roots and the graph can not cross the x-axis. When -4/3, the denominator equals zero. Therefore the graph of y = f(x) must have a vertical asymptote at -4/3. Use quadratic formula to solve x 3xy = 4y - for x. x - 3xy - 4y + = 0 a =, b = -3y, c = -4y + Observation: 3y ± 9y -4(-8y + 4) 4 3y ± 9y + 3y - 16 4 The + operator expresses x as two explicit functions of y. Therefore for every valid input value of y, there are two output values for x. As y changes from y min to y max, we can see two horizontal output values x for every valid element in the domain of y.

7. Consider the implicitly defined relation 0.5y (4x 5)y + x 1 = 0. Even though b and c are x- variable expressions, we can use the quadratic formula to solve the equation for y. Of course, we can also solve the equation for x. All three equations express the relationship between the x-coordinate and y-coordinate for every point (x,y) on the graph, but in a different way. It is truly amazing to see that the output graphs of all three equations are identical. Use the quadratic formula to solve 0.5y (4x 5)y + x 1 = 0 for y. Solve 0.5y (4x 5)y + x 1 = 0 for x. 0.5y - (4x - 5)y + x - 1 = 0 a = 0.5, b = -(4x - 5), c = x - 1 y = 4x - 5 ± (4x - 5) - 4(0.5( x -1)) y = 4x - 5 ± 16x - 40x + 5 - ( x -1) y = 4x - 5 ± 16x - 40x + 5 - x + 4 y = 4x - 5 ± 16x - 4x + 49 0.5y - (4x - 5)y + x - 1 = 0 0.5y - 4xy + 5y + x - 1 = 0 x - 4xy = -(0.5y + 5y -1) x(1-4 y) = -(0.5y + 5y -1) Observations: 0.5y + 5y -1 4y -1 The equation 0.5y + 5y 1 = 0 has roots y = and y = -1. Therefore the graph of x = f(y) has y-intercepts at (0, ) and (0, -1). When 4y 1 = 0, y = ¼. Therefore the graph of f(y) must have a horizontal asymptote at y = ¼.