Slide 2- 1

Chapter 2 Polynomial, Power, and Rational Functions

2.1 Linear and Quadratic Functions and Modeling

Slide 2- 4 Quick Review

Slide 2- 5 What you’ll learn about Polynomial Functions Linear Functions and Their Graphs Average Rate of Change Linear Correlation and Modeling Quadratic Functions and Their Graphs Applications of Quadratic Functions … and why Many business and economic problems are modeled by linear functions. Quadratic and higher degree polynomial functions are used to model some manufacturing applications.

Slide 2- 6 Polynomial Function

Slide 2- 7 Polynomial Functions of No and Low Degree NameFormDegree Zero Functionf(x) = 0Undefined Constant Functionf(x) = a (a ≠ 0)0 Linear Functionf(x)=ax + b (a ≠ 0)1 Quadratic Functionf(x)=ax 2 + bx + c (a ≠ 0)2

Slide 2- 8 Example Finding an Equation of a Linear Function

Slide 2- 9 Average Rate of Change

Slide Constant Rate of Change Theorem A function defined on all real numbers is a linear function if and only if it has a constant nonzero average rate of change between any two points on its graph.

Slide Characterizing the Nature of a Linear Function Point of ViewCharacterization Verbalpolynomial of degree 1 Algebraic f(x) = mx + b (m ≠ 0) Graphicalslant line with slope m and y-intercept b Analyticalfunction with constant nonzero rate of change m: f is increasing if m > 0, decreasing if m < 0; initial value of the function = f(0) = b

Slide Properties of the Correlation Coefficient, r ≤ r ≤ 1 2. When r > 0, there is a positive linear correlation. 3. When r < 0, there is a negative linear correlation. 4. When |r| ≈ 1, there is a strong linear correlation. 5. When |r| ≈ 0, there is weak or no linear correlation.

Slide Linear Correlation

Slide Regression Analysis 1. Enter and plot the data (scatter plot). 2. Find the regression model that fits the problem situation. 3. Superimpose the graph of the regression model on the scatter plot, and observe the fit. 4. Use the regression model to make the predictions called for in the problem.

Slide Example Transforming the Squaring Function

Slide Example Transforming the Squaring Function

Slide The Graph of f(x)=ax 2

Slide Vertex Form of a Quadratic Equation Any quadratic function f(x) = ax 2 + bx + c, a ≠ 0, can be written in the vertex form f(x) = a(x – h) 2 + k The graph of f is a parabola with vertex (h,k) and axis x = h, where h = -b/(2a) and k = c – ah 2. If a > 0, the parabola opens upward, and if a < 0, it opens downward.

Slide Find the vertex and the line of symmetry of the graph y = (x – 1) Domain Range (- ,  ) [2,  ) Vertex (1,2) x = 1

Slide Find the vertex and the line of symmetry of the graph y = -(x + 2) Domain Range (- ,  ) (- ,-3] Vertex (-2,-3) x = -2

Slide Let f(x) = x 2 + 2x + 4. (a) Write f in standard form. (b) Determine the vertex of f. (c) Is the vertex a maximum or a minimum? Explain f(x) = x 2 + 2x + 4 f(x) = (x + 1) Vertex (-1,3) opens up (-1,3) is a minimum

Slide Let f(x) = 2x 2 + 6x - 8. (a) Write f in standard form. (b) Determine the vertex of f. (c) Is the vertex a maximum or a minimum? Explain f(x) = 2(x + 3/2) /2 Vertex (-3/2,-25/2) opens up (-3/2,25/2) is a minimum + 9/4 - 9/2f(x) = 2(x 2 + 3x ) - 8

Slide If we perform completing the square process on f(x) = ax 2 + bx + c and write it in standard form, we get

Slide So the vertex is

Slide To get the coordinates of the vertex of any quadratic function, simply use the vertex formula. If a > 0, the parabola open up and the vertex is a minimum. If a < 0, the parabola opens down and the parabola is a maximum.

Slide Example Finding the Vertex and Axis of a Quadratic Function

Slide Characterizing the Nature of a Quadratic Function Point of View Characterization

Slide Vertical Free-Fall Motion

2.2 Power Functions and Modeling

Slide Quick Review

Slide What you’ll learn about Power Functions and Variation Monomial Functions and Their Graphs Graphs of Power Functions Modeling with Power Functions … and why Power functions specify the proportional relationships of geometry, chemistry, and physics.

Slide Power Function Any function that can be written in the form f(x) = k·x a, where k and a are nonzero constants, is a power function. The constant a is the power, and the k is the constant of variation, or constant of proportion. We say f(x) varies as the a th power of x, or f(x) is proportional to the a th power of x.

Slide Example Analyzing Power Functions

Slide Monomial Function Any function that can be written as f(x) = k or f(x) = k·x n, where k is a constant and n is a positive integer, is a monomial function.

Slide Example Graphing Monomial Functions

Slide Graphs of Power Functions For any power function f(x) = k·x a, one of the following three things happens when x < 0. f is undefined for x < 0. f is an even function. f is an odd function.

Slide Graphs of Power Functions

2.3 Polynomial Functions of Higher Degree with Modeling

Slide Quick Review

Slide What you’ll learn about Graphs of Polynomial Functions End Behavior of Polynomial Functions Zeros of Polynomial Functions Intermediate Value Theorem Modeling … and why These topics are important in modeling and can be used to provide approximations to more complicated functions, as you will see if you study calculus.

Slide The Vocabulary of Polynomials

Slide Example Graphing Transformations of Monomial Functions

Slide Cubic Functions

Slide Quartic Function

Slide Local Extrema and Zeros of Polynomial Functions A polynomial function of degree n has at most n – 1 local extrema and at most n zeros.

Slide Leading Term Test for Polynomial End Behavior

Slide Example Applying Polynomial Theory

Slide Example Finding the Zeros of a Polynomial Function

Slide Multiplicity of a Zero of a Polynomial Function

Slide Example Sketching the Graph of a Factored Polynomial

Slide Intermediate Value Theorem If a and b are real numbers with a < b and if f is continuous on the interval [a,b], then f takes on every value between f(a) and f(b). In other words, if y 0 is between f(a) and f(b), then y 0 =f(c) for some number c in [a,b].

2.4 Real Zeros of Polynomial Functions

Slide Quick Review

Slide What you’ll learn about Long Division and the Division Algorithm Remainder and Factor Theorems Synthetic Division Rational Zeros Theorem Upper and Lower Bounds … and why These topics help identify and locate the real zeros of polynomial functions.

Slide Remainder Check: Quotient * Divisor + Remainder = Dividend Division

Slide Division Algorithm for Polynomials

Slide Dividing Polynomials Long division of polynomials is similar to long division of whole numbers. dividend = (quotient divisor) + remainder The result is written in the form: quotient + When you divide two polynomials you can check the answer using the following:

Slide Example: Divide x 2 + 3x – 2 by x + 1 and check the answer. x x 2 + x 2x2x– 2 2x + 2 – 4– 4 remainder Check: correct (x + 2) quotient (x + 1) divisor + (– 4) remainder = x 2 + 3x – 2 dividend Answer: x – 4– 4

Slide Example: Divide 4x + 2x 3 – 1 by 2x – 2 and check the answer. Write the terms of the dividend in descending order. 1. x2x2 2. 2x 3 – 2x x22x2 + 4x 4. + x 5. 2x 2 – 2x 6. 6x6x – x – Check: (x 2 + x + 3)(2x – 2) + 5 = 4x + 2x 3 – 1 Answer: x 2 + x Since there is no x 2 term in the dividend, add 0x 2 as a placeholder.

Slide x x 2 – 2x – 3x + 6 – 3 – 3x Answer: x – 3 with no remainder. Check: (x – 2)(x – 3) = x 2 – 5x + 6 Example: Divide x 2 – 5x + 6 by x – 2.

Slide Example: Divide x 3 + 3x 2 – 2x + 2 by x + 3 and check the answer. x2x2 x 3 + 3x 2 0x20x2 – 2x – 2 – 2x – 6 8 Check: (x + 3)(x 2 – 2) + 8 = x 3 + 3x 2 – 2x + 2 Answer: x 2 – Note: the first subtraction eliminated two terms from the dividend. Therefore, the quotient skips a term. + 0x

Slide Synthetic division is a shorter method of dividing polynomials. This method can be used only when the divisor is of the form x – a. It uses the coefficients of each term in the dividend. Example: Divide 3x 2 + 2x – 1 by x – 2 using synthetic division. 3 2 – 1 2 Since the divisor is x – 2, a = Bring down 3 2. (2 3) = (2 + 6) = 8 4. (2 8) = (–1 + 16) = 15 coefficients of quotient remainder value of a coefficients of the dividend 3x + 8Answer: 15

Slide Example: Divide x 3 – 3x + 4 by x + 3 using synthetic division. Notice that the degree of the first term of the quotient is one less than the degree of the first term of the dividend. remainder a coefficients of quotient – 3– 3 Since, x – a = x + 3, a = – – – 3– 3 – 3– 3 9– 18 6– 14 coefficients of dividend = x 2 – 3x + 6 – 14 Insert zero coefficient as placeholder for the missing x 2 term.

Slide Remainder Theorem: The remainder of the division of a polynomial f (x) by x – a is f (a). Example: Using the remainder theorem, evaluate f(x) = x 4 – 4x – 1 when x = – 4 – The remainder is 68 at x = 3, so f (3) = 68. You can check this using substitution:f(3) = (3) 4 – 4(3) – 1 = 68. value of x

Slide Example: Using synthetic division and the remainder theorem, evaluate f (x) = x 2 – x at x = – – 1 0 – 2– 2 1 – 2– 2 – 3– 36 Then f (– 2) = 6 and (– 2, 6) is a point on the graph of f(x) = x 2 – x. f(x) = x 2 – x x y 2 4 (– 2, 6) remainder

Slide Example Using Polynomial Long Division

Slide Example Using Polynomial Long Division

Slide Remainder Theorem

Slide Example Using the Remainder Theorem

Slide Factor Theorem

Slide Example Using Synthetic Division

Slide Rational Zeros Theorem

Slide Upper and Lower Bound Tests for Real Zeros

Slide Show 2x - 3 is a factor of 6x 2 + x – 15 (x = 3/2) 3/ Note: Since the remainder is 0, 2x - 3 is a factor of 6x 2 + x – 15

Slide Factor x 4 – 3x 3 – 5x 2 + 3x + 4 are possible factors P(1) = x 3 – 2x 2 – 7x - 4 so x 4 – 3x 3 – 5x 2 + 3x + 4 = (x 3 – 2x 2 – 7x – 4)(x – 1)

Slide Factor x 3 – 2x 2 – 7x - 4 are possible factors P(-1) = x 2 – 3x – 4 so x 4 – 3x 3 – 5x 2 + 3x + 4 = (x 2 – 3x – 4)(x – 1)(x + 1)

Slide Factor x 3 – 2x 2 – 7x - 4 are possible factors P(-1) = x 2 – 3x – 4 so x 4 – 3x 3 – 5x 2 + 3x + 4 = (x 2 – 3x – 4)(x – 1)(x + 1)

Slide x 4 – 3x 3 – 5x 2 + 3x + 4 = (x 2 – 3x – 4)(x – 1)(x + 1) or (x – 4)(x + 1)(x – 1)(x + 1)

Slide Factor x 4 – 8x 3 +17x 2 + 2x - 24 are possible factors P(4) = x 3 – 4x 2 + x + 6 so x 4 – 8x x 2 + 2x - 24 = (x 3 – 4x 2 + x + 6)(x – 4)

Slide Factor x 3 – 4x 2 + x + 6 P(2) = x 2 – 2x – 3 so x 4 – 8x 3 +17x 2 + 2x - 24 = (x 2 – 2x – 3)(x – 4)(x - 2) are possible factors

Slide x 4 – 8x 3 +17x 2 + 2x - 24 = (x 2 – 2x – 3)(x – 4)(x - 2) or (x – 3)(x + 1)(x – 4)(x - 2)

Slide Show x 3 – 3x = 0 has no rational roots  1,  5 only possible roots P(1) = 1 3 – 3(1) = 3 P(-1) = (-1) 3 – 3(-1) = 1 P(5) = (5) 3 – 3(5) = 55 P(-5) = (-5) 3 – 3(-5) = -195 Since none of the possible roots give zero in the remainder theorem, there are no rational roots.

Slide Example Finding the Real Zeros of a Polynomial Function

Slide Example Finding the Real Zeros of a Polynomial Function

Slide Example Finding the Real Zeros of a Polynomial Function

2.5 Complex Zeros and the Fundamental Theorem of Algebra

Slide Quick Review

Slide What you’ll learn about Two Major Theorems Complex Conjugate Zeros Factoring with Real Number Coefficients … and why These topics provide the complete story about the zeros and factors of polynomials with real number coefficients.

Slide Fundamental Theorem of Algebra A polynomial function of degree n has n complex zeros (real and nonreal). Some of these zeros may be repeated.

Slide Linear Factorization Theorem

Slide Fundamental Polynomial Connections in the Complex Case The following statements about a polynomial function f are equivalent if k is a complex number: 1. x = k is a solution (or root) of the equation f(x) = 0 2. k is a zero of the function f. 3. x – k is a factor of f(x).

Slide Example Exploring Fundamental Polynomial Connections

Slide Complex Conjugate Zeros

Slide Example Finding a Polynomial from Given Zeros

Slide Factors of a Polynomial with Real Coefficients Every polynomial function with real coefficients can be written as a product of linear factors and irreducible quadratic factors, each with real coefficients.

Slide Find an equation of a polynomial with roots of 2i and 3. (x – 2i)(x + 2i)(x – 3) = 0 (x 2 – 2ix + 2ix – 4i 2 )(x – 3) = 0 (x 2 + 4)(x – 3) = 0 x 3 – 3x 2 + 4x – 12 = 0

Slide Find an equation of a polynomial with roots of 1- i and -2. (x – (1 - i))(x – (1 + i))(x + 2) = 0 (x 2 – x - ix – x i +ix -i - i 2 )(x + 2) = 0 (x 2 – 2x + 2)(x + 2) = 0 x 3 + 2x 2 –2x 2 - 4x + 2x + 4= 0 (x – 1 + i)(x – 1 - i)(x + 2) = 0 x 3 - 2x + 4 = 0

Slide Complex Zeros and the Fundamental Theorem of Algebra Page 230

Slide Complex Zeros and the Fundamental Theorem of Algebra (cont’d) Page 230

Slide Example Factoring a Polynomial

Slide Example Factoring a Polynomial

2.6 Graphs of Rational Functions

Slide Quick Review

Slide What you’ll learn about Rational Functions Transformations of the Reciprocal Function Limits and Asymptotes Analyzing Graphs of Rational Functions … and why Rational functions are used in calculus and in scientific applications such as inverse proportions.

Slide Rational Functions

Slide Example Finding the Domain of a Rational Function

Slide Rules for Asymptotes for Rational Functions

Slide Graph a Rational Function

Slide Graph a Rational Function

Slide Example Finding Asymptotes of Rational Functions

Slide Example Graphing a Rational Function  

Slide Sketch the graph of f(0) = -3 Solve x – 3 = 0; x = 3 Solve x + 1 = 0; x = -1 and sketch the asymptote. Horizontal asymptote, y =  Example Graphing a Rational Function

Slide Sketch the graph of f(0) = -2 Solve x + 2 = 0; x = -2 Solve x = 0; x = -1 x = 1 and sketch the asymptotes. Horizontal asymptote, y =   Example Graphing a Rational Function

Slide Graphs of Rational Functions Page 239

Slide Graphs of Rational Functions Page 239

Slide Graphs of Rational Functions Page 239

Slide Graphs of Rational Functions Page 239

2.7 Solving Equations in One Variable

Slide Quick Review

Slide What you’ll learn about Solving Rational Equations Extraneous Solutions Applications … and why Applications involving rational functions as models often require that an equation involving fractions be solved.

Slide Extraneous Solutions When we multiply or divide an equation by an expression containing variables, the resulting equation may have solutions that are not solutions of the original equation. These are extraneous solutions. For this reason we must check each solution of the resulting equation in the original equation.

Slide Example Solving by Clearing Fractions The LCD is x Multiply both sides by x Subtract 3x from both sides (x – 2)(x – 1) = 0 x = 2 or x = 1 Check Both solutions are correct

Slide Example Eliminating Extraneous Solutions The LCD is (x – 1)(x - 3) (x – 1)(1) + 2x(x - 3) = 2 2x 2 – 5x – 3 = 0 (2x + 1)(x - 3) = 0 x = -½ or x = 3 Check: x = 3 is not defined x = -½ is the only solution x – 1 + 2x 2 - 6x = 2

Slide Example Eliminating Extraneous Solutions The LCD is (x)(x + 2) (x + 2)(x - 3) + 3x + 6 = 0 x 2 + 2x = 0 x(x + 2) = 0 x = 0 or x = -2 Check: x = 0 is not defined x = -2 is not defined No Solution x 2 – x – 6 + 3x + 6 = 0 Solve the equation

Slide Example Finding a Minimum Perimeter

Slide Example Acid Problem a.Pure acid is added to 78 oz of a 63% acid solution. Let x be the amount (in ounces) of pure acid added. Find an algebraic representation for C(x), the concentration of acid as a function of x. Determine how much pure acid should be added so that the mixture is at least 83% acid. 1.00x + (.63)(78) = C(x)(x + 78) 1.00x >.83x x > 15.6x > 91.76

2.8 Solving Inequalities in One Variable

Slide Quick Review

Slide What you’ll learn about Polynomial Inequalities Rational Inequalities Other Inequalities Applications … and why Designing containers as well as other types of applications often require that an inequality be solved.

Slide Polynomial Inequalities

Slide Example Finding where a Polynomial is Zero, Positive, or Negative -34 (-)(-) 2 (+)(-) 2 (+)(+) 2 negative positive

Slide Example Solving a Polynomial Inequality Graphically

Slide Example Solving a Polynomial Inequality Graphically

Slide Example Creating a Sign Chart for a Rational Function -31 (-) (-)(-) negative positive (-) (+)(-) (+) (+)(+) (+) (+)(-) negative 0und. 0

Slide Example Solving an Inequality Involving a Radical 2 (-)(+)(+)(+) undefined positivenegative 00

Slide Chapter Test

Slide Chapter Test

Slide Chapter Test

Slide Chapter Test Solutions

Slide Chapter Test Solutions

Slide Chapter Test Solutions