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Chapter 2 Review

Chapter 2 Review. Polynomial Functions. First, some corrections on 2 slides……. CORRECTIONS on SLIDE 1. L2.1 – 2.2 Evaluating Polynomials. P(x) = a n x n + a n-1 x n-1 + … + a 2 x 2 + a 1 x + a 0 Evaluate by direct substitution or

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Chapter 2 Review

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  1. Chapter 2 Review Polynomial Functions First, some corrections on 2 slides……

  2. CORRECTIONS on SLIDE 1 L2.1 – 2.2 Evaluating Polynomials • P(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 • Evaluate by • direct substitution or • Remainder Theorem: When P(x) is divided by (x – a), the remainder is P(a). a.k.a Synthetic Substitution • Zeros and Factors • a is a zero of P(x), if P(a) = 0 [A root refers to the soln to an eqn] • (x – a) is a factor of P(x), if P(x) divided by (x – a) leaves a zero remainder. [Factor Theorem, (x – a) is factor of P(x),iffP(a) = 0] EXERCISES: Give real zeros of each function. • P(x) = (x – 7)2(2x + 1) • f(x) = x3 + 2x2 + x, Also evaluate f(-2) by direction & synthetic substitution • g(x) = 6x2 – 7x – 20 • h(x) = x3 – 5x2 + x – 5 • k(x) = 4x4 – 17x2 + 18

  3. CORRECTIONS on Last Slide Exercises: Answer Key L2.1–2.2: 1. 7, –1/2; 2. 0, –1,f(–2) = –2; 3. 5/2, –4/3; 4. 5; 5. L2.3a: 1. 2. 3. L2.3b: 1. y = (x – 1)(x – 3)2 2. y = 2(x – 1)(x – 3)2(x – 4) 3. y = ½(x – 2)3 4. y = –(x – 1)3(x – 3) L2.4: 1. The minimum product is –9 and the numbers are –3 and 3. 2. a) The base of the box is a square with length = width = 10 – 2x. The height is x. Volume, V(x) = x(10 – 2x)2 b) x(10 – 2x)2 > 0 so 0 < x < 5 L2.6: 1. , x = -5; 2. ; 3. x = -2, L2.7: 1. x3 – 10x2 + 33x – 34 = 0 2. other roots: , –2 3. other roots: , 0, –6 L2.4 L2.6 ??

  4. NOW LETS START THE PRESENTATION: Chapter 2 Review Polynomial Functions

  5. L2.1 – 2.2 Evaluating Polynomials • P(x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 • Evaluate by • direct substitution or • Remainder Theorem: When P(x) is divided by (x – a), the remainder is P(a). a.k.a Synthetic Substitution • Zeros and Factors • a is a zero of P(x), if P(a) = 0 [A root refers to the soln to an eqn] • (x – a) is a factor of P(x), if P(x) divided by (x – a) leaves a zero remainder. [Factor Theorem, (x – a) is factor of P(x),iffP(a) = 0] EXERCISES: Give real zeros of each function. • P(x) = (x – 7)2(2x + 1) • f(x) = x3 + 2x2 + x, Also evaluate f(-2) by direct & synthetic substitution • g(x) = 6x2 – 7x – 20 • h(x) = x3 – 5x2 + x – 5 • k(x) = 4x4 – 17x2 + 18

  6. L2.3 Graphing Polynomial Functions • Leading coefficient test: if an > 0 RHS , if an < 0 RHS • If degree of P(x) is even, both ends go in the same direction odd, the ends go in opposite directions • Multiplicity of zeros: (x – a)n is a factor of P(x) • If n is even, graph is tangent to x-axis at a (touches but doesn’t cross) • If n is odd, graph [flattens and ]crosses the x-axis at a. • To graph: identify zeros and create a sign graph to determine intervals where graph is above and below the x-axis. Verify by examining degree and sign of an. EXERCISES: • Sketch graphs. • y = –x2(x + 4)(x – 4) • y = x(x + 3)2 • y = x3(2 – x)

  7. 3. 1. 2. (2, 1) (2, -4) 4. L2.3 More Exercises • Write an equation for each graph

  8. x L2.4 Max’s & Min’s of Quadratic & Cubic Functions • Quadratic Functions – max or min occurs at vertex Value is • Cubic Functions – have “local” max’s and min’s. Odd degree functions are unbounded, so have min or max points only within an interval. Use a graphing calculator to approximate max’s or min’s. • EXERCISES: • Two numbers have a difference of 6. Find their minimum possible product. • An open box is to be formed by cutting squares from a 10cm square sheet of metal and then folding up the sides as shown. • Write a function, V(x), to describe the volume of • the resulting box. • Find the domain of V(x). x 10cm

  9. L2.6 Solving Polynomial Equations (finding roots) • Use Quadratic Techniques (Factor, CTS, Quad Formula) • Recast higher degree polynomials to Quadratic Form • Rational Root Theorem – potential rational roots are of the form , where p = factors of a0 and q = factors of an. Test roots using synthetic division and find subsequent roots using depressed polynomial. EXERCISES: Solve. • x3 + 5x2 – 4x – 20 = 0 • 2x4 – x2 – 3 = 0 • 3x4 + 13x3 + 15x2 – 4 = 0

  10. L2.7 General Results for Polynomial Functions • Fundamental Theorem of Algebra: a polynomial of degree n has exactly n roots in the complex number system. • Complex and irrational roots occur in conjugate pairs. • Odd degree polynomials have at least one real root and their graphs are unbounded. • anxn + an-1xn-1 + … + a2x2 + a1x + a0 = 0 Sum of Roots: Product of Roots: • Quadratics: sum = –b/a, product = c/a EXERCISES: • Find a cubic equation with integral coefficients that has 2 and 4 + i as roots. • A cubic equation has no quadratic term. What are its roots if one of them is . • A quartic (4th degree) equation has no cubic term and no constant term. What are its roots if one of them is .

  11. Exercises: Answer Key L2.1–2.2: 1. 7, –1/2; 2. 0, –1,f(–2) = –2; 3. 5/2, –4/3; 4. 5; 5. L2.3a: 1. 2. 3. L2.3b: 1. y = (x – 1)(x – 3)2 2. y = 2(x – 1)(x – 3)2(x – 4) 3. y = ½(x – 2)3 4. y = –(x – 1)3(x – 3) L2.4: 1. The minimum product is –9 and the numbers are –3 and 3. 2. a) The base of the box is a square with length = width = 10 – 2x. The height is x. Volume, V(x) = x(10 – 2x)2 b) x(10 – 2x)2 > 0 so 0 < x < 5 L2.6: 1. , x = -5; 2. ; 3. x = -2, L2.7: 1. x3 – 10x2 + 33x – 34 = 0 2. other roots: , –2 3. other roots: , 0, –6

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