Chapter 2 review
This presentation is the property of its rightful owner.
Sponsored Links
1 / 11

Chapter 2 Review PowerPoint PPT Presentation


  • 48 Views
  • Uploaded on
  • Presentation posted in: General

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

Download Presentation

Chapter 2 Review

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Chapter 2 review

Chapter 2 Review

Polynomial Functions

First, some corrections on 2 slides……


L2 1 2 2 evaluating polynomials

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


Exercises answer key

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

??


Chapter 2 review1

NOW LETS START THE PRESENTATION:

Chapter 2 Review

Polynomial Functions


L2 1 2 2 evaluating polynomials1

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


L2 3 graphing polynomial functions

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)


L2 3 more exercises

3.

1.

2.

(2, 1)

(2, -4)

4.

L2.3 More Exercises

  • Write an equation for each graph


L2 4 max s min s of quadratic cubic functions

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


L2 6 solving polynomial equations finding roots

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


L2 7 general results for polynomial functions

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 .


Exercises answer key1

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


  • Login