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5.1

5.1. A quadratic function f is a function of the form f(x) = ax 2 + bx + c where a, b and c are real numbers and a not equal to zero. The graph of the quadratic function is called a parabola. It is a "U" shaped curve that may open up or down depending on the sign of coefficient a.

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5.1

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  1. 5.1 • A quadratic function f is a function of the form f(x) = ax2 + bx + c where a, b and c are real numbers and a not equal to zero. The graph of the quadratic function is called a parabola. It is a "U" shaped curve that may open up or down depending on the sign of coefficient a. Jeff White

  2. Graphing A Quadratic Function Vertex is (2,-2) Then draw the axis of symmetry which is x=2 Then plot two points on one side of the axis of symmetry. Use symmetry to plot two more points. Jeff White

  3. Graphing A Quadratic Function In Vertex Form Vertex Form First plot the vertex (H,K) = (-3,4) Then draw the axis of symmetry X=-3 and plot two points on one side of it. Use symmetry to complete the graph. Jeff White

  4. Graphing A Quadratic Function In Intercept Form Intercept Form X-intercept occur at (-2,0) and (4,0) Axis of symmetry is 1 X-coordinate of the vertex is x=1. The y-coordinate of the vertex is Jeff White

  5. 5.4 Complex Numbers • Imaginary unit is called i • i= √ (-1) • r is a positive real number √ (–r)= i √ (r) Complex number written in standard form is a+bi a & b real numbers a real part of complex number b imaginary part of complex number • b ≠0 a+bi is imaginary • a=0, b≠0 a=bi is pure imaginary number • z=a+bi is complex number

  6. 1. Solve 3. Write the (2+3i)+(7+i) as a complex number in standard form. 5. Divide 2. Write (8+5i)-(1+2i) as a complex number in standard form. 4. Multiply i(3+i). 6. Find the absolute value of 3-4i Sample Problems

  7. Helpful Hints • 1. Take the square root of x squared and -4. • 2. Distribute the minus sign to 1 and 2i. Combine like terms. • 3. Distribute the plus sign to 7 and i. Combine like terms. • 4. Distribute the i to 3 and i. • 5. Divide 8 by 1 and 8 by i. • 6. Consult the formula on the first page: a=3, b=-4

  8. 1. 2. Answers

  9. 3. 4.

  10. 5. 6.

  11. Chapter 5.5: Completing the Square Goal 1: Solving Quadratic Equations by Completing the Square Completing the square is a process that allows you to write an expression of the form x2 + bx as the square of a binomial. To complete the square for x2 + bx, you need to add (b/2) 2. The following is a rule for completing the square: X2 + bx +(b/2)2 = (x+[b/2])2 Example 1: Completing the Square: Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial. x2 + 18x + c Write the equation out b=18 Use the formula to find b c = (b/2)2 = (18/2)2 = 92 = 81 Find the value of c that makes the expression a perfect square trinomial x2 + 18x + 81 Substitute the C value in the expression. (x+9)2Factor to get your answer Matt

  12. Chapter 5.5: Completing the Square Example 2: Solving a Quadratic Equation if the Coefficient of x2 Is 1: Solve the equation by completing the square. X2 + 2x = 9 Write out original equation X2+ 2x + 1 = 10 Add (2/2)2 = 12 = 1 to each side (x+1)2 = 10 Write the left side as a binomial squared X + 1 = √10 Take the square roots of each side X = -1 + √10 Solve for x Example 3: Solving a Quadratic Equation if the Coefficient of x2 Is Not 1: Solve the Equation by Completing the Square. 6x2 +84x +300 = 0 Write the original equation X2 +14x +50 = 0 Divide both sides by the coefficient of x2 X2 + 14x = -50 Write the left side in the form of x2 + bx X2 + 14x + 49 = -1 Add (14/2)2 = 72 = 49 to each side (X + 7)2 = -1 Write left side as a binomial squared X + 7 = √-1 Take the square roots of each side X = -7 ± √-1 Solve for x X = -7 ± iWrite in terms of the imaginary unit i

  13. Chapter 5.5: Completing the Square Goal 2: Writing Quadratic Functions in Vertex Form: Given a quadratic function in standard form, y = ax2 + bx + c, you can use completing the square to write the function in vertex form, y = a(x – h)2 + k. Example 4: Writing a Quadratic Function in Vertex Form: Write the quadratic function in vertex form and identify the vertex. Y = x2 – 6x + 11 Write out the original function Y + 9 = (x2 – 6x + 9) + 11 Complete the square of x2 – 6; add (-6/2)2 = -32 = 9 Y + 9 = (x – 3)2 + 11 Write x2 – 6x + 9 as a binomial squared Y = (x – 3)2 +2 Solve for y Vertex = (3,2)

  14. Chapter 5.5: Completing the Square Practice Problems for Completing the Square: Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial. X2 – 44x + c Answer: C = 484; (x – 22)2 Practice Problems for Solving a Quadratic Equation if the Coefficient of x2 Is 1: Solve the equation by completing the square. X2 + 20x + 104 = 0 Answer: -10 + 2i, -10 – 2i Practice Problems for Solving a Quadratic Equation if the Coefficient of x2 Is Not 1: Solve the Equation by Completing the Square. 2x2 – 12x = -14 Answer: 3 ± √2 Practice Problems for Writing a Quadratic Function in Vertex Form: Write the quadratic function in vertex form and identify the vertex. Y = x2 – 3x – 2 Answer: y = (x – [3/2])2 – (17/4) Vertex: ([3/2], [-17/4])

  15. Algebra II Section 5.6 A presentation by: Elise Couillard; Block 5 June 9, 2010

  16. Solving Equations With The Quadratic Formula By completing the square once for the general equation , you can develop a formula that gives the solutions of any quadratic equation. The formula for the solutions is called the quadratic formula.

  17. The Quadratic Formula The Quadratic Formula: *Let a, b, and c be real numbers such that a does not equal 0. The solutions of the quadratic equation are:

  18. Solving a Quadratic Equation With 2 Real Solutions Solve The solutions are x=1.35 and x=-1.85

  19. Number and Type of Solutions of a Quadratic Equation Consider the quadratic equation *If >0, then the equation has two real solutions. *If =0, then the equation has one real solution. *If <0, then the equation has two imaginary solutions.

  20. Sources: Algebra II Textbook The End!

  21. Quadratic Inequality in 2 Variables Quadratic Inequality in 1 Variable y < ax2 + bx +c y < ax2 + bx +c ax2 + bx +c < 0 ax2 + bx +c < 0 y < ax2 + bx +c y > ax2 + bx +c ax2 + bx +c > 0 ax2 + bx +c > 0 Example 1 1. Graph y < x2 + 8x + 16 2. Test the point (0,0) y<x2 + 8x+ 16 0<02 + 8(0)+ 16 0 < 16 3. Shade outside region because 0 < 16 Lesson 5.7 Jessica Semmelrock

  22. Example 2: Graph the system of quadratic inequalities y > x2 y < x2 + 3 Only difference to this problem is when the shading overlaps that is your answer. Lesson 5.7 Jessica Semmelrock

  23. Example 3: Solve x2 + x -2 < 0 by graphing • Step 1: x2 + x -2 = 0 • Find graph intercepts by replacing 0 for x. • Step 2: x = -1 + 1 – 4(1)(-2) • 2(1) • Use quadratic formula to solve for x. Answer: x ≤ –1.37 or x ≥ .37 Lesson 5.7 Jessica Semmelrock

  24. Example 4: Solve x2 + 3 -18 > 0 algebraically x2 + 3 -18 > 0 x2 + 3 -18 = 0 (x-3) (x+6) = 0 Test an x-value in each interval to see if it satisfies the inequality. Test these points with the arrows. Answer x = 3 or x = -6 Lesson 5.7 Jessica Semmelrock

  25. Travis Deskus 6.1 / 7.2

  26. - Using Properties of Exponents

  27. Evaluating Numerical Expressions • 1. Product of like bases: Example: x5 x3 = x5+3 = x8 • To multiply powers with the same base, add the exponents and keep the common base. • =32 • 2.Evaluating numerical expresions a.

  28. Simplifying Algebraic Expressions • a. • b. • c. • Scientific Notation

  29. 7.2 Properties of Rational Exponents • Example 1.   • If m= for some integer n greater than 1, the third and sixth properties can be written using radical notation as following: • product property • Quotient property • a. • b. • Using Properties of Radicals

  30. Writing Radicals in Simplest Form • a. steps- factor out perfect cube, product property, simplify • Adding and Subtracting Roots and Radicles • Book Example • The properties of rational exponent and radicals can also be applied to expressions involving variables. Because a variable can be positive, negative or zero, sometimes absolute value is needed when simplifying a variable expression. • =x when n is odd = when n is even • Simplifying expressions involving varribles

  31. Section 6.2

  32. IDENTIFYING POLYNOMIAL FUNCTIONS • A polynomial is a function if it’s in standard form and the exponent is a whole number • ex: f(x)= 3 • If the polynomial has an exponent that is not a whole number it’s not a function • ex: f(x)= 3x1/2 – 2x2 +5

  33. Using Synthetic Substitution • Write the polynomial in standard form • Insert terms with coefficients of 0 for missing terms • Then write the coefficients of f(x) in a row • Bring down the leading coefficients and multiply them by 1 • Write results in the next column and bring down your results • Continue until you reach the end of the row • EX:

  34. Graphing Polynomials Functions • Begin by making a table of values, including positive, negative, and zero values for x • Plot the points and connect them with a smooth curve. Then check the behavior • The Degree is odd and the leading coefficients is positive, so • f(x)→ + - • As x → - • f(x) → + • As x → +

  35. Examples of Graphs

  36. Work • Graph the polynomial function • 1.) f(x)= x4+3 • 2.) g(x)= x3-5 • 3.) h(x)= 2+ x2-x4 • Synthetic substitution • 1.) f(x)=x3+ 5x2+4x+6, x=2 • 2.) f(x)= x3-x5 +3, x=-1 • 3.) f(x)= 5x3-4x2-2, x=0 • Functions yes or no • 1.) f(x)= x4+3 • 2.) f(x)=5x3/4-5x2+3

  37. Adding, Subracting, and Multiplying Polynomials 6.3 By: Robert Johnson

  38. How to Solve To add or subtract polynomials, add or subtract the coefficients of LIKE terms. You can do this by using a vertical or horizontal format To Multiply two polynomials, each term of the first polynomial must be multiplied by each term of the second polynomial. Then combine LIKE terms

  39. Adding Polynomials Add 2x3-5x2+3x-9 and x3+6x2+11 in vertical format Add 3x3+2x2-x-7 and x3-10x2+8 in horizontal format

  40. Subtracting Polynomials Subtract 3x3+2x2-x+7 from 8x3-x2-5x+2 in vertical format Subract 8x3-3x2-2x+9 from2x3+6x2-x+1 in horizontal format

  41. Multiplying Polynomials Multiply -2y2+3y-6 and y-2 in vertical format Multiply -x2+2x+4 and x-3 in horizontal format

  42. Special Product Patterns Sum and difference (a+b)(a-b)= a2 - b2 Square of a Binomial (a+b)2 = a2 + 2ab + b2 (a-b)2 = a2 - 2ab + b2 Cube of a Binomial (a+b)3 = a3 + 3a2b + 3ab2 + b3 (a-b)3 = a3 - 3a2b + 3ab2 - b3

  43. Using Special Product Patterns (x + 2)(3x2 - x – 5) (a – 5)(a + 2)(a + 6) (xy - 4)3

  44. 6.4 Factoring and solving polynomials Anjy Grasso

  45. Step One Step Two x² + 8x + 12 = 0 x² + 8x = −12 Now that you have the equation you're solving, find the first factor Put the equation into Standard form In other words, make it equal zero. x² + 8x + 12 = 0 ( ) ( ) = 0 x² + 8x = −12 +12 +12 the only factors of x2 are x * x, you now have the first factors. x² + 8x + 12 = 0 x² + 8x + 12 = 0 (x ) (x ) = 0 Anjy Grasso

  46. Step Three Now that we have the first factors, the x2 goes away, and we're left with this: (x 6) (x 2) = 0 Now we find the factors of 12. Since the entire equation was positive, Both of these should be positive too. 1 and 12 wont work, neither will 4 and 3, so lets use 2 and 6 (x + 6) (x + 2) = 0 8x + 12 = 0 8x + 12 = 0 (x ) (x ) = 0 (x ) (x ) = 0 Anjy Grasso

  47. Checking Your Work So how exactly do we know this is right? Lets use FOIL (first, outer, inner, last) multiplication to test it out. (x + 6) (x + 2) = 0 F - First x * x = x2 O - Outer x * 2 = + 2x x2 + 2x + 6x + 12 I - Inner 6 * x = + 6x Now Simplify… x2 + 8x + 12 L - Last 6 * 2 = + 12 There's Your original equation! Anjy Grasso

  48. 6.5 Remainder and Factor Theorems Polynomial Long Division – when dividing a polynomial f(x) by a divisor d(x), you get a quotient polynomial -q(x) and a remainder polynomial r(x) f(x) = q(x) + r(x) d(x) d(x) Remainder Theorem– If polynomial f(x) is divided by x-k, then the remainder is r = f(x) Synthetic Division – only use the Coefficients of the polynomial and the x – k must be in the form of a divisor. Factor Theorem – A polynomial f(x) has a factor x-k if and only if f(k) = 0 Rachael SKinner

  49. Brief Refresher Long Division f(x) = 3x² + 4x -3 by x² - 3x + 5 *Don't forget to add in exponents if needed exponents must go in numerical order 3x² + 4x -3 x² - 3x + 5 3x⁴ - 5x³ 0x² + 4x – 6 - 3x⁴- 9x³+ 15x² 4x² - 15x² + 4x - 4x² - 12x² - 20x *Remember to subtract – which means the signs will change -3x² - 16x-6 -3x² + 9x -15 25x+9 25x+9 Remainder 3x² + 4x -3 x² - 3x + 5 Rachael SKinner

  50. Brief Review Synthetic Division f(x) = 2x³ +x² - 8x +5 by x + 3 Factoring Completely f(x) = 3x³ - 4x² - 28x – 16 x +2 is a factor x + 3 = 0 x – 3 = - 3 x = - 3 x = 2 3 -4 -28 -16 2 -6 20 16 2 1 -8 5 *Use only coefficients 3 -10 -8 0 no remainder -3 -6 15 -21 f(x) (2 + x) (3x² - 10x -8) Factor 2 -5 7 -16 Remainder f(x) (x+2) (3x+2) (x – 4) Solve finding the 0's of f(x) (x+2) = -2 (3x+2) = -2/3 (x – 4) = 4 2x³ +x² - 8x +5 -16 x+3 Rachael SKinner

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