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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**• 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**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**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**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**5.4 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**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**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**1.**2. Answers**3.**4.**5.**6.**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**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**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)**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])**Algebra II Section 5.6**A presentation by: Elise Couillard; Block 5 June 9, 2010**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.**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:**Solving a Quadratic Equation With 2 Real Solutions**Solve The solutions are x=1.35 and x=-1.85**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.**Sources: Algebra II Textbook**The End!**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**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**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**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**Travis Deskus**6.1 / 7.2**-**Using Properties of Exponents**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.**Simplifying Algebraic Expressions**• a. • b. • c. • Scientific Notation**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**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**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**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:**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 → +**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**Adding, Subracting, and Multiplying Polynomials**6.3 By: Robert Johnson**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**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**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**Multiplying Polynomials**Multiply -2y2+3y-6 and y-2 in vertical format Multiply -x2+2x+4 and x-3 in horizontal format**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**Using Special Product Patterns**(x + 2)(3x2 - x – 5) (a – 5)(a + 2)(a + 6) (xy - 4)3**6.4**Factoring and solving polynomials Anjy Grasso**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**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**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**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**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**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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