Ch. 5 Notes

5.1 Warm-Up Graph y = -x2 + 2x + 3 1) First find the x-coordinate of the vertex: x = - b/2a y = -(1)2 + 2(1) + 3 –1 –5 –4 –3 –2 1 2 5 4 3 6 9 8 7 10 x = - 2/2(-1) y = -1 + 5 y = 4 x = 1 Vertex = (1,4) 2) Draw the axis of symmetry x = 1 3) Plot two points on one side of the axis of symmetry x y Use symmetry to mirror those two points -1 3

5.1 Modeling Data With Quadratic Functions CA State Standard 9.0 Students explain the effect that a, b, and c, have on the parabola in the equation y = a(x – b)2 + c. CA State Standard 10.0 Students graph quadratic functions and determine the maxima, minima, and zeros of the function. CA State Standard 17.0 Given ax2 + by2 + cx + dy + e = 0, students can use the method for completing the square to put the equation in standard form and recognize what type of graph it is.

5.1 FOIL METHOD F O I L x2 + 5x + 3x + 15 = x2 + 8x + 15 by Jason L. Bradbury F O I L x2 + 5x + 3x + 15 = x2 + 8x + 15 (x + 3)(x + 5) = F = First Terms O = Outer Terms I = Inner Terms L = Last Terms

Extra Example 1a y = -(x + 4)(x – 9) y = -(x2 – 9x + 4x – 36) y = -(x2 – 5x – 36) y = -x2 + 5x + 36

Extra Example 1b y = 3(x – 1)2 + 8 y = 3(x – 1)(x – 1) + 8 y = 3(x2 –x – x + 1) + 8 y = 3(x2 –2x + 1) + 8 y = 3x2 –6x + 3 + 8 y = 3x2 –6x + 11

5.2 Warm-Up Graph y = x2 by choosing values for x and solving for y. Tell me where the vertex is. –1 –5 –4 –3 –2 1 2 5 4 3 6 9 8 7 10 x y Axis of Symmetry -3 -2 -1 1 2 3 9 4 1 Vertex = (0,0) 1 4 9

5.2 Properties of Parabolas CA State Standard 9.0 Students explain the effect that a, b, and c, have on the parabola in the equation y = a(x – b)2 + c. CA State Standard 10.0 Students graph quadratic functions and determine the maxima, minima, and zeros of the function. CA State Standard 17.0 Given ax2 + by2 + cx + dy + e = 0, students can use the method for completing the square to put the equation in standard form and recognize what type of graph it is.

5.2 Properties of Parabolas Objective – To be able to graph quadratic equations and identify the vertex by using –b/2a What shape does a quadratic have? A Parabola or a “U” shaped graph. What is different about the equation from what we have dealt with so far this year? It has an x2 value.

The Graph of A Quadratic Function The graph of y = ax2 + bx + c is a parabola with these characteristics: The x coordinate of the vertex is – b/2a The axis of symmetry goes through the vertex vertically. The graph opens upward if a is positive. Opens downward if a is negative. The graph is wider if |a| < 1, and narrower if |a| > 1. Compared to the parent graph y = x2.

Parabola's in REAL Life! Man Made Objects

Parabola's in REAL Life! Man Made Objects

Parabola's in REAL Life! Man Made Objects

Parabola's in REAL Life! Man Made Objects

Parabola's in REAL Life! How about in Nature?

Parabola's in REAL Life! How about in Nature?

Parabola's in REAL Life! How about in Sports?

Don't forget the Home Run King! Parabola's in REAL Life! Don't forget the Home Run King!

Extra Example 2a Graph y = -x2 + 4x + 1 1) First find the x-coordinate of the vertex: x = - b/2a y = -(2)2 + 4(2) + 1 –1 –5 –4 –3 –2 1 2 5 4 3 6 9 8 7 10 x = - 4/2(-1) y = -4 + 8 + 1 y = 5 x = 2 Vertex = (2,5) 2) Draw the axis of symmetry x = 2 3) Plot two points on one side of the axis of symmetry x y Use symmetry to mirror those two points 1 4 1

Extra Example 2b Graph y = x2 + 4x + 3 x = - b/2a x = - 4/2(1) x = -2 1) First find the x-coordinate of the vertex: x = - b/2a y = (-2)2 + 4(-2) + 3 –1 –5 –4 –3 –2 1 2 5 4 3 6 9 8 7 10 x = - 4/2(1) y = 4 – 8 + 3 y = -1 x = -2 Vertex = (-2,-1) 2) Draw the axis of symmetry x = -2 3) Plot two points on one side of the axis of symmetry x y Use symmetry to mirror those two points -1 3

5.3 Warm-Up Graph y = x2 – 2x + 1 x = - b/2a y = (1)2 – 2(1) + 1 –1 –5 –4 –3 –2 1 2 5 4 3 6 9 8 7 10 x = - -2/2(1) y = 1 – 2 + 1 y = 0 x = 1 Vertex = (1,0) x y -1 1 4

5.3 Transforming Parabolas CA State Standard 9.0 Students explain the effect that a, b, and c, have on the parabola in the equation y = a(x – b)2 + c. CA State Standard 10.0 Students graph quadratic functions and determine the maxima, minima, and zeros of the function. CA State Standard 17.0 Given ax2 + by2 + cx + dy + e = 0, students can use the method for completing the square to put the equation in standard form and recognize what type of graph it is.

If a > 0 the graphs open upward, if a < 0 it opens downward. Vertex And Intercept Forms Of A Quadratic Function Form Of Quadratic Function Characteristics: Vertex Form y = a(x - h)2 + k vertex = (h,k) axis of symmetry is x = h. Intercept Form y = a(x – p)(x – q) x-intercepts are p and q. axis of symmetry is half- way between (p,0), (q,0). If a > 0 the graphs open upward, if a < 0 it opens downward.

Extra Example 1 (Using Vertex Form!) Graph y = 2(x – 1)2 + 3 1) First find the vertex: vertex = (h,k) –1 –5 –4 –3 –2 1 2 5 4 3 6 9 8 7 10 vertex = (1, 3) 2) Plot the axis of symmetry. 3) Plot two points on one side of the axis of symmetry x y Use symmetry to mirror those two points 2 3 5 11

Example 2 (Using Intercept Form!) Graph y = 2(x – 3)(x + 1) 1) First find the x-intercepts: (p,0) = (3,0) –10 –5 –4 –3 –2 –1 1 2 5 4 3 -5 -8 -7 -6 -4 -3 -9 -1 -2 (q,0) = (-1,0) Plot the axis of symmetry. (Which lies half-way between the x-intercepts.) 3) Since the vertex is at x = 1, substitute 1 in for x and solve for y x y 1 -8

Example 3 Graph y = –(x + 2)(x – 4) 1) First find the x-intercepts: –1 –5 –4 –3 –2 1 2 5 4 3 6 9 8 7 10 (p,0) = (-2,0) (q,0) = (4,0) Plot the axis of symmetry. (Which lies half-way between the x-intercepts.) 3) Since the vertex is at x = 1, substitute 1 in for x and solve for y x y 1 9

5.4a Warm-Up Solve the equation. 3) -3x – 5 = 2x 1) 3x – 4 = 0 +5 +5 +4 +4 -3x = 2x + 5 3x = 4 -2x -2x 3 3 -5x = 5 -5 -5 x = 4/3 x = -1 2(x – 3) = 6 2) 2x – 11 = -15 +11 +11 2 2 x – 3 = 3 2x = -4 +3 +3 2 2 x = -2 x = 6

5.4a Factoring Quadratic Expressions CA State Standard 4.0 Students factor polynomials. CA State Standard 8.0 Students solve and graph quadratic equations by factoring. Students apply this technique in solving word problems. CA State Standard 10.0 Students graph quadratic functions and determine the maxima, minima, and zeros of the function. CA State Standard 11.2 Students judge the validity of an argument according to whether the properties of real #’s, exponents have been applied correctly at each step. CA State Standard 25.0 Students use properties from # systems to justify steps in combining and simplifying functions.

5.4a Factoring Quadratic Expressions x2 + bx + c = (x + m)(x + n) = x2 + (m + n)x + mn Example 1 Factor x2 – 2x – 48 So what two things: When multiplied together equal –48 And added together equal –2. -1 · 48 or 1 · -48 -2 · 24 or 2 · -24 -4 · 12 or 4 · -12 -6 · 8 or 6 · -8 (x – 8)(x + 6)

(x – 7)(x – 7) Example 2 Factor x2 – 14x + 49 FACTORING PATTERNS So what two things: When multiplied together equal 49 But when added together equal –14. -1 · -49 -7 · -7 (x – 7)(x – 7) FACTORING PATTERNS a2 – b2 = (a + b)(a – b) x2 – 9 = a2 + 2ab + b2 = (a + b)2 x2 + 12x + 36 = a2 – 2ab + b2 = (a – b)2 x2 – 8x + 16 = (x + 3)(x – 3) (x + 6)2 (x – 4)2

(4y + 15)(4y – 15) 2(7x – 6)(x + 1) 3v(v – 6) Example 3 Factor 16y2 – 225 14x2 + 2x – 12 3v2 – 18v Remember: a2 – b2 = (a + b)(a – b) (4y)2 – (15) 2 (4y + 15)(4y – 15) Look at what is in common with each #. 2(7x2 + x – 6) 2(7x ? )(x ? ) 2(7x – 6)(x + 1) 3v(v – 6)

5.4b Warm-Up Factor: 3) x2 + 8x + 15 1) x2 + 4x – 21 (x + 7)(x – 3) (x + 5)(x + 3) x2 + 9x + 14 2) x2 + 6x – 2 Not Possible (x + 7)(x + 2)

5.4b Factoring Quadratic Expressions CA State Standard 4.0 Students factor polynomials. CA State Standard 8.0 Students solve and graph quadratic equations by factoring. Students apply this technique in solving word problems. SAME AS 5.4a! CA State Standard 10.0 Students graph quadratic functions and determine the maxima, minima, and zeros of the function. CA State Standard 11.2 Students judge the validity of an argument according to whether the properties of real #’s, exponents have been applied correctly at each step. CA State Standard 25.0 Students use properties from # systems to justify steps in combining and simplifying functions.

5.4b Factoring Quadratic Expressions But now we have an equal sign. Extra Example 1 Factor x2 + 3x – 18 = 0 (x + 6)(x – 3) = 0 x + 6 = 0 and x – 3 = 0 -6 -6 +3 +3 x = -6 and x = 3

-6 = x2 – 3x -10 x2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 and x = -1 Example 2 Factor 3x – 6 = x2 – 10 -3x -3x -6 = x2 – 3x -10 +6 +6 x2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x – 4 = 0 and x + 1 = 0 x = 4 and x = -1

The zeros of the function are 2 and 5. Example 3 Find the zeros of: y = x2 – 7x + 10 Remember the Intercept Form where: y = a(x – p)(x – q) 0 = (x – 5)(x – 2) p = 5 and q = 2 The zeros of the function are 2 and 5.

5.5 Warm-Up 3) Find the value of y when x = 2 1) 5x – 3 = 17 +3 +3 Solve the equation: 3) Find the value of y when x = 2 1) 5x – 3 = 17 +3 +3 y = -6x2 + 24 5x = 20 x = 4 y = -6(2)2 + 24 y = -24 + 24 2) 0 = -12 + 3t +12 +12 y = 0 12 = 3t t = 4

(Solve by Finding Square Roots) 5.5 Quadratic Equations (Solve by Finding Square Roots) CA State Standard 8.0 Students solve and graph quadratic equations by factoring. Students apply this technique in solving word problems. CA State Standard 11.2 Students judge the validity of an argument according to whether the properties of real #’s, exponents have been applied correctly at each step. CA State Standard 25.0 Students use properties from # systems to justify steps in combining and simplifying functions.

(Solve by Finding Square Roots) 5.5 Quadratic Equations (Solve by Finding Square Roots) PROPERTIES OF SQUARE ROOTS: ab = a · b a a √ b = 36 11 Example 1 500 b) 25 c) 2 √ 100 · 5 √ √ 25 36 = 5 6 11 · 11 · √ √ 2 11 = 22 11 √ 10 5 √

2x2 = 16 x2 = 8 x2 = +/- 8 x = +/- 4 · 2 x = +/- 2 2 Example 2 Solve 2x2 + 1 = 17 -1 -1 2x2 = 16 2 2 x2 = 8 x2 = +/- 8 √ x = +/- 4 · 2 √ x = +/- 2 2 √

(x – 2)2 = 7 (x – 2)2 = ± 7 x – 2 = ± 7 x = 2 ± 7 Example 3 Solve 3(x – 2)2 = 21 3 3 (x – 2)2 = 7 (x – 2)2 = ± 7 √ x – 2 = ± 7 √ +2 +2 x = 2 ± 7 √

5.6 Warm-Up √ √ 100 · 2 √ 4 · 5 √ √ √ √ √ 25 · 3 √ 49 · 2 √ √ √ Simplify. 20 98 √ 1) 200 √ 100 · 2 √ 4 · 5 √ 2 5 √ 10 2 √ 2) 75 √ √ 25 · 3 √ 49 · 2 √ 5 3 √ 7 2 √

5.6 Complex Numbers Objective: Students will identify and graph complex numbers, and they will be able to add, subtract, multiply, and divide complex numbers. CA State Standard 5.0 Students demonstrate knowledge of how real and complex numbers are related both arithmetically and graphically. In particular, they can plot complex numbers as points in the plane. CA State Standard 6.0 Students add, subtract, multiply, and divide complex numbers.

5.6 Complex Numbers imaginary unit - i, i = -1 √ i2 = -1 THE SQUARE ROOT OF A NEGATIVE #: (i 5 )2 = i2 · 5 = (-1)(5) = -5 √ -5 = i 5

Example 1a Simplify √ -27 i 27 √ √ i 9 3 √ 3i 3

2x2 = -36 x2 = -18 x2 = +/- -18 x =+/- i 9 · 2 x = +/- 3i 2 Example 1b Solve 2x2 + 26 = -10 -26 -26 2x2 = -36 2 2 x2 = -18 x2 = +/- -18 √ x =+/- i 9 · 2 √ x = +/- 3i 2 √

Example 2 Plot a) -4 – i b) 5 c) 1 + 3i imaginary (1, 3i) (5, 0) real –5 –4 –3 –2 –1 1 2 5 4 3 (1, 3i) (5, 0) real (-4, -i)

4 + 3 – i + 2i 2 – 3i – 3 + 7i 7 + i 2 – 3 – 3i + 7i –1 + 4i Example 4 (Adding and Subtracting) Simplify a) (4 – i) + (3 + 2i) b) (2 – 3i) – (3 – 7i) 4 + 3 – i + 2i 2 – 3i – 3 + 7i 7 + i 2 – 3 – 3i + 7i –1 + 4i

-12 – 4i – 18i – 6i2 -12 – 22i – 6(-1) -12 – 22i + 6 -6 – 22i Example 5 (Multiplying) Simplify (2 + 3i)(–6 – 2i) -12 – 4i – 18i – 6i2 -12 – 22i – 6(-1) -12 – 22i + 6 -6 – 22i

Example 5 (Dividing) Simplify (1 + 2i) (1 + 2i) (1 – 2i) (1 + 2i) 1 + 2i – 2i – 4i2 1 + 2i + 2i + 4i2 1 – 4(-1) 1 + 4i + 4(-1) 1 + 4 1 + 4i – 4 5 4i – 3

imaginary unit - i, -i 1 i = -1 √ i2 = -1 i2 · i = i3 = i2 · i2 = i4 =

5.7 Warm-Up Solve the equation. 1) (x – 2)2 = 16 √ 11(x –7)2 = 22 x – 2 = +/- 4 (x – 7)2 = 2 + 2 + 2 (x – 7)2 = +/- 2 √ x = -2 or 6 2) 3(x + 5)2 = 24 x – 7 = +/- 2 √ (x + 5)2 = 8 x = 7 +/- 2 √ (x + 5)2 = +/- 8 √ x + 5 = +/- 2 2 √ x = – 5 +/- 2 2 √

5.7 Completing the Square CA State Standard 8.0 Students solve and graph quadratic equations by completing the square. Students apply this technique in solving word problems. CA State Standard 9.0 Students explain the effect that a, b, and c, have on the parabola in the equation y = a(x – b)2 + c. CA State Standard 10.0 Students graph quadratic functions and determine the maxima, minima, and zeros of the function. CA State Standard 17.0 Given ax2 + by2 + cx + dy + e = 0, students can use the method for completing the square to put the equation in standard form and recognize what type of graph it is.

5.7 Completing the Square is a process that allows you to write an expression of the form x2 + bx as the square of a binomial. Completing the Square - To complete the square for x2 + bx, you need to add (b/2)2. x2 + bx + (b/2)2 = (x + b/2)2

c = ( b/2 )2 = ( -3/2 )2 = 9/4 x2 – 3x + c = x2 – 3x + 9/4 Extra Example 1 Find the value for “c” that makes x2 – 3x + c a perfect square. Then write the expression as the square of a binomial. c = ( b/2 )2 = ( -3/2 )2 = 9/4 x2 – 3x + c = x2 – 3x + 9/4 Perfect square trinomial = ( x – 3/2 )2 Square of a binomial: (x + b/2)2

Extra Example 2 Solve x2 + 6x – 8 = 0 by completing the square. x2 + 6x = 8 x2 + 6x + ( b/2 )2 = 8 + ( b/2 )2 x2 + 6x + ( 6/2 )2 = 8 + ( 6/2 )2 x2 + 6x + 9 = 8 + 9 √ +/- (x + 3)2 = 17 x + 3 = 17 √ +/- x = – 3 17 √ +/-

Extra Example 3 Solve 5x2 – 10x + 30 = 0 by completing the square. 5x2 – 10x + 30 = 0 x2 – 2x + 6 = 0 x2 – 2x = – 6 x2 – 2x + ( b/2 )2 = – 6 + ( b/2 )2 x2 – 2x + ( -2/2 )2 = – 6 + ( -2/2 )2 x2 – 2x + 1 = – 6 + 1 x – 1 = i 5 +/- √ √ +/- (x – 1)2 = -5 x = 1 i 5 +/- √

5.8 Warm-Up Evaluate the expression b2 – 4ac for the given values of a, b, and c. 1) a = 1, b = 3, c = -1 a = -1, b = 0, c = 5 a = -2, b = 2, c = -3 b2 – 4ac b2 – 4ac 32 – 4(1)(-1) 02 – 4(-1)(5) 9 + 4 0 + 20 13 20 2) a = 2, b = -2, c = 0 b2 – 4ac b2 – 4ac (-2)2 – 4(2)(0) 22 – 4(-2)(-3) 4 + 0 4 – 24 4 –20

5.8 The Quadratic Formula CA State Standard 8.0 Students solve and graph quadratic equations by factoring. Students apply this technique in solving word problems. http://www.youtube.com/watch?v=YJpGiBIGoME

5.8 The Quadratic Formula Discriminant THE QUADRATIC FORMULA: Let a, b, and c be real numbers such that a  0. The solutions of the quadratic equation ax2 + bx + c = 0 are: -b  b2 – 4ac x = 2a √ Discriminant What does the Quadratic Equation tell us? 2 Solutions 1 Solution 2 Imaginary Solutions

x = -5 and x = 7/3 Solve 3x2 + 8x = 35 -8  484 x = 6 -35 -35 Extra Example 1 Solve 3x2 + 8x = 35 -8  484 √ x = 6 -35 -35 484 √ 3x2 + 8x –35 = 0 √ 4 121 √ -b  b2 – 4ac x = 2a -8  (2)(11) x = 6 √ -8  82 – 4(3)(-35) x = 2(3) -8  22 x = 6 √ -8  64 + 4(105) x = 6 -30 x = and x = 6 14 -8  64 + 420 √ x = 6 x = -5 and x = 7/3

Solve 2x2 - 4x = - 20 2  -36 x = 2 2  i 36 x = 2 x2 - 2x + 10 = 0 Extra Example 2 Solve 2x2 - 4x = - 20 2  -36 √ x = 2 +20 +20 2  i 36 √ x = 2 x2 - 2x + 10 = 0 √ -b  b2 – 4ac x = 2a 2  6i x = 2 -(-2)  (-2)2 – 4(1)(10) √ x = 2(1) 1  3i x = 2  4 – 40 √ x = 2 2  36 √ x = 2