Do Now! Evaluate each expression for a = -1, b = 3, and c = -2. 1) 2a – b2 + c 2) b2 – 4ac 2a Graph each function. 3) y = x2 Factor each expression. 4) 4x2 + 4x + 1 5) m2 – 7m – 18

Chapter 9: Quadratic Functions and Equations SWBAT solve, graph, compare, and identify Quadratic Functions and Equations.

9-1 Quadratic Graphs and Their Properties Vocabulary: Quadratic Function Standard Form of a Quadratic Function Quadratic Parent Function Parabola Axis of Symmetry Vertex Minimum Maximum

9-1 Quadratic Graphs and Their Properties Vocabulary: Quadratic Function: a function that can be written in the form of y = ax2 + bx + c Standard Form of a Quadratic Function: y = ax2 + bx + c Quadratic Parent Function: The simplest quadratic function f(x) = x2 or y = x2 Parabola: The U-Shaped curve of a quadratic function Axis of Symmetry: The fold or line that divides the parabola into 2 matching halves Vertex: the highest or lowest point of a parabola Minimum: Lowest point of the parabola Maximum: Highest point of the parabola

Standard Form of a Quadratic Function Examples: y = 3x2 y = x2 + 9 y = x2 – x - 2

Identifying a Vertex What are the coordinates of the vertex of each graph? Is it a minimum of maximum? Which direction does the parabola open? DOWNWARD What is the highest point of the parabola? (0, 3) Is this vertex a maximum or minimum? Maximum

You Do! What is the vertex of the graph? Is it a minimum or maximum?

Graphing y = ax2 Graph the function y = 1/3 x2 . Make a table of values. What are the domain and range? X y = 1/3 x2 (x, y) 1/3(0)2 = 0 (0, 0) 3 1/3(3)2 = 3 (3,3) 6 1/3(6)2 = 12 (6, 12) The domain is all real numbers. The range is y ≥ 0.

You Do! Graph the function y = -3x2. What are the domain and range? All real numbers, Range: y ≤ 0.

Comparing Widths of Parabolas What is the order, from widest to narrowest, of the graphs of the quadratic functions: f(x) = -4x2 f(x) = ¼ x2 f(x) = x2

Comparing Widths of Parabolas What is the order, from widest to narrowest, of the graphs of the quadratic functions: f(x) = -4x2 f(x) = ¼ x2 f(x) = ¼ x2 is the widest f(x) = x2 f(x) = -4x2 is the narrowest So the order from widest to narrowest is: f(x) = -x2, f(x) = 3x2, and f(x) = -1/3 x2

You Do! What is the order, from widest to narrowest, of the graphs of the functions: f(x) = -x2, f(x) = 3x2, and f(x) = -1/3 x2 f(x) = -1/3 x2 , f(x) = -x2 , f(x) = 3x2

Graphing y = ax2 + c How is the graph of y = 2x2 + 3 different from the graph of y = 2x2 ? X y = 2x2 y = 2x2 + 3 -2 8 11 -1 2 5 3 1 The graph of y = 2x2 + 3 has the same shape as the graph of y = 2x2 but is shifted up 3 units.

You Do!! Graph y = x2 and y = x2 – 3. How are the graphs related? The 2 graphs have the shape But the second parabola is Shifted down 3 units.

Falling Object As an object falls, its speed continues to increase, so its height above ground decreases at a faster rate. You can model the object’s height with the function: h = -16t2 + c. The height h is in feet, the time t is in seconds, and the object’s initial height c is in feet.

Using the Falling Object Model An acorn drops from a tree branch 20 ft above the ground. The function h = -16t2 + 20 gives the height h of the acorn (in feet) after t seconds. What is the graph of this quadratic function? At about what time does the acorn hit the ground?

Using the Falling Object Model An acorn drops from a tree branch 20 ft above the ground. The function h = -16t2 + 20 gives the height h of the acorn (in feet) after t seconds. What is the graph of this quadratic function? At about what time does the acorn hit the ground? Plan: Use a table of values to graph the function. Use the graph to estimate when the acorn hits the ground. Know: The function for the acorn’s height The initial height is 20 ft Need: The function’s graph and the time the acorn hits the ground

Using the Falling Object Method h = -16t2 + 20 20 0.5 16 1 4 1.5 -16 The acorn hits the ground when its height above the ground is 0 ft. From the graph you can see that the acorn hits the ground after slightly more than 1 s.

You Do! In the previous example, suppose the acorn drops from a tree branch 70 ft above the ground. The function h = -16t2 + 70 gives the height h of the acorn (in feet) after t seconds. What is the graph of this function? At about what time does the acorn hit the ground? About 2 seconds

Homework Workbook Pages: 9-1 pg. 255-256 1 – 27 odd

9-2 Quadratic Functions The graph of y = ax2 + bx + c, where a ≠ 0, has the line x = -b/2a as its axis of symmetry. The x-coordinate of vertex is –b/2a.

Why is it important that a is not equal to zero? If a = 0, then the equation does not have a quadratic term and is not a parabola. If a and b are both positive, will the axis of symmetry be to the left or to the right of the y-axis? It will be to the left of the y-axis. If a and b are both negative, will the axis of symmetry be to the left or to the right of the y axis? It will be to the left of the y-axis

Graphing y = ax2 + bx + c What is the graph of the function: y = ax2 + bx + c Step 1: Find the axis of symmetry and the coordinates of the vertex. x = -b/2a = -(-6)/2(1) = 3 Use the equation to find the axis of symmetry The axis of symmetry is x = 3. So the x-coordinate of the vertex is 3. y = x2 – 6x + 4 Rewrite function = 32 – 6(3) + 4 Substitute 3 for x y = -5 Simplify The vertex is (3, -5)

Graphing y = ax2 + bx + c Step 2: Find two other points on the graph. Find the y-intercept. When x = 0, y = 4, so one point is (0, 4). Find another point by choosing a value for x on the same side of the vertex as the y-intercept. Let x = 1. y = x2 – 6x + 4 Rewrite function. = 12 – 6 (1) + 4 = -1 Substitute 1 and simplify When x = 1, y = - 1, so another point is (1, -1).

Graphing y = ax2 + bx + c Step 2: Find two other points on the graph. Find the y-intercept. When x = 0, y = 4, so one point is (0, 4). Find another point by choosing a value for x on the same side of the vertex as the y-intercept. Let x = 1. y = x2 – 6x + 4 Rewrite function. = 12 – 6 (1) + 4 = -1 Substitute 1 and simplify When x = 1, y = - 1, so another point is (1, -1).

Graphing y = ax2 + bx + c Step 3: Graph the vertex and the points you found in Step 2, (0, 4) and (1, -1). Reflect the points from Step 2 across the axis of symmetry (x = 3) to get two more points on the graph. Then connect the points with a parabola.

Graphing y = ax2 + bx + c How do you determine the reflection of the point (1, -1) across the axis of symmetry? You determine that the horizontal distance from (1, -1) to the axis of symmetry is 2 units left, and then count 2 horizontal units right from the axis of symmetry to get to point (5, -1).

You Do! What is the graph of the function: y = -x2 + 4x – 2? Label the Vertex and tell if it’s a minimum or maximum. Vertex: (2,2) Maximum

Note! In lesson 9-1 we used h = -16t2 + c to find he height h above the ground of an object falling from an initial eight c at a time t. if an object projected into the air given an initial upward velocity v continues with no additional force of its own, the formula: h = -16t2 + vt + c gives its approximate height above the ground.

Using the Vertical Motion Model During halftime of a basketball game, a sling shot launches T-shirts at the crowd. A T-shirt is launched with an initial upward velocity of 72 ft/s. The T-shirt is caught 35 ft above the court. How long will it take the T-shirt to reach its maximum height? What is the maximum height? What is the range of the function that models the height of the T-shirt over time?

Using the Vertical Motion Model The function h = -16t2 + 72t + 5 gives the T-shirt’s height h, in feet, after t seconds. Since the coefficient of t2 is negative, the parabola opens downward, and the vertex is the maximum point. Method 1: Use a formula. t = -b/2a = -72/2(-16) = 2.25 Find t h= -16(2.25)2 + 72(2.25) + 5 = 86 Find h The T-shirt will reach its maximum height of 86 ft after 2.25 s. The range describes the height of the T-shirt during its flight. The T-shirt starts at 5ft, peaks at 86 ft, and then is caught at 35 ft. The height of the T-shirt at any time is between 5 ft and 86 ft, inclusive, so the range is 5 ≤ h ≤ 86.

Using the Vertical Motion Model Method 2: Use a graphing calculator

You do! In the previous example, suppose a T-shirt is launched with an initial upward velocity of 64 ft/s and is caught 35 ft above the court. How long will it take the T-shirt to reach its maximum height? How far above court level will it be? What is the range of the function that models the height of the T-shirt over time? 2 seconds; 69 feet; 5≤ h ≤ 69

Homework! Workbook pages: 9-2 pg. 259-260 1 – 27 odd pg. 261 1 – 6 all

Do Now! In your composition notebook!! Explain how you can use the y- intercept, vertex, and axis of symmetry to graph a quadratic function. Assume the vertex is not on the y – axis.

9-3 Solving Quadratic Equations Vocabulary: Quadratic equation Standard Form of a Quadratic Equation Root of an Equation Zero of a Function

9-3 Solving Quadratic Equations Vocabulary: Quadratic equation: is an equation that can be written in the form of ax2 + bx + c = 0, where a≠0. Standard Form of a Quadratic Equation: ax2 + bx + c = 0, where a≠0. Root of an Equation: the solutions of quadratic equation and the x-intercepts of the graph of ht related function are called Root of an Equation. Zero of a Function: the solutions of quadratic equation and the x-intercepts of the graph of ht related function are called Zero of a Function

The x-intercepts show the solutions of the equation. Solving by Graphing What are the solutions of each equation? Use a graph of the related function. x2 – 1 = 0 Graph y = x2 - 1 There are 2 solutions, + 1. (+1, -1) The x-intercepts show the solutions of the equation.

Solving by Graphing What are the solutions of each equation? Use a graph of the related function. b) x2 = 0 Graph y = x2. There is one solution, 0.

Solving by Graphing What are the solutions of each equation? Use a graph of the related function. c) x2 + 1 = 0 Graph y = x2 + 1 There is no real- number solution.

You Do! What are the solutions of each expression? Use a graph of the related function. x2 – 16 = 0 + 4 2) x2 – 25 = -25

Solving Using Square Roots What are the solutions of 3x2 – 75 = 0 ? 3x2 – 75 = 0 3x2 = 75 x2 = 25 x = +√25 x = + 5

You Do! What are the solutions of each equation? m2 – 36 = 0 3x2 + 15 = 0 4d2 + 16 = 16 + 6 no solution

Choosing Reasonable Solution An aquarium is designing a new exhibit to showcase tropical fish. The exhibit will include a tank that is a rectangular prism with a length l that is twice the width w. The volume of the tank is 420 ft3. What is the width of the tank to the nearest tenth of a foot?

Choosing Reasonable Solution An aquarium is designing a new exhibit to showcase tropical fish. The exhibit will include a tank that is a rectangular prism with a length l that is twice the width w. The volume of the tank is 420 ft3. What is the width of the tank to the nearest tenth of a foot? V = lwh Use Volume Formula 420 = (2w)(w)(3) Substitute 420 = 6w2 Simplify 70 = w2 Divide by √6 +√70 = w Find square root + 8.366600265 = w Use calculator

Choosing Reasonable Solution An aquarium is designing a new exhibit to showcase tropical fish. The exhibit will include a tank that is a rectangular prism with a length l that is twice the width w. The volume of the tank is 420 ft3. What is the width of the tank to the nearest tenth of a foot? + 8.366600265 = w Use calculator A tank cannot have a negative width, so only the positive square root make sense. The tank will have a width of about 8.4 ft.

You Do! Suppose the tank in the previous example will have a height of 4 ft and a volume of 500 ft3. What is the width of the tank to the nearest tenth of a foot? 7.9 ft

Homework! Workbook: 9-3 pg. 263 – 263 1-23 odd; 31 – 35 odd

Do Now What are the solutions of each equation? m2 – 36 = 0 3x2 + 15 = 0 4d2 + 16 = 16 + 6 no solution

9-4 Factoring to Solve Quadratic Equations Vocabulary: Zero – Product Property : For any real numbers a and b, if ab = 0, then a = 0 or b = 0. Example: If (x + 3)(x +2) = 0, then x + 3 = 0 or x + 2 = 0

Using the Zero-Product Property What are the solutions of the equation (4t + 1)(t – 2) = 0? (4t + 1)(t – 2) = 0 4t + 1 = 0 or t – 2 = 0 Zero Product Prop 4t = -1 or t = 2 Solve for t t = -1/4 or t = 2

You Do! What are the solutions of each equation? (x + 1)(x – 5) = 0 -1, 5 2) (7n – 2)(5n – 4) = 0 2/7, 4/5

Solving by Factoring What are the solutions of the equation x2 + 8x + 15 = 0? -5, -3 -5, 3 -3, 5 3, 5 x2 + 8x + 15 = 0 Rewrite Equation (x + 3)(x + 5) = 0 Factor x2 + 8x + 15 = 0 x + 3 = 0 or x + 5 = 0 Zero Product Prop x = -3 or x = -5 Solve for x

You Do! What are the solutions of each equation? m2 – 5m – 14 -2, 7 b) p2 + p – 20 = 0 -3/2, 4

Writing in Standard Form First What are the solutions of 4x2 – 21x = 18? 4x2 – 21x = 18 4x2 – 21x – 18 = 0 Subtract 19 (4x + 3)(x – 6) = 0 Factor 4x + 3 = 0 or x – 6 = 0 Zero Product Prop x = -3/4 or x = 6 Solve for x The solutions are -3/4 and 6.

You Do! What are the solutions of x2 + 14x = -49 ? -7

Using Factoring to Solve a Real-World Problem You are constructing a frame for the rectangle photo shown. You want the frame to be the same width all the way around and the total area of the frame and photo to be 315 in2. What should the outer dimensions of the frame be?

Using Factoring to Solve a Real-World Problem You are constructing a frame for the rectangle photo shown. You want the frame to be the same width all the way around and the total area of the frame and photo to be 315 in2. What should the outer dimensions of the frame be? (2x + 11)(2x + 17) = 315 Width × Length = Area 4x2 + 56x + 187 = 315 Find the product 4x2 + 56x – 128 = 0 Subtract 315 4(x2 + 14x -32) = 0 Factor out 4 4(x2 + 16)(x – 2) = 0 Factor x2 + 14x – 32 x + 16 = 0 or x – 2 = 0 Zero Product Prop x= -16 or x = 2 Solve for x

Using Factoring to Solve a Real-World Problem You are constructing a frame for the rectangle photo shown. You want the frame to be the same width all the way around and the total area of the frame and photo to be 315 in2. What should the outer dimensions of the frame be? x= -16 or x = 2 Solve for x The only reasonable solution is 2. So the outer dimensions of the frame are: 2(2) + 11 in. by 2(2) + 17 in., or 15 in. by 21 in.

You Do! In the previous example, suppose the total area of the frame and photo were 391 in.2. What would the outer dimensions of the frame be? 17 in. by 23 in.

Homework Workbook pages: 9-4 pg. 267-268 1- 33 odd Pg. 269 1 - 5

9-5 Completing the Square Vocabulary: Completing the Square You can change the expression x2 + bx into a perfect-square trinomial by adding (b/2)2 to x2 + bx. This process is called completing the square.

Finding c to Complete the Square. What is the value of c such that x2 – 16x + c is a perfect-square trinomial? The value of b is -16. The term to add to x2 – 16x is (-16/2)2, or 64. So c = 64.

You Do! What is the value of c such that x2 + 20x + c is a perfect-square trinomial? 100

Solving x2+bx = c What are the solutions of the equation x2 + 6x = 216? x2 + 6x = 216 x2 + 6x + 9 = 216 + 9 Add (6/2)2, or 9 (x + 3)2 = 216 + 9 Write x2+6x+9 as square (x+3)2 = 225 Simplify x+3 = +√225 Find square root x+3 = +15 Simplify x+3 = 15 or x+3 = -15 Write as 2 equations x=12 or x = -18 Subtract 3

You Do! What are the solutions of the equation: t2 – 6t = 247? -13, 19

Solving x2 + bx + c = 0 What are the solutions of the equation x2 – 14x + 16 = 0? x2 – 14x + 16 = 0 x2 – 14x = -16 Subtract 16 x2 – 14x + 49 = -16 + 49 Add (-14/2)2, or 49 (x – 7)2 = 33 Write x2 – 14x + 16 as a square x – 7 = + √33 Find square Root x – 7 = + 5.74 Use calculator x – 7 ≈ 5.74 or x – 7 ≈ -5.74 Write 2 equations x ≈ 5.74 + 7 or x ≈ -5.74 + 7 Add 7 to each side x≈ 12.74 or x ≈ 1.26 Simplify

You Do! What are the solutions to the equation x2 + 9x + 15 = 0? -2.21 , -6.79

Completing the Square When a ≠ 1 What are the Solutions of : 4a2 -8a = 24 ? a2 – 2a = 6 Divide each side by 4 a2 – 2a + 1 = 6 + 1 Add (2/2)2 or 1 (a - 1)2 = 7 Write as square a – 1 = + √7 Find Square Root a – 1 = + 2.65 Simplify Using Calc. a – 1 = 2.65 or a – 1 = -2.65 a = 3.65 or a = -1.65 Solve for x

You Do! Solve each equation: 3r2 + 18r = 21 -7, 1

Homework Workbook: Pg. 271 – 272 1 – 35 odd

Do Now! Solve by factoring: x2 + 11x + 10 = 0 s2 – 14s + 45 = 0 Solve each equation by graphing. x2 – 9 = 0 Solve each equation by finding square roots. k2 – 196 = 0

9-6 The Quadratic Formula and the Discriminant Vocabulary: Quadratic formula: Discriminant

9-6 The Quadratic Formula and the Discriminant Vocabulary: Quadratic formula: If ax2 + bx + c = 0, and a ≠ 0, then: x = -b + √(b2 – 4ac) 2a Discriminant: is the expression under the radical sign in the quadratic formula. The discriminant

Quadratic Formula If ax2 + bx + c = 0, and a ≠ 0, then: x = -b + √(b2 – 4ac) 2a Example: Suppose 2x2 + 3x – 5 = 0. Then a = 2, b = 3, and c = -5. Therefore: x = -(3) +√(3)2 – 4(2)(-5) 2(2)

Using the Quadratic Formula What are the solutions x2 – 8 = 2x? Use the quadratic formula. x2 – 2x – 8 = 0 Write in Standard Form x = -b + √(b2 – 4ac) Use Quadratic Formula 2a x = -(-2) + √(-2)2 – 4(1)(-8) Substitute 2(1) x = 2 + √36 Simplify 2

Using the Quadratic Formula What are the solutions x2 – 8 = 2x? Use the quadratic formula. x = 2 + √36 Simplify 2 x = 2 + 6 or x = 2 – 6 Write 2 eqn. 2 2 x = 4 or x = -2 Simplify

You Do! What are the solutions of: x2 – 4x = 21? USE QUADRATIC FORMULA! -3, 7 -0.04x2 + 0.84x + 2? USE QUADRATIC FORMULA! -2.16, 23.16

Which Method Should I Use? Method When to Use Graphing Use if you have a graphing calculator Square Roots Use if equation has no x – term Factoring Use if can factor easily Completing the Square Use if coefficient of x2 is 1, but you cannot easily factor the equation Quadratic Formula Use if equation cannot be factored easily or at all

Choosing an Appropriate Method Which method(s) would you choose to solve each equation? Explain your reasoning. 3x2 – 9 = 0 Square roots; no x- term x2 – x – 30 = 0 Factoring 6x2 + 13x – 17 = 0 Quadratic Formula, Graphing x2 – 5x + 3 = 0 Quadratic Formula, Completing the Square, or Graphing -16x2 – 50x + 21 = 0 Quadratic Formula, Graphing

You Do! Which method would you choose to solve each equation? Justify. x2 – 8x + 12 = 0 169x2 = 36 5x2 + 13x – 1 = 0 Factoring; equation easily factorable Square roots; there is no x-term Quadratic Formula; graphing, the equation cannot be factored.

Using the Discriminant

Using the Discriminant How many real number solutions does: 2x2 – 3x = -5 have? 2x2 – 3x + 5 = 0 Write in Standard Form b2 – 4ac = (-3)2 – 4(2)(5) = -31 Evaluate the discriminant by substituting 2 for a, -3 for b, and 5 for c Because the discriminant is negative, the equation has no real-number solutions.

You Do! How many real-number solutions does: 6x2 – 5x = 7 have? 2 9x2 + 12x + 4 = 0 have? 1 x2 + 2x = 0 have?

Homework Workbook: Pg. 275 – 276 1 – 41 odd

9-7 Linear, Quadratic, and Exponential Models

Choosing a Model by Graphing Graph each set of points. Which model is most appropriate for each set? (1,3), (0,0), (-3,3), (-1, -1), (-2,0) Quadratic Model

Choosing a Model by Graphing Graph each set of points. Which model is most appropriate for each set? b) (0,2), (-1, 4), (1, 1), (2, 0.5) Exponential Model

Choosing a Model by Graphing Graph each set of points. Which model is most appropriate for each set? c) (-1, -2), (0, -1), (1, 0), (3, 2) Linear Model

You do! Graph each set of points. Which model is most appropriate for each set? (0,0), (1,1), (-1, -0.5), (2, 3) Exponential b)(-2, 11), (-1, 5) , (0, 3), (1, 5) Quadratic

Choosing a Model Using Differences or Ratios If the y- values have a common difference. A linear model fits the data. The y-values have a common difference of 3. A linear model fits the data.

Choosing a Model Using Differences or Ratios If the first difference of the y-values does not produce a common difference, find the 2nd difference of the y-values and if it produces a common difference, then a quadratic model fits the data. The second difference of the y-values are all 4, so a quadratic model fits the data.

Choosing a Model Using Differences or Ratios c) If the y-values have a common ratio, an exponential model fits the data. The y-values have a common ratio of 2. An exponential model fits the data.

Choosing a Model Using Differences or Ratios Which type of function best models the data? Use differences or ratios. b) The 1st differences are The 2nd differences Constant, so a linear are constant, so a Function models the data. Quadratic function models the data.

You do! Which type of function best models the ordered pairs: (-1, 0.5), (0, 1), (1,2) , (2, 4), and (3, 8)? Use differences or ratios. Exponential

Modeling Data Which type of function best models the data in the table? Write an equation to model the data.

Modeling Data Which type of function best models the data in the table? Write an equation to model the data. Step 1: Graph the data. What 2 models does this graph resemble? Quadratic and Exponential

There is a common 2nd difference of 1. Modeling Data Which type of function best models the data in the table? Write an equation to model the data. Step 2: The data appear to be quadratic. Test for a common second difference? There is a common 2nd difference of 1.

Modeling Data Which type of function best models the data in the table? Write an equation to model the data. Step 3: The graph appears to be a parabola with vertex at (0, 0), so use y = ax2, form for the equation. y = ax2 2 = a(2)2 Use a point other than (0,0) to find a. 2 = 4a Simplify 0.5 = a Divide by 4 y = 0.5x2 Write a quadratic Function

Modeling Data Which type of function best models the data in the table? Write an equation to model the data. Step 4: Test 2 points in the data set other than (2,2) and (0, 0) Test (3, 4.5): Test (4, 8): y = 0.5x2 y = 0.5x2 y = 0.5(3)2 y = 0.5(4)2 y = 4.5 y = 8 The points (3, 4.5) and (4, 8) both satisfy y = 0.5x2. The equation y = 0.5x2 models the data.

You Do! Which type of function best models the data in the table? Write an equation to model the data. Exponential; y = 6(0.2)x

You Do Pt. 2 What type of function best models the data? Write an equation to model the data. Exponential; y = 12,575 ∙ (0.88)x

Homework Workbook: Pg. 279 – 280 1 – 25 odd

9-8 Systems of Linear and Quadratic Equations

Solving by Graphing What are the solutions of the system? Solve by graphing. y = x2 – x – 2 y = -x + 2 Step 1: Graph both equations in the same coordinate plane. Step 2: Identify the point(s) of intersection, if any. The points of intersection are (-2, 4) and (2, 0) The solutions of the system are (-2, 4) and (2, 0).

You Do! What are the solutions of each system? Solve by graphing. y = 2x2 + 1 b) y = x2 + x + 3 y = -2x + 5 y = -x (-2, 9), (1, 3) no solution

Using Elimination What are the solutions of the system, use elimination. y = 20x + 124 y = -x2 + 39x + 64 Step 1: Eliminate y. -(y = 20x + 124) Subtract 0 = -x2 + 19x – 60

Using Elimination What are the solutions of the system, use elimination. y = 20x + 124 y = -x2 + 39x + 64 Step 2: Factor and Solve for x. 0 = -x2 + 19x – 60 0 = -1 (x2 – 19x + 60) Factor out -1 0 = -(x – 4)(x – 15) Factor x – 4 = 0 or x – 15 = 0 Zero Product x = 4 or x = 15 Solve for x

Using Elimination What are the solutions of the system, use elimination. y = 20x + 124 y = -x2 + 39x + 64 x = 4 or x = 15 Solve for x Step 3: Find the corresponding y – values. y = 20x + 124 y = 20x + 124 y = 20(4) + 124 y = 20(15) + 124 y = 204 y = 424 The solutions of the system are: (4, 204) and (15, 424)

You Do! Solve each system using elimination. y = -x + 3 y = x2 + 1 (1, 2) , (-2, 5)

Using Substitution What are the solutions of the system? y = x2 – 6x + 10 y = 4 – x Step 1: Write a single equation containing only one variable. 4 – x = x2 – 6x + 10 Substitute 4 – x 4 – x – (4 – x) = x2 – 6x + 10 – (4 – x) Subtract (4 – x) 0 = x2 – 5x + 6 Write in Stand. Form

Using Substitution What are the solutions of the system? y = x2 – 6x + 10 y = 4 – x Step 2: Factor and Solve for x. 0 = x2 – 5x + 6 0 = (x – 2)(x – 3) Factor. x – 2 = 0 or x – 3 = 0 Zero Product Prop. x = 2 or x = 3 Solve for x.

Using Substitution What are the solutions of the system? y = x2 – 6x + 10 y = 4 – x Step 3: Find corresponding y –values. Use either original equation. y = 4 – x = 4 – 2 = 2 y = 4 – x = 4 – 3 = 1 The solutions are (2, 2) and (3, 1).

You Do! What are the solutions of the system? y – 30 = 12x y = x2 + 11x – 12 (-6, -42), (7, 114)

Solving With Graphing Calculator What are the solutions of the system? y = -x + 5 y = -x2 + 4x + 1 Step 1: Enter the equations on the Y = screen. Press to display the system. Step 2: Use CALC feature. Select INTERSECT. Move the cursor close to a point of intersection. Press 3 times to find the point of intersection.

Solving With Graphing Calculator What are the solutions of the system? y = -x + 5 y = -x2 + 4x + 1 Step 3: Repeat Step 2 to find the second intersection point. The solutions are (1, 4) and (4, 1)

You Do! What are the solutions of the system? Use Graphing Calculator. y = x2 – 2 y = -x (-2, 2) (1, -1)

Homework Workbook: Pg. 283 – 284 1 – 25 odd Chapter 9 Cornell Notes Study for Test on Monday, December 10, 2012

Answer Key from Chapter 8 B 19. B 37. D C 20. A 38. D B 21. D 39. C A 22. A 40. D D 23. D 41. A A 24. A 42. C D 25. A 43. B A 26. A 44. C B 27. C 45. B D 28. A 46. D C 29. B 47. D A 30. C 48. C C 31. A 49. A D 32. C 50. C B 33. D 51. 2qr2(9q – 1)(11q – 9) B 34. A 52. 3(2q4 + 3r2)(4q3 – 7r) A 35. C B 36. D