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Lesson 9-1 Graphing Quadratic Functions Lesson 9-2 Solving Quadratic Equations by Graphing Lesson 9-3 Solving Quadratic Equations by Completing the Square Lesson 9-4 Solving Quadratic Equations by Using the Quadratic Formula Lesson 9-5 Exponential Functions Lesson 9-6 Growth and Decay Five-Minute Check (over Chapter 8) Main Ideas and Vocabulary California Standards Key Concept: Quadratic Function Example 1: Graph Opens Upward Example 2: Real-World Example: Graph Opens Downward Key Concept: Axis of Symmetry of a Parabola Example 3: Vertex and Axis of Symmetry Example 4: Standards Example: Match Equations and Graphs • Graph quadratic functions. • Find the equation of the axis of symmetry and the coordinates of the vertex of a parabola. • quadratic function • vertex • parabola • symmetry • minimum • axis of symmetry • maximum Standard 21.0 Students graph quadratic functions and know that their roots are the x-intercepts. Graph Opens Upwards Use a table of values to graph y = x2 – x – 2. Graph these ordered pairs and connect them with a smooth curve. Answer: Use a table of values to graph y = x2 + 2x + 3. A. B. A. A C. D. B. B 0% C. 0% C 0% 0% D. D A B C D Graph Opens Downward A. ARCHERY The equation y = –x2 + 6x + 4 represents the height y of an arrow x seconds after it is shot into the area. Use a table of values to graph y = –x2 + 6x + 4. Graph these ordered pairs and connect them with a smooth curve. Answer: Graph Opens Downward B. ARCHERY The equation y = –x2 + 6x + 4 represents the height y of an arrow x seconds after it is shot into the area. What are the mathematical domain and range of the function? Answer: domain: all real numbers; range: {y | y ≤ 13} Graph Opens Downward C. ARCHERY The equation y = –x2 + 6x + 4 represents the height y of an arrow x seconds after it is shot into the area. Describe reasonable domain and range values for this situation. Answer: The arrow is in the air for about 6.6 seconds, so a reasonable domain is D: {x | 0 < x < 6.6}. The height of the arrow ranges from 0 to 13 feet, so a reasonable range is R: {y | 0 < y < 13}. A. Use a table of values to graph y = –x2 + 4. A. B. 1. A 0% C. D. 2. B 3. C 4. D A B C D B. What are the domain and range of the function y = –x2 + 4? A. D: {x | x is a real number.} R: {y | y 4} B. D: {x | x is a real number.} 0% R: {y | –2 y 2} 1. A 2. B C. D: {x | –2 x 2} 3. C R: {y | y is a real number.} 4. D D. D: {x | x 4} A B C D R: {y | y is a real number.} Vertex and Axis of Symmetry A. Consider the graph of y = –2x2 – 8x – 2. Write the equation of the axis of symmetry. In y = –2x2 – 8x – 2, a = –2 and b = –8. Equation for the axis of symmetry of a parabola a = –2 and b = –8 Answer: The equation of the axis of symmetry is x = –2. Vertex and Axis of Symmetry B. Consider the graph of y = –2x2 – 8x – 2. Find the coordinates of the vertex. Since the equation of the axis of symmetry is x = –2 and the vertex lies on the axis, the x-coordinate for the vertex is –2. y = –2x2 – 8x – 2 Original equation y = –2(–2)2 – 8(–2) – 2 x = –2 y = –8 + 16 – 2 Simplify. y=6 Add. Answer: The vertex is (–2, 6). Vertex and Axis of Symmetry C. Consider the graph of y = –2x2 – 8x – 2. Identify the vertex as a maximum or minimum. Answer: Since the coefficient of the x2 term is negative, the parabola opens downward and the vertex is a maximum point. Vertex and Axis of Symmetry D. Consider the graph of y = –2x2 – 8x – 2. Graph the function. You can use the symmetry of the parabola to help you draw its graph. On a coordinate plane, graph the vertex and the axis of symmetry. Choose a value for x other than –2. For example, choose –1 and find the y-coordinate that satisfies the equation. y = –2x2 – 8x – 2 Original equation y = –2(–1)2 – 8(–1) – 2 x = –1 y=4 Simplify. Vertex and Axis of Symmetry D. Consider the graph of y = –2x2 – 8x – 2. Graph the function. Graph (–1, 4). Since the graph is symmetrical about its axis of symmetry x = –2, you can find another point on the other side of the axis of symmetry. The point at (–1, 4) is 1 unit to the right of the axis. Go 1 unit to the left of the axis and plot the point (–3, 4). Vertex and Axis of Symmetry D. Consider the graph of y = –2x2 – 8x – 2. Graph the function. Repeat this for several other points. Then sketch the parabola. Answer: A. Consider the graph of y = 3x2 – 6x + 1. Write the equation of the axis of symmetry. A. x = –6 0% B. x = 6 1. A 2. B C. x = –1 3. C 4. D A B C D D. x = 1 B. Consider the graph of y = 3x2 – 6x + 1. Find the coordinates of the vertex. A. (–1, 10) 0% B. (1, –2) 1. A 2. B C. (0, 1) 3. C 4. D A B C D D. (–1, –8) C. Consider the graph of y = 3x2 – 6x + 1. Identify the vertex as a maximum or minimum. A. minimum 0% B. maximum 1. A 2. B C. neither 3. C 4. D A B C D D. cannot be determined D. Consider the graph of y = 3x2 – 6x + 1. Graph the function. A. B. 1. A0% 2. B C. D. 3. C 4. D A B C D Match Equations and Graphs Which is the graph of y = –x2 – 2x –2? A B C D Match Equations and Graphs Read the Item You are given a quadratic function, and you are asked to choose its graph. Solve the Item Find the axis of symmetry of the graph y = –x2 – 2x – 2. Equation for the axis of symmetry a = –1 and b = –2 Match Equations and Graphs The axis of symmetry is –1. Look at the graphs. Since only choices C and D have x = –1 as their axis of symmetry, you can eliminate choices A and B. Since the coefficient of the x2 term is negative, the graph opens downward. Eliminate choice C. Answer: D Which is the graph of y = –x2 + 2x? A. B. C. D. A. A B. B 0% 0% 0% 0% C. C A D. D B C D Five-Minute Check (over Lesson 9-1) Main Ideas and Vocabulary California Standards Example 1: Two Roots Example 2: A Double Root Example 3: No Real Roots Example 4: Factoring Example 5: Rational Roots Example 6: Real-World Example • Solve quadratic equations by graphing. • Estimate solutions of quadratic equations by graphing. • quadratic equation • roots • zeros • double root Standard 21.0 Students graph quadratic functions and know that their roots are the x-intercepts. Standard 22.0 Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points. Two Roots Solve x2 – 3x – 10 = 0 by graphing. Graph the related function f(x) = x2 – 3x – 10. Two Roots Make a table of values to find other points to sketch the graph. To solve x2 – 3x – 10 = 0 you need to know where the value of f(x) is 0. This occurs at the x-intercepts. The x- intercepts of the parabola appear to be –2 and 5. Two Roots Check Solve by factoring. x2 – 3x – 10 = 0 Original equation (x – 5)(x + 2) = 0 Factor. x – 5 = 0 or x + 2 = 0 Zero Product Property x = 5 x = –2 Solve for x. Answer: The solutions of the equation are –2 and 5. Animation: Solving Quadratic Equations By Graphing Solve x2 – 2x – 8 = 0 by graphing. A. {–2, 4} B. {2, –4} C. {2, 4} A. A B. B D. {–2, –4} 0% 0% 0% 0% C. C D. D A B C D A Double Root Solve x2 + 8x = –16 by graphing. First, rewrite the equation so one side is equal to zero. x2 + 8x = –16 Original equation x2 + 8x + 16 = –16 + 16 Add 16 to each side. x2 + 8x + 16 = 0 Simplify. (x + 4)(x + 4) = 0 Factor. x + 4 = 0 or x + 4 = 0 Zero Product Property x = –4 x = –4 A Double Root Answer: The solution is –4. Solve x2 + 2x = –1 by graphing. A. {1} B. {–1} 0% 1. A C. {–1, 1} 2. B 3. C D. 4. D A B C D No Real Roots Solve x2 + 2x + 3 = 0 by graphing. Graph the related function f(x) = x2 + 2x + 3. Answer: The graph has no x-intercept. Thus, there are no real number solutions for the equation. Solve x2 + 4x + 5 = 0 by graphing. A. {1, 5} B. {–1, 5} 0% 1. A C. {5} 2. B 3. C 4. D D. A B C D Factoring Use factoring to determine how many times the graph of f(x) = x2 + 3x – 10 intersects the x-axis. Identify each root. The graph intersects the x-axis when f(x) = 0. x2 + 3x – 10 = 0 Original equation (x – 2)(x + 5) = 0 Factor. Answer: Since the trinomial factors into two distinct factors, the graph of the function intersects the x-axis 2 times. The roots are x = 2 and x = –5. Use factoring to determine how many times the graph of f(x) = x2 + 2x – 15 intersects the x-axis. Identify each root. A. 2 times; x = 3 and x = 5 B. 2 times; x = 3 and x = –5 C. 2 times; x = –3 and x = 5 A. A D. 2 times; x = –3 and x = –5 B. B 0% C. 0% C 0% 0% D. D A B C D Rational Roots Solve x2 – 4x + 2 = 0 by graphing. If integral roots cannot be found, estimate the roots by stating the consecutive integers between which the roots lie. Graph the related function f(x) = x2 – 4x + 2. Notice that the value of the function changes from negative to positive between the x values of 0 and 1 and between 3 and 4. Rational Roots The x-intercepts of the graph are between 0 and 1 and between 3 and 4. Answer: One root is between 0 and 1, and the other root is between 3 and 4. Solve x2 – 2x – 5. A. One root is between 0 and 1, and the other root is between 4 and 5. B. One root is between –1 and 0, and the other root is between 3 and 4. C. One root is between –1 and –2, and the other root is A. A between 3 and 4. B. B D. 0% C. 0% C 0% 0% D. D A B C D MODEL ROCKETS Shelly built a model rocket for her science project. The equation y = –16t2 + 250t models the flight of the rocket, launched from ground level at a velocity of 250 feet per second, where y is the height of the rocket in feet after t seconds. For how many seconds was Shelly’s rocket in the air? You need to find the solution of the equation 0 = –16t2 + 250t. Use a graphing calculator to graph the related function y = –16t2 + 250t. The x-intercept is between 15 and 16 seconds. Answer: between 15 and 16 seconds GOLF Martin hits a golf ball with an upward velocity of 120 feet per second. The function y = –16t2 + 120t models the flight of the golf ball, hit at ground level, where y is the height of the ball in feet after t seconds. How long was the golf ball in the air? A. between 7 and 8 seconds A. A B. between 6 and 7 seconds 0% 0% B. 0% B 0% C. between 8 and 9 seconds C. C A B C D D. between 0 and 1 second D. D Five-Minute Check (over Lesson 9-2) Main Ideas and Vocabulary California Standards Example 1: Irrational Roots Key Concept: Completing the Square Example 2: Complete the Square Example 3: Solve an Equation by Completing the Square Example 4: Real-World Example: Solve a Quadratic Equation in Which a ≠ 1 • Solve quadratic equations by finding the square root. • Solve quadratic equations by completing the square. • completing the square Standard 14.0 Students solve a quadratic equation by factoring or completing the square. (Key) Irrational Roots Solve x2 + 6x + 9 = 5 by taking the square root of each side. Round to the nearest tenth if necessary. x2 + 6x + 9 = 5 Original equation (x + 3)2 = 5 x2 + 6x + 9 is a perfect square trinomial. Take the square root of each side. Take the square root of each side. Definition of absolute value Irrational Roots Subtract 3 from each side. Simplify. Use a calculator to evaluate each value of x. or Answer: The solution set is {–5.2, –0.8}. Solve x2 + 8x + 16 = 3 by taking the square root of each side. Round to the nearest tenth if necessary. A. {–4} B. {–2.3, –5.7} C. {2.3, 5.7} A. A 0% 0% B. 0% B 0% C. C D. Ø A B C D D. D Complete the Square Find the value of c that makes x2 – 12x + c a perfect square. Method 1 Use algebra tiles. Arrange the tiles To make the for x2 – 12x + c so figure a that the two sides square, add of the figure are 36 positive congruent. 1-tiles. x2 – 12x + 36 is a perfect square. Complete the Square Method 2 Complete the square. Step 1 Step 2 Square the result (–6)2 = 36 of Step 1. Step 3 Add the result of x2 –12x + 36 Step 2 to x2 – 12x. Answer: c = 36 Notice that x2 – 12x + 36 = (x – 6)2. Animation: Completing the Square Find the value of c that makes x2 + 14x + c a perfect square. A. 7 B. 14 0% 1. A C. 156 2. B 3. C D. 49 4. D A B C D Solve an Equation by Completing the Square Solve x2 – 18x + 5 = –12 by completing the square. Isolate the x2 and x terms. Then complete the square and solve. x2 – 18x + 5 = –12 Original equation x2 + 18x – 5 – 5 = –12 – 5 Subtract 5 from each side. x2 – 18x = –17 Simplify. x2 – 18x + 81 = –17 + 81 Solve an Equation by Completing the Square (x – 9)2 = 64 Factor x2 –18x + 81. (x – 9) = ±8 Take the square root of each side. x – 9 + 9 = ±8 + 9 Add 9 to each side. x=9±8 Simplify. x = 9 + 8 or x = 9 – 8 Separate the solutions. = 17 =1 Simplify. Answer: The solution set is {1, 17}. Solve x2 – 8x + 10 = 30. A. {–2, 10} 0% B. {2, –10} 1. A C. {2, 10} 2. B 3. C 4. D D. Ø A B C D Solve a Quadratic Equation in Which a ≠ 1 CANOEING Suppose the rate of flow of an 80-foot-wide river is given by the equation r = –0.01x2 + 0.8x where r is the rate in miles per hour, and x is the distance from the shore in feet. Joacquim does not want to paddle his canoe against a current faster than 5 miles per hour. At what distance from the river bank must he paddle in order to avoid a current of 5 miles per hour? Explore You know the function that relates distance from shore to the rate of the river current. You want to know how far away from the river bank he must paddle to avoid the current. Solve a Quadratic Equation in Which a ≠ 1 Plan Find the distance when r = 5. Use completing the square to solve –0.01x2 + 0.8x = 5. Solve –0.01x2 + 0.8x = 5 Equation for the current Divide each side by –0.01. x2 – 80x = –500 Simplify. Solve a Quadratic Equation in Which a ≠ 1 x2 – 80x + 1600 = –500 + 1600 (x – 40)2 = 110 Factor x2 – 80x + 1600. Take the square root of each side. Add 40 to each side. Simplify. Solve a Quadratic Equation in Which a ≠ 1 Use a calculator to evaluate each value of x. Examine The solutions of the equation are about 7 ft and about 73 ft. The solutions are distances from one shore. Since the river is about 80 ft wide, 80 – 73 = 7. Answer: He must stay within about 7 feet of either bank. CANOEING Suppose the rate of flow of a 60-foot- wide river is given by the equation r = –0.01x2 + 0.6x where r is the rate in miles per hour, and x is the distance from the shore in feet. Joacquim does not want to paddle his canoe against a current faster than 5 miles per hour. At what distance from the river bank must he paddle in order to avoid a current of 5 miles per hour? A. 6 feet A. A B. 5 feet B. B C. C C. 1 foot 0% 0% 0% 0% D. D D. 10 feet A B C D Five-Minute Check (over Lesson 9-3) Main Ideas and Vocabulary California Standards Key Concept: The Quadratic Formula Example 1: Solve Quadratic Equations Concept Summary: Solving Quadratic Equations Example 2: Real-World Example: Use the Quadratic Formula to Solve a Problem Key Concept: Using the Discriminant Example 3: Use the Discriminant • Solve quadratic equations by using the Quadratic Formula. • Use the discriminant to determine the number of solutions for a quadratic equation. • Quadratic Formula • discriminant Standard 19.0 Students know the quadratic formula and are familiar with its proof by completing the square. (Key) Standard 20.0 Students use the quadratic formula to find the roots of a second-degree polynomial and to solve quadratic equations. (Key) Standard 22.0 Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points Solve Quadratic Equations A. Solve x2 – 2x – 35 = 0. Round to the nearest tenth if necessary. Method 1 Factoring x2 – 2x – 35 = 0 Original equation (x –7)(x + 5) = 0 Factor x2 –2x – 35. x –7 = 0 or x + 5 = 0 Zero Product Property x=7 x = –5 Solve for x. Solve Quadratic Equations Method 2 Quadratic Formula Quadratic Formula a = 1, b = –2, and c = –35 Multiply. Solve Quadratic Equations Add. Simplify. or Separate the solutions. =7 = –5 Answer: The solution set is {–5, 7}. Solve Quadratic Equations B. Solve 15x2 – 8x = 4. Round to the nearest tenth if necessary. Step 1 Rewrite the equation in standard form. 15x2 – 8x = 4 Original equation 15x2 – 8x – 4 = 4 – 4 Subtract 4 from each side. 15x2 – 8x – 4 = 0 Simplify. Solve Quadratic Equations Step 2 Apply the Quadratic Formula. Quadratic Formula a = 15, b = –8, and c = –4 Multiply. Solve Quadratic Equations Add. or Separate the solutions. Simplify. Solve Quadratic Equations Check the solutions by using the CALC menu on a graphing calculator to determine the zeros of the related quadratic function. Answer: To the nearest tenth, the set is {–0.3, 0.8}. A. Solve x2 + x – 30 = 0. Round to the nearest tenth if necessary. A. {6, –5} B. {–6, 5} C. {6, 5} A. A 0% 0% B. 0% B 0% D. Ø C. C A B C D D. D B. Solve 20x2 – 4x = 8. Round to the nearest tenth if necessary. A. {0.5, 0.7} B. {–0.5, –0.7} C. {–0.5, 0.7} A. A 0% 0% B. 0% B 0% D. Ø C. C A B C D D. D Use the Quadratic Formula to Solve a Problem SPACE TRAVEL Two possible future destinations of astronauts are the planet Mars and a moon of the planet Jupiter, Europa. The gravitational acceleration on Mars is about 3.7 meters per second squared. On Europa, it is only 1.3 meters per second squared. Using the information and equation from Example 2 in the textbook, find how much longer baseballs thrown on Mars and on Europa will stay above the ground than similarly thrown baseballs on Earth. In order to find when the ball hits the ground, you must find when H = 0. Write two equations to represent the situation on Mars and on Europa. Use the Quadratic Formula to Solve a Problem Baseball Thrown on Mars Baseball Thrown on Europa These equations cannot be factored, and completing the square would involve a lot of computation. Use the Quadratic Formula to Solve a Problem To find accurate solutions, use the Quadratic Formula. Use the Quadratic Formula to Solve a Problem Since a negative number of seconds is not reasonable, use the positive solutions. Answer: A ball thrown on Mars will stay aloft 5.6 – 2.2 or about 3.4 seconds longer than the ball thrown on Earth. The ball thrown on Europa will stay aloft 15.6 – 2.2 or about 13.4 seconds longer than the ball thrown on Earth. SPACE TRAVEL The gravitational acceleration on Venus is about 8.9 meters per second squared, and on Callisto, one of Jupiter’s moons, it is 1.2 meters per second squared. Suppose a baseball is thrown on Callisto with an upward velocity of 10 meters per second from two meters above the ground. Find how much longer the ball will stay in air than a similarly-thrown ball on Venus. Use the equation where H is the height of an object t seconds after it is thrown upward, v is is the initial velocity, g is the gravitational pull, and h A the initial 1. height. 2. B A. about 10 seconds 0% 3. C B. about 25.2 seconds 4. D C. about 12.5 seconds A B C D D. about 14.5 seconds Use the Discriminant A. State the value of the discriminant for 3x2 + 10x = 12. Then determine the number of real roots of the equation. Step 1 Rewrite the equation in standard form. 3x2 + 10x = 12 Original equation 3x2 + 10x – 12 = 12 – 12 Subtract 12 from each side. 3x2 + 10x – 12 = 0 Simplify. Use the Discriminant Step 2 Find the discriminant. b2 – 4ac = (10)2 – 4(3)(–12) a = 3, b = 10, and c = –12 = 244 Simplify. Answer: The discriminant is 244. Since the discriminant is positive, the equation has two real roots. Use the Discriminant B. State the value of the discriminant for 4x2 – 2x + 14 = 0. Then determine the number of real roots of the equation. b2 – 4ac = (–2)2 – 4(4)(14) a = 4, b = –2, and c = 14 = –220 Simplify. Answer: The discriminant is –220. Since the discriminant is negative, the equation has no real roots. A. State the value of the discriminant for the equation x2 + 2x + 2 = 0. Then determine the number of real roots for the equation. A. –4; no real roots 0% B. 4; 2 real roots 1. A 2. B 3. C C. 0; 1 real root 4. D A B C D D. cannot be determined B. State the value of the discriminant for the equation –5x2 + 10x = –1. Then determine the number of real roots for the equation. A. –120; no real roots 0% B. 120; 2 real roots 1. A 2. B 3. C C. 0; 1 real root 4. D A B C D D. cannot be determined Five-Minute Check (over Lesson 9-4) Main Ideas and Vocabulary California Standards Key Concept: Exponential Function Example 1: Graph an Exponential Function with a>1 Example 2: Graph an Exponential Function with 0<a<1 Example 3: Real-World Example: Use Exponential Functions to Solve Problems Example 4: Identify Exponential Behavior • Graph exponential functions. • Identify data that displays exponential behavior. • exponential function Preparation for Algebra II Standard 12.0 Students know the laws of fractional exponents, understand exponential functions, and use these functions in problems involving exponential growth and decay. (Key) Graph an Exponential Function with a > 1 A. Graph y = 3x. State the y-intercept. Graph the ordered pairs and connect the points with a smooth curve. Answer: The y-intercept is 1. Graph an Exponential Function with a > 1 B. Use the graph to determine the approximate value of 31.5. The graph represents all real values of x and their corresponding values of y for y = 3x. Answer: The value of y is about 5 when x = 1.5. Use a calculator to confirm this value. 31.5 ≈ 5.196 A. Graph y = 5x. A. B. C. D. 0% 0% 0% 0% A. A A B C D B. B C. C D. D B. Use the graph to determine the approximate value of 50.25. A. about 2.5 B. about 5 C. about 2 A. A B. B D. about 1.5 0% 0% 0% 0% C. C D. D A B C D Graph Exponential Functions with 0 < a < 1 A. Graph the ordered pairs and connect the points with a smooth curve. Answer: The y-intercept is 1. Graph Exponential Functions with 0 < a < 1 B. Answer: The value of y is about 8 when x = –1.5. Use a calculator to confirm this value. A. Graph A. B. C. D. 0% 0% A. 0% A 0% A B C D B. B C. C D. D B. Use the graph to determine the approximate value of A. about 1 B. about 3 1. A 0% 2. B C. about 2 3. C 4. D D. about 0.1 A B C D Use Exponential Functions to Solve Problems DEPRECIATION Some people say that the value of a new car decreases as soon as it is driven off the dealer’s lot. The function V = 25,000 ● 0.82t models the depreciation of the value of a new car that originally cost $25,000. V represents the value of the car and t represents the time in years from the time the car was purchased. Graph the function. What values of V and t are meaningful in the function? Use a graphing calculator to graph the function. Use Exponential Functions to Solve Problems Answer: Only the values of 0 ≤ V ≤ 25,000 and t ≥ 0 are meaningful in the context of the problem. Use Exponential Functions to Solve Problems B. What is the value of the car after one year? V = 25,000 ● 0.82t Original equation V = 25,000 ● 0.821 t=1 V = 20,500 Use a calculator. Answer: After one year, the car's value is about $20,500. Use Exponential Functions to Solve Problems C. What is the value of the car after five years? V = 25,000 ● 0.82t Original equation V = 25,000 ● 0.825 t=5 V = 9268.50 Use a calculator. Answer: After five years, the car’s value is about $9,270. A. Depreciation The function V = 22,000 ● 0.82t models the depreciation of the value of a new car that originally cost $22,000. V represents the value of the car and t represents the time in years from the time the car was purchased. Graph the function. A. B. 0% 1. A 2. B 3. C C. D. 4. D A B C D B. What is the value of the car after one year? A. $21,000 B. $23,600 0% 1. A C. $18,040 2. B 3. C 4. D D. $20,000 A B C D C. What is the value of the car after three years? A. $12,130 B. $25,120 0% 1. A C. $10,000 2. B 3. C 4. D D. $15,000 A B C D Identify Exponential Behavior Determine whether the set of data displays exponential behavior. Explain why or why not. Method 1 Look for a Pattern The domain values are at regular intervals of 10. Look for a common factor among the range values. 10 25 62.5 156.25 × 2.5 × 2.5 × 2.5 Identify Exponential Behavior Answer: Since the domain values are at regular intervals and the range values have a common factor, the data are probably exponential. The equation for the data may involve (2.5)x. Method 2 Graph the Data Answer: The graph shows a rapidly increasing value of y as x increases. This is a characteristic of exponential behavior. Determine whether the set of data displays exponential behavior. A. no 0% B. yes C. cannot be determined 1. A 2. B A B C 3. C Five-Minute Check (over Lesson 9-5) Main Ideas and Vocabulary California Standards Key Concept: General Equation for Exponential Growth Example 1: Real-World Example: Exponential Growth Example 2: Real-World Example: Compound Interest Key Concept: General Equation for Exponential Decay Example 3: Real-World Example: Exponential Decay • Solve problems involving exponential growth. • Solve problems involving exponential decay. • exponential growth • compound interest • exponential decay Preparation for Algebra II Standard 12.0 Students know the laws of fractional exponents, understand exponential functions, and use these functions in problems involving exponential growth and decay. (Key) Exponential Growth A. POPULATION In 2005 the town of Flat Creek had a population of about 280,000 and a growth rate of 0.85% per year. Write an equation to represent the population of Flat Creek since 2005. The rate 0.85% can be written has 0.0085. y = C(1 + r)t General equation for exponential growth y = 280,000(1 + 0.0085)t C = 280,000 and r = 0.0085 y = 280,000(1.0085)t Simplify. Answer: An equation to represent the population of Flat Creek is y = 280,000(1.0085)t, where y is the population and t is the number of years since 2005. Exponential Growth B. POPULATION In 2005 the town of Flat Creek had a population of about 280,000 and a growth rate of 0.85% per year. According to the equation, what will be the population of Flat Creek in the year 2015? In 2015, t will equal 2015 – 2005 or 10. y = 280,000(1.0085)t Equation for population of Flat Creek y = 280,000(1.0085)10 t = 10 y ≈ 304,731 Use a calculator. Answer: In 2015, there will be about 304,731 people in Flat Creek. A. POPULATION In 2000, Scioto School District had a student population of about 4500 students, and a growth rate of about 0.15% per year. Write an equation to represent the student population of the Scioto School District since the year 2000. A. y = 4500(1.0015) B. y = 4500(1.0015)t A. A t C. y = 4500(0.0015) B. B 0% 0% 0% 0% C. C D. y = (1.0015)t A B C D D. D B. POPULATION In 2000, Scioto School District had a student population of about 4500 students, and a growth rate of about 0.15% per year. According to the equation, what will be the student population of the Scioto School District in the year 2006? A. about 9000 students B. about 4600 students A. A C. about 4540 students B. B 0% 0% 0% 0% C. C D. about 4700 students A B C D D. D Compound Interest COLLEGE When Jing May was born, her grandparents invested $1000 in a fixed rate savings account at a rate of 7% compounded annually. The money will go to Jing May when she turns 18 to help with her college expenses. What amount of money will Jing May receive from the investment? Compound interest equation P = 1000, r = 7% or 0.07, n = 1, and t = 18 Compound Interest A = 1000(1.07)18 Compound interest equation A = 3379.93 Simplify. Answer: She will receive about $3380. COMPOUND INTEREST When Lucy was 10 years old, her father invested $2500 in a fixed rate savings account at a rate of 8% compounded semiannually. When Lucy turns 18, the money will help to buy her a car. What amount of money will Lucy receive from the investment? A. about $4682 0% 1. A B. about $5000 2. B 3. C C. about $4600 4. D A B C D D. about $4500 Exponential Decay A. CHARITY During an economic recession, a charitable organization found that its donations dropped by 1.1% per year. Before the recession, its donations were $390,000. Write an equation to represent the charity’s donations since the beginning of the recession. y = C(1 + r)t General equation for exponential growth y = 390,000(1 – 0.011)t C = 390,000 and r = 1.1% or 0.011 y = 390,000(0.989)t Simplify. Answer: y = 390,000(0.989)t Exponential Decay B. CHARITY During an economic recession, a charitable organization found that its donations dropped by 1.1% per year. Before the recession, its donations were $390,000. Estimate the amount of the donations 5 years after the start of the recession. y = 390,000(0.989)t General equation for exponential growth y = 390,000(0.989)5 t=5 y = 369,016.74 Answer: The amount of donations should be about $369,017. A. CHARITY A charitable organization found that the value of its clothing donations dropped by 2.5% per year. Before this downturn in donations, the organization received clothing valued at $24,000. Write an equation to represent the value of the charity’s clothing donations since the beginning of the downturn. 1. A t 0% A. y = (0.975) 2. B 3. C B. y = 24,000(0.975)t 4. D C. y = 24,000(1.975)t D. y = 24,000(0.975) A B C D B. CHARITY A charitable organization found that the value of its clothing donations dropped by 2.5% per year. Before this downturn in donations, the organization received clothing valued at $24,000. Estimate the value of the clothing donations 3 years after the start of the downturn. A. about $23,000 1. A 0% 2. B B. about $21,000 3. C 4. D C. about $22,245 D. about $24,000 A B C D Five-Minute Checks Image Bank Math Tools Animation Menu Exploring Quadratic Functions Lesson 9-1 (over Chapter 8) Lesson 9-2 (over Lesson 9-1) Lesson 9-3 (over Lesson 9-2) Lesson 9-4 (over Lesson 9-3) Lesson 9-5 (over Lesson 9-4) Lesson 9-6 (over Lesson 9-5) To use the images that are on the following three slides in your own presentation: 1. Exit this presentation. 2. Open a chapter presentation using a full installation of Microsoft® PowerPoint® in editing mode and scroll to the Image Bank slides. 3. Select an image, copy it, and paste it into your presentation. 9-1 Families of Quadratic Functions 9-2 Solving Quadratic Equations By Graphing 9-3 Completing the Square (over Chapter 8) Factor the polynomial a2 – 5a + 9, if possible. If the polynomial cannot be factored using integers, write prime. A. (a + 3)(a – 3) B. (a + 3)(a + 3) A. A C. (a – 3)(a – 3) B. 0% B 0% 0% 0% C. C A B C D D. prime D. D (over Chapter 8) Factor the polynomial 6z2 – z – 1, if possible. If the polynomial cannot be factored using integers, write prime. A. (2z – 1)(3z – 1) 0% B. (2z – 1)(3z + 1) 1. A 2. B C. (2z + 1)(3z – 1) 3. C 4. D A B C D D. prime (over Chapter 8) Solve the equation 5x2 = 125. A. {–5, 5} 0% B. {–25, 25} 1. A 2. B C. 3. C 4. D D. A B C D (over Chapter 8) Solve the equation 2x2 + 11x – 21 = 0. A. B. C. A. A 0% 0% B. 0% B 0% D. C. C A B C D D. D (over Chapter 8) A certain basketball player’s hang time can be described by 4t2 = 1, where t is time in seconds. How long is this player’s hang time? A. 4 seconds 0% B. 2 seconds 1. A 2. B C. 3. C 4. D A B C D D. (over Chapter 8) The area of a rectangle is given by 6y2 + 5y – 6 and the width is given by 3y – 2. What is the length? A. 2y + 3 0% 1. A B. y + 3 2. B 3. C C. 2y – 3 4. D A B C D D. y – 3 (over Lesson 9-1) Which of the following options shows a graph and the solution of y = x2 + 2x – 1? A. B. C. D. A. A B. B 0% 0% 0% 0% C. C A B C D D. D (over Lesson 9-1) For y = –x2 + 2, identify the equation of axis of symmetry, and the coordinates of the vertex of the graph of the equation. Identify the vertex as a maximum or a minimum. A. 0% 1. A B. 2. B 3. C C. x = 0; (0, 2); maximum 4. D A B C D D. x = 0; (0, 2); minimum (over Lesson 9-1) For y = x2 – 5x, identify the equation of the axis of symmetry, and the coordinates of the vertex of the graph of the equation. Identify the vertex as a maximum or a minimum. 0% A. x = –2.5; (–2.5, 18.75); maximum 1. A B. x = –2.5; (–2.5, 18.75); 2. B minimum 3. C C. x = 2.5; (2.5, –6.25); 4. D maximum A B C D D. x = 2.5; (2.5, –6.25); minimum (over Lesson 9-1) What is the maximum height of a rocket fired straight up if the height in feet is described by h = –16t2 + 64t + 1, where t is time in seconds? A. 66 ft B. 65 ft A. A C. 64 ft B. B 0% 0% 0% 0% C. C A B C D D. 63 ft D. D (over Lesson 9-1) The graph of which of the following equations has a vertex at (2, –1)? A. y = x2 – 4x + 2 0% B. y = 2x2 + 4x + 2 1. A 2. B C. y = x2 – 4x + 3 3. C 4. D A B C D D. y = x – 4x – 2 2 (over Lesson 9-2) Which of the following options shows a graph and the solution of m2 – 2m – 3 = 0? A. B. C. D. A. A B. B 0% 0% 0% 0% C. C A B C D D. D (over Lesson 9-2) Which choice shows the graph of w2 + 5w – 1 = 0, and its roots or the consecutive integers between which the roots lie? A. B. 0% 1. A 2. B C. D. 3. C 4. D A B C D (over Lesson 9-2) Use a quadratic equation to find two numbers whose difference is 3 and whose product is 10. A. 0% B. –5, 2 or 5, –2 1. A 2. B C. –5, –2 or 5, 2 3. C 4. D D. A B C D (over Lesson 9-2) Which of the following are the roots of x2 + 6x + 8 = 0? A. 2 and 4 B. 8 and 1 C. –8 and –1 A. A 0% 0% B. 0% B 0% C. C D. –4 and –2 A B C D D. D (over Lesson 9-3) Solve the equation x2 + 8x + 16 = 16. A. –8, 0 B. 8, 0 C. 2, 16 A. A 0% 0% B. 0% B 0% C. C D. –2, 16 A B C D D. D (over Lesson 9-3) Solve the equation x2 – 6x – 2 = 5. A. 1, –7 0% B. –1, –7 1. A 2. B C. –1, 7 3. C 4. D D. 1, 7 A B C D (over Lesson 9-3) Find the value of c that makes z2 + z + c a perfect square. A. 0% B. 1. A 2. B 3. C C. 4. D A B C D D. (over Lesson 9-3) The area of a square can be tripled by increasing the length and width by 10 inches. What is the original length of the square? Round to the nearest tenth of an inch. A. 2.24 in. B. 3.7 in. A. A C. 10 in. 0% 0% B. 0% B 0% C. C A B C D D. 13.7 in. D. D (over Lesson 9-3) Use the roots of x2 – 14x + 49 = 81 to determine which statement is true. A. The sum of the roots is greater than the product of the roots. 0% B. The product of the roots is 1. A greater than the sum of the roots. 2. B C. The product of the roots is 3. C positive. 4. D A B C D D. The sum of the roots is negative. (over Lesson 9-4) Solve the equation x2 – 6x – 7 = 0 by using the Quadratic Formula. Round to the nearest tenth if necessary. A. 1, 7 B. –1, 7 A. A C. –7, 1 0% 0% B. 0% B 0% C. C D. Ø A B C D D. D (over Lesson 9-4) Solve the equation y2 – 11y = –30 by using the Quadratic Formula. Round to the nearest tenth if necessary. A. 5, 6 0% B. –5, 6 1. A 2. B C. –5, –6 3. C 4. D A B C D D. Ø (over Lesson 9-4) Solve the equation 5z2 + 16z + 3 = 0 by using the Quadratic Formula. Round to the nearest tenth if necessary. A. 0% 1. A B. 2. B 3. C C. 4. D A B C D D. Ø (over Lesson 9-4) Solve the equation 4n2 = –19n – 25 by using the Quadratic Formula. Round to the nearest tenth if necessary. A. B. A. A C. 0% 0% B. 0% B 0% C. C D. Ø A B C D D. D (over Lesson 9-4) Without graphing, determine the number of x-intercepts of the graph of f(x) = 3x2 + x + 7. A. 0 0% B. 1 1. A C. 2 2. B 3. C 4. D D. 3 A B C D (over Lesson 9-4) The work shown is the first step to finding the roots of which equation? A. 3x2 – 2x + 4 = 0 0% 1. A B. 4x – 3x – 4 = 0 2 2. B 3. C C. 2x2 – 3x + 4 = 0 4. D D. 2x2 + 3x – 4 = 0 A B C D (over Lesson 9-5) Which choice shows the graph of y = 4x, and also states the y-intercept and the approximate value of 40.6? Use a calculator to confirm the value. A. B. C. D. A. A 0% 0%B. 0%B 0% C. C A B C D D. D (over Lesson 9-5) Which choice states and explains whether or not the data in the given table display exponential behavior? A. Yes, the domain values are at a regular intervals and the range values have a common difference. B. Yes, the domain values are at regular intervals and the range values have a 1. A common ratio of 3. 2. 0% B C. No, the domain values are at regular 3. C intervals and the range values have a common ratio of 3. 4. D D. No, the domain values are at regular A B C D intervals and the range values have a common difference. (over Lesson 9-5) A tournament begins with 64 teams. After each round, only of the teams remain. Write an exponential function for the number of teams remaining after x rounds of play. How many teams remain after 3 rounds? A. 8 teams 1. A 0% B. 16 teams 2. B 3. C C. 8 teams 4. D A B C D D. 16 teams (over Lesson 9-5) What is the y-intercept of the graph of y = 2x – 5? A. –5 B. –4 C. 3 A. A 0% 0% B. 0% B 0% D. 11 C. C A B C D D. D This slide is intentionally blank.