Solving Quadratic Equations Real World Problem Solving 8

Solving Quadratic Equations Real World Problem Solving 8 -6 8 by Factoring Quadratic Equations Warm Up Lesson Presentation Lesson Quiz Holt Mc. Dougal Algebra 1 Algebra 11 Holt Mc. Dougal

Solving Quadratic Equations 8 -6 by Factoring Example 5: Application The graph of f(x) = – 0. 06 x 2 + 0. 6 x + 10. 26 can be used to model the height in meters of an arch support for a bridge, where the xaxis represents the water level and x represents the horizontal distance in meters from where the arch support enters the water. Can a sailboat that is 14 meters tall pass under the bridge? Explain. The vertex represents the highest point of the arch support. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Example 5 Continued Step 1 Find the x-coordinate. a = – 0. 06, b = 0. 6 Identify a and b. Substitute – 0. 06 for a and 0. 6 for b. Step 2 Find the corresponding y-coordinate. Use the function rule. f(x) = – 0. 06 x 2 + 0. 6 x + 10. 26 Substitute 5 for x. = – 0. 06(5)2 + 0. 6(5) + 10. 26 = 11. 76 Since the height of each support is 11. 76 m, the sailboat cannot pass under the bridge. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 5 The height of a small rise in a roller coaster track is modeled by f(x) = – 0. 07 x 2 + 0. 42 x + 6. 37, where x is the distance in feet from a supported pole at ground level. Find the greatest height of the rise. Step 1 Find the x-coordinate. a = – 0. 07, b= 0. 42 Identify a and b. Substitute – 0. 07 for a and 0. 42 for b. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 5 Continued Step 2 Find the corresponding y-coordinate. f(x) = – 0. 07 x 2 + 0. 42 x + 6. 37 = – 0. 07(3)2 + 0. 42(3) + 6. 37 = 7 ft The height of the rise is 7 ft. Holt Mc. Dougal Algebra 1 Use the function rule. Substitute 3 for x.

Solving Quadratic Equations 8 -6 by Factoring Example 2: Application The height in feet of a basketball that is thrown can be modeled by f(x) = – 16 x 2 + 32 x, where x is the time in seconds after it is thrown. Find the basketball’s maximum height and the time it takes the basketball to reach this height. Then find how long the basketball is in the air. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Example 2 Continued 1 Understand the Problem The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the ground. List the important information: • The function f(x) = – 16 x 2 + 32 x models the height of the basketball after x seconds. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Example 2 Continued 2 Make a Plan Find the vertex of the graph because the maximum height of the basketball and the time it takes to reach it are the coordinates of the vertex. The basketball will hit the ground when its height is 0, so find the zeros of the function. You can do this by graphing. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Example 2 Continued 3 Solve Step 1 Find the axis of symmetry. Use x = . Substitute – 16 for a and 32 for b. Simplify. The axis of symmetry is x = 1. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Example 2 Continued Step 2 Find the vertex. f(x) = – 16 x 2 + 32 x The x-coordinate of the vertex is 1. = – 16(1)2 + 32(1) Substitute 1 for x. = – 16(1) + 32 = – 16 + 32 = 16 The vertex is (1, 16). Holt Mc. Dougal Algebra 1 Simplify. The y-coordinate is 16.

Solving Quadratic Equations 8 -6 by Factoring Example 2 Continued Step 3 Find the y-intercept. f(x) = – 16 x 2 + 32 x + 0 Identify c. The y-intercept is 0; the graph passes through (0, 0). Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Example 2 Continued Step 4 Graph the axis of symmetry, the vertex, and the point containing the y-intercept. Then reflect the point across the axis of symmetry. Connect the points with a smooth curve. (1, 16) (0, 0) Holt Mc. Dougal Algebra 1 (2, 0)

Solving Quadratic Equations 8 -6 by Factoring Example 2 Continued The vertex is (1, 16). So at 1 second, the basketball has reached its maximum height of 16 feet. The graph shows the zeros of the function are 0 and 2. At 0 seconds the basketball has not yet been thrown, and at 2 seconds it reaches the ground. The basketball is in the air for 2 seconds. (1, 16) (0, 0) Holt Mc. Dougal Algebra 1 (2, 0)

Solving Quadratic Equations 8 -6 by Factoring Example 2 Continued 4 Look Back Check by substitution (1, 16) and (2, 0) into the function. ? 16 = – 16(1)2 + 32(1) ? 16 = – 16 + 32 16 = 16 ? 0 = – 16(2)2 + 32(2) ? 0 = – 64 + 64 0 = 0 Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Remember! The vertex is the highest or lowest point on a parabola. Therefore, in the example, it gives the maximum height of the basketball. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 As Molly dives into her pool, her height in feet above the water can be modeled by the function f(x) = – 16 x 2 + 16 x + 12, where x is the time in seconds after she begins diving. Find the maximum height of her dive and the time it takes Molly to reach this height. Then find how long it takes her to reach the pool. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 Continued 1 Understand the Problem The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the pool. List the important information: • The function f(x) = – 16 x 2 + 16 x + 12 models the height of the dive after x seconds. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 Continued 2 Make a Plan Find the vertex of the graph because the maximum height of the dive and the time it takes to reach it are the coordinates of the vertex. The diver will hit the water when its height is 0, so find the zeros of the function. You can do this by graphing. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 Continued 3 Solve Step 1 Find the axis of symmetry. Use x = . Substitute – 16 for a and 16 for b. Simplify. The axis of symmetry is x = 0. 5. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 Continued Step 2 Find the vertex. f(x) = – 16 x 2 + 16 x + 12 = – 16(0. 5)2 + 16(0. 5) +12 = – 16(0. 25) + 8 +12 The x-coordinate of the vertex is 0. 5. Substitute 0. 5 for x. Simplify. = – 4 + 20 = 16 The vertex is (0. 5, 16). Holt Mc. Dougal Algebra 1 The y-coordinate is 9.

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 Continued Step 3 Find the y-intercept. f(x) = – 16 x 2 + 16 x + 12 Identify c. The y-intercept is 12; the graph passes through (0, 12). Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 Continued Step 4 Find another point on the same side of the axis of symmetry as the point containing the y-intercept. Since the axis of symmetry is x = 0. 5, choose an x-value that is less than 0. 5. Let x = 0. 25 f(x) = – 16(0. 25)2 + 16(0. 25) + 12 = – 1 + 4 + 12 = 15 Another point is (0. 25, 15). Holt Mc. Dougal Algebra 1 Substitute 0. 25 for x. Simplify.

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 Continued Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and the other point. Then reflect the points across the axis of symmetry. Connect the points with a smooth curve. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 Continued The vertex is (0. 5, 16). So at 0. 5 seconds, Molly's dive has reached its maximum height of 16 feet. The graph shows the zeros of the function are – 0. 5 and 1. 5 seconds. At – 0. 5 seconds the dive has not begun, and at 1. 5 seconds she reaches the pool. Molly reaches the pool in 1. 5 seconds. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 Continued 4 Look Back Check by substituting (0. 5, 16) and (1. 5, 0) into the function. ? 16 = – 16(0. 5)2 + 16(0. 5) + 12 ? 16 = – 4 + 8 +12 16 = 16 ? 0 = – 16(1. 5)2 + 16(1. 5) +12 ? 0 = – 36 + 24 + 12 0 = 0 Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 What if…? A dolphin jumps out of the water. The quadratic function y = – 16 x 2 + 32 x models the dolphin’s height above the water after x seconds. How long is the dolphin out of the water? When the dolphin leaves the water, its height is 0, and when the dolphin reenters the water, its height is 0. So solve 0 = – 16 x 2 + 32 x to find the times when the dolphin leaves and reenters the water. Step 1 Write the related function 0 = – 16 x 2 + 32 x y = – 16 x 2 + 32 x Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Check It Out! Example 2 Continued Step 2 Graph the function. Use a graphing calculator. Step 3 Use to estimate the zeros. The zeros appear to be 0 and 2. The dolphin leaves the water at 0 seconds and reenters at 2 seconds. The dolphin is out of the water for 2 seconds. Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Example 3: Application The height in feet of a diver above the water can be modeled by h(t) = – 16 t 2 + 8 t + 8, where t is time in seconds after the diver jumps off a platform. Find the time it takes for the diver to reach the water. h = – 16 t 2 + 8 t + 8 0 = – 8(2 t 2 – t – 1) The diver reaches the water when h = 0. Factor out the GFC, – 8. 0 = – 8(2 t + 1)(t – 1) Factor the trinomial. 0= – 16 t 2 Holt Mc. Dougal Algebra 1 + 8 t + 8

Solving Quadratic Equations 8 -6 by Factoring Example 3 Continued Use the Zero Product – 8 ≠ 0, 2 t + 1 = 0 or t – 1= 0 Property. 2 t = – 1 or t = 1 It takes the diver 1 second to reach the water. Solve each equation. Since time cannot be negative, does not make sense in this situation. Check 0 = – 16 t 2 + 8 t + 8 0 – 16(1)2 + 8(1) + 8 Substitute 1 into the original equation. 0 – 16 + 8 0 0 Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Lesson Quiz 2. The height in feet of a fireworks shell can be modeled by h(t) = – 16 t 2 + 224 t, where t is the time in seconds after it is fired. Find the maximum height of the shell, the time it takes to reach its maximum height, and length of time the shell is in the air. 784 ft; 7 s; 14 s Holt Mc. Dougal Algebra 1

Solving Quadratic Equations 8 -6 by Factoring Lesson Quiz: Part II 7. A rocket is shot straight up from the ground. The quadratic function f(t) = – 16 t 2 + 96 t models the rocket’s height above the ground after t seconds. How long does it take for the rocket to return to the ground? 6 s 8. The height of a rocket launched upward from a 160 foot cliff is modeled by the function h(t) = – 16 t 2 + 48 t + 160, where h is height in feet and t is time in seconds. Find the time it takes the rocket to reach the ground at the bottom of the cliff. 5 s Holt Mc. Dougal Algebra 1