Chapter 9 Section 6 9 6 More About

Chapter 9 Section 6

9. 6 More About Parabolas and Their Applications Objectives 1 • Find the vertex of a vertical parabola. 2 • Graph a quadratic function. 3 • Use the discriminant to find the number of x-intercepts of a parabola with a vertical axis. 4 • Use quadratic functions to solve problems involving maximum or minimum value. 5 • Graph parabolas with horizontal axes. Copyright © 2012, 2008, 2004 Pearson Education, Inc.

Objective 1 Find the vertex of a vertical parabola. Copyright © 2012, 2008, 2004

Find the vertex of a vertical parabola. When the equation of a parabola is given in the form f(x) = ax 2 + bx + c, we need to locate the vertex to sketch an accurate graph. There are two ways to do this: 1. Complete the square. 2. Use a formula derived by completing the square. Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 4

CLASSROOM EXAMPLE 1 Completing the Square to Find the Vertex (a = 1) Find the vertex of the graph of f(x) = x 2 + 4 x – 9. Solution: We need to complete the square. f(x) = x 2 + 4 x – 9 = (x 2 + 4 x )– 9 = (x 2 + 4 x + 4 – 4 ) – 9 = (x 2 + 4 x + 4) – 4 – 9 = (x + 2)2 – 13 The vertex of the parabola is (– 2, – 13). Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 5

CLASSROOM EXAMPLE 2 Completing the Square to Find the Vertex (a ≠ 1) Find the vertex of the graph of f(x) = 2 x 2 – 4 x + 1. Solution: We need to complete the square, factor out 2 from the first two terms. f(x) = 2 x 2 – 4 x + 1 f(x) = 2(x 2 – 2 x) + 1 = 2(x 2 – 2 x + 1 – 1 ) + 1 = 2(x 2 – 2 x + 1) + 2(– 1) + 1 = 2(x 2 – 2 x + 1) – 2 + 1 = 2(x – 1)2 – 1 The vertex of the parabola is (1, – 1). Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 6

Find the vertex of a vertical parabola. Vertex Formula The graph of the quadratic function defined by f(x) = ax 2 + bx + c (a ≠ 0) has vertex and the axis of the parabola is the line Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 7

CLASSROOM EXAMPLE 3 Using the Formula to Find the Vertex Use the vertex formula to find the vertex of the graph of f(x) = – 2 x 2 + 3 x – 1. Solution: a = – 2, b = 3, and c = – 1. The x-coordinate: The y-coordinate: The vertex of the parabola is Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 8

Objective 2 Graph a quadratic function. Copyright © 2012, 2008, 2004 Pearson Education, Inc.

Graph a quadratic function. Graphing a Quadratic Function y = f(x) Step 1 Determine whether the graph opens up or down. If a > 0, the parabola opens up; if a < 0, it opens down. Step 2 Find the vertex. Use either the vertex formula or completing the square. Step 3 Find any intercepts. To find the x-intercepts (if any), solve f(x) = 0. To find the y-intercept, evaluate f(0). Step 4 Complete the graph. Plot the points found so far. Find and plot additional points as needed, using symmetry about the axis. Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 10

CLASSROOM EXAMPLE 4 Graphing a Quadratic Function Graph f(x) = x 2 – 6 x + 5. Give the vertex, axis, domain, and range. Solution: Step 1 The graph opens up since a = 1, which is > 0. Step 2 Find the vertex. Complete the square. f(x) = x 2 – 6 x + 5 = x 2 – 6 x + 9 – 9 + 5 = (x – 3)2 – 4 The vertex is at (3, – 4). Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 11

CLASSROOM EXAMPLE 4 Graphing a Quadratic Function (cont’d) Graph f(x) = x 2 – 6 x + 5. Step 3 Find any x-intercepts. Let f(x) = 0 The x-intercepts are (5, 0) and (1, 0). Find the y-intercept. Let x = 0. The y-intercept is (0, 5). Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 12

CLASSROOM EXAMPLE 4 Graphing a Quadratic Function (cont’d) Graph f(x) = x 2 – 6 x + 5. Step 4 Plot the points. Vertex: (3, – 4) x-intercepts: (5, 0) and (1, 0) y-intercept. (0, 5) axis of symmetry: x = 3 By symmetry (6, 5) is also another point on the graph. domain: ( , ) Copyright © 2012, 2008, 2004 Pearson Education, Inc. range: [ 4, ) Slide 9. 6 - 13

Objective 3 Use the discriminant to find the number of x-intercepts of a parabola

Use the discriminant to find the number of x-intercepts of a parabola with a vertical axis. You can use the discriminant to determine the number x-intercepts. Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 15

CLASSROOM EXAMPLE 5 Using the Discriminant to Determine the Number of x- Intercepts Find the discriminant and use it to determine the number of x-intercepts of the graph of the quadratic function. Solution: a = 3, b = – 1, c = 2 The discriminant is positive, the graph has two x-intercepts. Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 16

CLASSROOM EXAMPLE 5 Using the Discriminant to Determine the Number of x- Intercepts (cont’d) Find the discriminant and use it to determine the number of x-intercepts of the graph of each quadratic function. Solution: a = 1, b = – 1, c = 1 The discriminant is negative, the graph has no x-intercepts. a = 1, b = – 8, c = 16 The discriminant is 0, the graph has one x-intercept. Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 17

Objective 4 Use quadratic functions to solve problems involving maximum or minimum value. Copyright

Use quadratic functions to solve problems involving maximum or minimum value. PROBLEM-SOLVING HINT In many applied problems we must find the greatest or least value of some quantity. When we can express that quantity in terms of a quadratic function, the value of k in the vertex (h, k) gives that optimum value. Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 2. 3 - 19

CLASSROOM EXAMPLE 6 Finding the Maximum Area of a Rectangular Region A farmer has 100 ft of fencing to enclose a rectangular area next to a building. Find the maximum area he can enclose, and the dimensions of the field when the area is maximized. Solution: Let x = the width of the field x + length = 100 2 x + length = 100 – 2 x Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 20

CLASSROOM EXAMPLE 6 Finding the Maximum Area of a Rectangular Region (cont’d) Area = width length A(x) = x(100 – 2 x) = 100 x – 2 x 2 Determine the vertex: a = – 2, b = 100, c = 0 y-coordinate: f(25) = – 2(25)2 + 100(25) = – 2(625) + 2500 = – 1250 + 2500 = 1250 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Vertex: (25, 1250) Slide 9. 6 - 21

CLASSROOM EXAMPLE 6 Finding the Maximum Area of a Rectangular Region (cont’d) Parabola opens down with vertex (25, 1250). The vertex shows that the maximum area will be 1250 square feet. The area will occur if the width, x, is 25 feet and the length is 100 – 2 x, or 50 feet. Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 22

CLASSROOM EXAMPLE 7 Finding the Maximum Height Attained by a Projectile A toy rocket is launched from the ground so that its distance above the ground after t seconds is s(t) = – 16 t 2 + 208 t Find the maximum height it reaches and the number of seconds it takes to reach that height. Solution: Find the vertex of the function. a = – 16, b = 208 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 23

CLASSROOM EXAMPLE 7 Finding the Maximum Height Attained by a Projectile (cont’d) Find the y-coordinate. The toy rocket reaches a maximum height of 676 feet in 6. 5 seconds. Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 24

Objective 5 Graph parabolas with horizontal axes. Copyright © 2012, 2008, 2004 Pearson Education,

Graph parabolas with horizontal axes. Graph of a Horizontal Parabola The graph of x = ay 2 + by + c or x = a(y – k)2 + h is a parabola. q. The vertex of the parabola is (h, k). q. The axix is the horizontal line y = k. q. The graph opens to the right if a > 0 and to the left if a < 0. Copyright © 2012, 2008, 2004 Pearson Education, Inc. Slide 9. 6 - 26

CLASSROOM EXAMPLE 8 Graphing a Horizontal Parabola (a = 1) Graph x = (y + 1)2 – 4. Give the vertex, axis, domain, and range. Solution: Vertex: ( 4, 1) Opens: right since a > 1 Axis: y = 1 x f(x) 3 0 3 2 0 1 0 3 Domain: [ 4, ) Copyright © 2012, 2008, 2004 Pearson Education, Inc. Range: ( , ) Slide 9. 6 - 27

CLASSROOM EXAMPLE 9 Completing the Square to Graph a Horizontal Parabola (a ≠ 1) Graph x = y 2 + 2 y + 5. Give the vertex, axis, domain, and range. Solution: Complete the square. x = (y 2 – 2 y) + 5 = (y 2 – 2 y + 1) – ( 1) + 5 = (y – 1)2 + 6 Vertex: (6, 1) Opens: left, since a < 1 Axis: y = 1 Domain: [ , 6] Copyright © 2012, 2008, 2004 Pearson Education, Inc. Range: ( , ) Slide 9. 6 - 28