Document Sample

Properties of Parabolas LESSON 5-2 Additional Examples 1 Graph y = 3 x2 + 1. Step 1: Graph the vertex, which is the y-intercept (0, 1). Step 2: Make a table of values to find some points on one side of the axis of symmetry x = 0. Graph the points. x 1 2 3 4 y 11 3 213 4 613 Step 3: Graph corresponding points on the other side of the axis of symmetry. Step 4: Sketch the curve. ALGEBRA 2 Properties of Parabolas LESSON 5-2 Additional Examples 1 Graph y = 2 x2 + x + 3. Label the vertex and axis of symmetry. Step 1: Find and graph the axis of symmetry. b 1 x=– = – 1 = –1 2a 2( 2 ) Step 2: Find and graph the vertex. The x-coordinate of the vertex is –1. The y-coordinate is y = 1 (–1)2 + 1 2 (–1) + 3 = 2 . So the vertex is 2 1 (–1, 2 ). 2 Step 3: Find and graph the y-intercept and its reflection. Since c = 3, the y-intercept is (0, 3) and its reflection is (–2, 3). ALGEBRA 2 Properties of Parabolas LESSON 5-2 Additional Examples (continued) Step 4: Evaluate the function for another value of x, such as 1 y = 2 (2)2 + (2) + 3 = 7. Graph (2, 7) and its reflection (–4, 7). Step 5: Sketch the curve. ALGEBRA 2 Properties of Parabolas LESSON 5-2 Additional Examples 1 Graph y = – 4 x2 + 2x – 3. What is the maximum value of the function? Since a < 0, the graph of the function opens down, and the vertex represents the maximum value. Find the coordinates of the vertex. x = – b = – 21 = 4 Find the x-coordinate of the vertex. 2a 2(– 4 ) 1 y = – 4 (4)2 + 2(4) – 3 = 1 Find the y-coordinate of the vertex. Graph the vertex and the axis of symmetry x = 4. Graph two points on one side of the axis of symmetry, such as (6, 0) and (8, –3). Then graph corresponding points (2, 0) and (0, –3). The maximum value of the function is 1. ALGEBRA 2 Properties of Parabolas LESSON 5-2 Additional Examples The number of weekend get-away packages a hotel can sell is modeled by –0.12p + 60, where p is the price of a get-away package. The revenue is the product of the price and the number of packages sold. What price will maximize revenue? What is the maximum revenue? Relate: revenue equals price times number of get-away packages sold Define: Let R = revenue. Let p = price of a get-away package. Let –0.12p + 60 = number of a get-away packages sold. Write: R = p ( –0.12p + 60 ) = –0.12p2 + 60p Write in standard form. ALGEBRA 2 Properties of Parabolas LESSON 5-2 Additional Examples (continued) Find the maximum value of the function. Since a < 0, the graph of the function opens down, and the vertex represents a maximum value. b 60 p = – 2a = – 2(–0.12) = 250 Find p at the vertex. R = –0.12(250)2 + 60(250) Evaluate R for p = 250 = 7500 Simplify. A price of $250 will maximize revenue. The maximum revenue is $7500. ALGEBRA 2

DOCUMENT INFO

Shared By:

Categories:

Tags:

Stats:

views: | 0 |

posted: | 2/20/2014 |

language: | Unknown |

pages: | 6 |

How are you planning on using Docstoc?
BUSINESS
PERSONAL

By registering with docstoc.com you agree to our
privacy policy and
terms of service, and to receive content and offer notifications.

Docstoc is the premier online destination to start and grow small businesses. It hosts the best quality and widest selection of professional documents (over 20 million) and resources including expert videos, articles and productivity tools to make every small business better.

Search or Browse for any specific document or resource you need for your business. Or explore our curated resources for Starting a Business, Growing a Business or for Professional Development.

Feel free to Contact Us with any questions you might have.