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Properties of Parabolas.ppt

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					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

				
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Lingjuan Ma Lingjuan Ma
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