ADVANCED EXERCISES

(1) Find the equations for the parabolas described below:

 (a) vertex is 2 units from the x-axis, opens downward, symmetric about x = 1, 
     y-intercept = 23/12

 (b) has vertical axis of symmetry and passing through (-2,3), (0,3), and 
     (1,9).

 (c) Opens upward, passing through (-2,7), vertex inon the positive 
     y-axis and is 5 units from the origin.

 (d) Find an equation that represents a parabola which opens to the 
     right, has its vertex at (2,2) and passes through the point (5,0).

 (a) y = Ax2 + Bx + C  (since it opens down)

     23/12 = A(0)2 + B(0) + C  ==>  C = 23/12

     Since the vertex is (1,2) we get

        (-B/(2A), C - B2/(4A)) = (1,2)

        ==>  -B/2A = 1  ==>  B = -2A
             C - B2/(4A) = 2
               ==> 23/12 - B2/(4A) = 2
                   
                   -B2/(4A) = 1/12

                   -(-2A)2/(4A) = 1/12

                   -A = 1/12  ==>  A = -1/12

        ==>  B = 1/6

     y = -x2/12 + x/6 + 23/12

     12y = -x2 + 2x + 23


 (b) y = Ax2 + Bx + c

     3 = A(0)2 + B(0) + C  ==> C = 3

     3 = A(-2)2 + B(-2) + 3
       ==>  0 = 4A -2B  ==> B = 2A

     9 = A(1)2 + B(1) + 3
       ==> 9 = A + 2A + 3
           6 = 3A  ==>  A = 2

     ==> B = 4

     y = 3x2 + 6x + 3


 (c) (y - 5) = A(x - 0)2
             = Ax2

      7 - 5 = A(-2)2

      2 = 4A  ==>  A = 1/2

      y - 5 = x2/2

      2y - 10 = x2


 (d)  x - 2 = A(y - 2)2

      5 - 2 = A(0 - 2)2

      3 = 4A  ==>  A = 3/4

      x - 2 = (3/4)(y - 2)2

      4x - 8 = 3(y - 2)2