Question 1138
 Since the quadratic equation with vertex (a,b) should be as:
  y - b = c (x-a)^2,
 Now (a,b) = (3,5) and (5,1)is on the graph, we have
 1 - 5 = c (5 -3)^2 = 4 c 
 and so c = -1, the quadratic equation as
  y - 5 = - (x - 3)^2

 Next, let the equation of the parabola be y = a x^2 + bx + c
 Since it passing through (-1,4), (1, -2) and (2,1).
 We have
   a - b + c = 4 ...(1)
   a + b + c = -2 ...(2)
  4a + 2b + c = 1 ...(3)

  By (2)-(1), we have 2b = -6, so b=-3.
  Replace b by -3 in (1),we have a + c = 1...(4)
  Replace b by -3 in (3),we have 4a + c = 7...(5)
  By (5)-(4), we have 3a=6, a =2.
  
  Replace a by 2 in (4),we have c= 1-2 = -1.

  Hence,the equation of the parabola be y = 2 x^2 - 3x -1.








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