Find a+b+c if the graph of the equation y=ax^2+bx+c is a parabola with vertex (5,3), vertical axis of symmetry, and contains the point (2,0).
+33661 The symmetry of the parabola dictates that if the vertex is at (5, 3) and it goes through (2, 0) then it must also go through (8, 0).
So there are three conditions from which to find the three unknowns:
3 = 25a + 5b + c. (1)
0 = 4a + 2b + c. (2)
0 = 64a + 8b + c. (3)
From (2) we get. c = -4a - 2b. (4)
Put this into (1) and (2).
3 = 21a + 3b. (5)
0 = 60a + 6b. (6)
Divide (5) by 3 and (6) by 6:
1 = 7a + b. (7)
0 = 10a + b. (8)
From (8) we get. b = -10a. (9)
Put this into (7)
1 = -3a. or: a = -1/3
Put this into (9): b = 10/3
Put these into (4): c = -16/3
+33661 The symmetry of the parabola dictates that if the vertex is at (5, 3) and it goes through (2, 0) then it must also go through (8, 0).
So there are three conditions from which to find the three unknowns:
3 = 25a + 5b + c. (1)
0 = 4a + 2b + c. (2)
0 = 64a + 8b + c. (3)
From (2) we get. c = -4a - 2b. (4)
Put this into (1) and (2).
3 = 21a + 3b. (5)
0 = 60a + 6b. (6)
Divide (5) by 3 and (6) by 6:
1 = 7a + b. (7)
0 = 10a + b. (8)
From (8) we get. b = -10a. (9)
Put this into (7)
1 = -3a. or: a = -1/3
Put this into (9): b = 10/3
Put these into (4): c = -16/3
