A parabola ax^2+bx+c contains the points (-1,0), (0,5), and (5,0). Find the value of 100a + 10b + c.
+33616 (-1,5): a(-1)2+b(-1)+c = 0 or a - b + c = 0
(0,5): a(0)2+b(0)+c = 5 or c = 5
(5,0): a(5)2+b(5)+c = 0 or 25a + 5b +c = 0
So:
a - b + 5 = 0. (1)
25a + 5b + 5 = 0. (2)
Multiply (1) by 5 and add to (2):
30a + 30 = 0. So a = -1
Hence, From (1):
-1 - b + 5 = 0. So b = 4
a = -1, b = 4, c = 5
I’ll leave you to calculate 100a + 10b + c
+33616 (-1,5): a(-1)2+b(-1)+c = 0 or a - b + c = 0
(0,5): a(0)2+b(0)+c = 5 or c = 5
(5,0): a(5)2+b(5)+c = 0 or 25a + 5b +c = 0
So:
a - b + 5 = 0. (1)
25a + 5b + 5 = 0. (2)
Multiply (1) by 5 and add to (2):
30a + 30 = 0. So a = -1
Hence, From (1):
-1 - b + 5 = 0. So b = 4
a = -1, b = 4, c = 5
I’ll leave you to calculate 100a + 10b + c
