The graph of y=ax^2+bx+c passes through points (0,5)$, $(1,10), and (2,19). Find a+b+c
+36916 When x = 0 y = 5
5 = a(0^2) + b(0) + c so c = 5
when x = 1 y = 10
10 = a(1^2) + b(1) + 5 or a + b -5 = 0
when x = 2 y= 19
19 = a(4) + 2b + 5 or 4a + 2b - 14 = 0
you know 'c' and now you have two equations in red that you can use to find 'a' and 'b'
