The graph of y=x^2+bx+c passes through points (0, 5),(1, 10), (2, 19) and . Find a+b+c.
When x = 0 you find c = 5
x^2 + bx + 5 now, when x = 1
1 + b + 5 = 10 so b = 4
x^2 + 4x + 5
Hmmmmm something seems amiss when x = 2 y should be 17 , not 19
did you enter your Q correctly?
+128475 I'm assuming that you left off the "a" in your original question
Guest is correct....c = 5
And we have this system
a + b + 5 =10 ⇒ a + b = 5 ⇒ b = 5 - a (1)
4a + 2b + 5 = 19 ⇒ 4a + 2b = 14 (2)
Sub (1) into (2) and we have
4a + 2 (5 - a) = 14
4a + 10 - 2a = 14
2a = 4
a = 2
And b = 5 - 2 = 3
So
a + b + c = 2 + 3 + 5 = 10
