+1206 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 \((2,0)\).
+128408 y = ax^2 + bx + c
The x coordinate of the vertex = -b / (2a) = 5 ⇒ -b = (2a)*5 ⇒ b = -10a
So we have this system
a(5)^2 - 10a(5) + c = 3 ⇒ 25a - 50a + c = 3 ⇒ -25a + c = 3 (1)
a(2)^2 - 10a(2) + c = 0 ⇒ 4a - 20a + c = 0 ⇒ - 16a + c = 0 ⇒ 16a - c = 0 (2)
Add (1) and (2) and we get that
-9a = 3
a = -1/3
b = -10 (-1/3) = 10/3
And using (2) 16 (-1/3) - c = 0 ⇒ -16/3 = c
So....
y = (-1/3)x^2 + (10/3)x - 16/3
a = -1/3 b = 10/3 c = -16/3
And their sum is -7/3
