The graph of \(y=ax^2 + bx + c\) is a parabola with vertical axis of symmetry. The vertex of this parabola is \((2,3)\) and the parabola contains the point \((4,4)\). Find the value of \(y\) when \(x=6\).
y = ax^2 + bx + c
The x coordinate of the vertex = -b / [2a] = 2 ⇒ b = - 4a
We know that
a(2)^2 + b(2) + c = 3 ⇒ 4a + 2b + c = 3 ⇒ 4a - 8a + c = 3 ⇒ -4a + c = 3 (1)
a(4)^2 + b(4) + c = 4 ⇒ 16a + 4b + c = 4 ⇒ 16a - 16a + c = 4 ⇒ c = 4 (2)
Subbing (2) into (1) we have that
-4a + 4 = 3
-4a = -1
a = -1/-4 = 1/4
And b = -4(1/4) = -1
So......the function is
y = (1/4)x^2 - x + 4
So....when x = 6
y = (1/4)6^2 - (6) + 4 = 9 - 6 + 4 = 7