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The zeros are 2 and 4 so the equation now looks like
(x-2)(x-4) = x^2 -6x +8
f(1) = 6? 1^2 -6 +8 = 3 which needs to double to equal 6
so 2x^2 - 12x +16 = 0 fits the bill! Graph to follow ....
The vertex is at x = 3
And the x coordinate of the vertex = -b / [2a] = 3 → b = -6a
The product of the roots = c /a = 8 → c = 8a
And we know that
a(1)^2 - 6a(1) + 8a = 6
3a = 6 →
a = 2 b = -12 c = 16
Here'a a graph : https://www.desmos.com/calculator/flafi02koq
+37084 
