Answer:
 Minimum at (2, -2)
Step-by-step explanation:
Factor the leading coefficient from the first two terms:
 f(x) = 2(x^2 -4x) +6
Inside parentheses, add the square of half the x-coefficient. Add the opposite of that amount outside parentheses.
 f(x) = 2(x^2 -4x +4) +6 -8
Rewrite in vertex form.
 f(x) = 2(x -2)^2 -2
The positive leading coefficient tells you the vertex is a minimum. Comparing this to the vertex form for a quadratic with vertex (h, k)
 f(x) = a(x -h)^2 +k
we find the vertex to be ...
 (h, k) = (2, -2)
The function has a minimum at (2, -2).
