The equation is
[tex]y=2x^2-32x+56[/tex].
Factoring 2, we get
[tex]y=2(x^2-16x+28)[/tex].
We notice that [tex](x-8)^2=x^2-16x+64[/tex], so [tex]x^2-16x=(x-8)^2-64[/tex].
Substituting in the previous equation, we have:
[tex]y=2(x^2-16x+28)=2[(x-8)^2-64+28]=2[(x-8)^2-36][/tex],
distributing 2 over the two terms inside the brackets, we finally get:
[tex]y=2(x-8)^2-72.[/tex]
This is a parabola opening upwards, since the coefficient of x^2 in the original equation is positive, and whose vertex is (8, -72), which is the lowest point of this parabola.
Answer: [tex]y=2(x-8)^2-72.[/tex]; x-coordinate of the minimum: 8.