There exist constants a, h and k such that x^2 + 12x + 4 = a(x - h)^2 + k for all real numbers x. Enter the ordered triple (a,h,k).
Quadratics aren't exactly my strong spot... but I would play around with the equation a bit. Factor the left side?
In x^2 + 12x + 4, what two numbers multiply to 4 and add up to 12?
