Obviously, \(k=0 \) is an answer. Let's try to find some more (or prove that k=0 is the only answer). Let the base of the triangle be the segment from (1, 0) to (0, 1). The length of this segment will then be \(\sqrt{2}\). We can see that the height of the triangle be \(\frac{\sqrt{2}}{2}\). Since the third point is (k, k), the point must lie on the line x=y which means that k can be 0 or 1.
First, we can simplify f(x): \(f(x)=-3x^2+6x+27\). Then, we factor out -3 for \(f(x)=-3(x^2-2x-9)=-3(x-1)^2+30\).
From this equation, we get that the vertex of f(x) is (x, y) = (1, 30).
