which function has a vertex at the origin?\n○ $f(x)=(x + 4)^2$\n○ $f(x)=x(x - 4)$\n○ $f(x)=(x - 4)(x +…

Explanation:

Step1: Recall vertex - form of a quadratic function

The vertex - form of a quadratic function is (y=a(x - h)^2+k), where ((h,k)) is the vertex of the parabola.

Step2: Analyze (f(x)=(x + 4)^2)

Here, (h=-4) and (k = 0), so the vertex is ((-4,0)).

Step3: Expand (f(x)=x(x - 4))

(f(x)=x^2-4x). Completing the square: (f(x)=(x - 2)^2-4), vertex is ((2,-4)).

Step4: Expand (f(x)=(x - 4)(x + 4))

(f(x)=x^2-16), in vertex - form (y = 1(x-0)^2-16), vertex is ((0,-16)).

Step5: Analyze (f(x)=-x^2)

In the form (y=a(x - h)^2+k), (a=-1), (h = 0), (k = 0). So the vertex is ((0,0)) (the origin).

Answer:

(f(x)=-x^2)