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)

For (y=(x + 4)^2=1\times(x-(-4))^2+0), the vertex is ((-4,0)) since (h=-4) and (k = 0).

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

Expand (y=x(x - 4)=x^{2}-4x). Complete the square: (y=x^{2}-4x=(x - 2)^{2}-4), so the vertex is ((2,-4)).

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

Expand (y=(x - 4)(x + 4)=x^{2}-16). In vertex - form (y = 1\times(x-0)^2-16), the vertex is ((0,-16)).

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

For (y=-x^{2}=-1\times(x - 0)^2+0), here (h = 0) and (k = 0), so the vertex is ((0,0)) which is the origin.

Answer:

(f(x)=-x^{2})