which function has a vertex at the origin?\no (f(x)=(x + 4)^2)\no (f(x)=x(x - 4))\no (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 $f(x)=(x + 4)^2=(x-(-4))^2+0$, the vertex is $(-4,0)$.

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

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

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

$f(x)=x^2-16$. In vertex - form $f(x)=(x - 0)^2-16$, the vertex is $(0,-16)$.

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

For $f(x)=-x^2=-(x - 0)^2+0$, the vertex is $(0,0)$ which is the origin.

Answer:

$f(x)=-x^2$