Which function has a vertex at the origin f(x)= (x+4)^2, f(x) x(x-4), f(x)=(x-4)(x+4), f(x)=-x^2

First of all, we need to know what is the vertex means which is the maximum or minimum point of a parabola and the formula will be:
x=-b/2a
Where b and a from
f(x)=ax^2+bx+c
So do find which function has a vertex of origin. Let's find the vertex of all the function that we had:
f(x)=(x+4)^2
f(x)=(x+4)(x+4)
f(x)=x^2+8x+16
x=-b/2a
x=-8/2(1)
x=-8/2
x=-4
Not the right answer because the vertex needs to be origin which is x=0

f(x)=x(x-4)
f(x)=x^2-4x
x=-b/2a
x=-(-4)/2(2)
x=4/4
x=1
Not the right answer

f(x)=(x-4)(x+4)
f(x)=x^2-16
x=-0/2(1)
x=0
Yay! This is the right answer. As a result, f(x)=(x-4)(x+4) is your final answer. Hope it help!