Question 131783
I presume you mean that {{{f(x)=(1/4)x^2}}} as opposed to {{{f(x)=1/(4x^2)}}}.


Given that assumption, you can re-write your equation as:


{{{f(x)=(1/4)x^2+0x+0}}} to put it in the {{{ax^2+bx+c}}} form, with a = 1/4, b = 0, and c = 0.


Now recall that the x-coordinate of the parabola vertex is {{{-b/2a}}}.  In this case:  {{{-0/(2(1/4))=0}}}.


And the y-coordinate of the vertex is at {{{f(0)=(1/4)0^2+0*0+0=0}}}.


Now we know that the vertex is at (0,0).


We know that the axis of symmetry is a vertical line that intersects the vertex, so the equation of the axis is {{{x=0}}}



{{{drawing(600,600,-3,3,-3,3,
grid(1),
graph(600,600,-3,3,-3,3,x^2/4),
locate(.1,-.1,V(0,0)),
circle(0,0,.05),
green(line(0,-3,0,3))
)}}}