Question 321273
The equation of the axis of symmetry of {{{f(x) = (1/5)x^2}}} can be found by:
{{{x = -b/2a}}} and in your equation, {{{a = (1/5)}}} and {{{b = 0}}} so...
{{{highlight(x = 0)}}} This is the equation of the line of symmetry.
The vertex can be found by substituting {{{x = 0}}} into the given equation.
Replace f(x) with y.
{{{y = (1/5)x^2}}} Substitute {{{x = 0}}}
{{{y = 0}}}
The vertex is at (0, 0)
{{{graph(400,400,-5,5,-5,5,(1/5)x^2)}}}