Question 422003
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Hi
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
f(x)=1/5x^2   Vertex(0,0) Line of symmentry the y a-xis (or x = 0)

Note the Coefficient determines 2 things:
1) whether the parabola opens upward (>0 or downward(<0)
2) the broadness of the parabola
It has NO effect of the axis of symmetry. 
Note: blue graph is:  y = -5x^2
{{{drawing(300,300,   -6, 6, -6, 6, grid(1),
graph( 300, 300, -6, 6, -6, 6,0,.2x^2,-5x^2))}}}