Question 1046925
Depending how you have the equation,  general form y=ax^2+bx+c or standard form y=a(x-h)^2+k;  {{{a>0}}} will make vertex to be a minimum.  Vertex of (0,0) will make the standard form equation take the format  y=ax^2.  The vertex would be the point,  (h,k).



{{{y=x^2}}} is a reference equation for a parabola.  Vertex will be the point  (0,0).  Setup a data table to find some point, graph the points, and sketch this and you will find a graph like this:
{{{graph(400,400,-10,10,-10,10,x^2)}}}


If you shift the graph leftward or rightward, using subtraction of some value h, then in standard form, you would have an equation like {{{y=(x-h)^2}}}.  This would be a shift TO THE RIGHT, IF {{{h>0}}}; or a SHIFT TO THE LEFT IF {{{h<0}}}.  This means that the vertex, still touching the x-axis, will be at  (h,0).
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Imagine that h=5.
The equation could become {{{y=(x-5)^2}}} and the graph is this:
{{{graph(400,400,-10,10,-10,10,(x-5)^2)}}}



Imagine that h=-2.
The equation could become {{{y=(x-(-2))^2}}}, or {{{y=(x+2)^2}}} and this shifts the model reference graph two units LEFTWARD, still with vertex touching the x-axis.
The graph:
{{{graph(400,400,-10,10,-10,10,(x+2)^2)}}}.
and vertex is at   (-2,0).



Looking again at standard form, {{{y=(x-h)^2+k}}}, and not yet discussing how k value will contribute to the shifting of the graph,  vertex is  (h,k); and if the vertex is (0,0), the Origin, then putting these coordinates values into the standard form model,  {{{y=(x-0)^2+0}}}, which simplifies to {{{y=x^2}}}.


I have used in these examples, a=1.