Question 366131
I think you mean the zeros of the function.
The average of the zeros would be the x-coordinate of the vertex.
.
.
.
Example:
{{{f(x)=(x-2)(x-4)}}}
has zeros at {{{x=2}}} and {{{x=4}}}.
When you convert to vertex form, {{{y=a(x-h)^2+k}}},
{{{f(x)=x^2-4x-2x+8}}}
{{{f(x)=x^2-6x+8}}}
{{{f(x)=(x^2-6x+9)+8-9}}}
{{{f(x)=(x-3)^2-1}}}
The vertex is located at (3,-1).
.
.
{{{drawing(300,300,-2,8,-4,6,blue(line(3,-10,3,10)),circle(3,-1,0.2),circle(2,0,0.2),circle(4,0,0.2),grid(1),graph(300,300,-2,8,-4,6,0,(x-2)(x-4)))}}}