The vertex of the given parabola [tex]$x^{2}=8 \mathrm{y}$[/tex] exists (0, 0).
The vertex of a parabola exists as the point at the intersection of the parabola and its line of symmetry. The vertex of the parabola exists as the minimum point on the graph for a positive right-handed parabola.
Given the equation of parabola exists [tex]$x^{2}=8 y$[/tex].
[tex]$\mathrm{y}=x^{2} / 8[/tex]
General vertex form for any given parabola exists [tex]$\mathrm{y}=\mathrm{a}(x-a)^{2}+\mathrm{b}$[/tex]
where (a, b) exists the coordinates of the vertex.
For this function it exists [tex]$\mathrm{y}=1 / 8(x-0)^{2}+0$[/tex]
The vertex of the given parabola exists at (0, 0).
Therefore, the vertex of the given parabola [tex]$x^{2}=8 \mathrm{y}$[/tex] exists (0, 0).
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