Question 196127
If the equation has the form
{{{ax^2 + bx + c}}}, then the 
vertex is where {{{x = -b/(2a)}}}
In your equation, {{{y = x^2 + 5}}},
there is no {{{x}}} term and no {{{b}}},
so {{{b=0}}} and {{{-b/(2a) = 0}}},
{{{x=0}}} is the axis of symmetry
That is the y-axis.
I already know that {{{x=0}}} is the
x-coordinate of the vertex. What is
the y-coordinate?
{{{y = x^2 + 5}}}
{{{y = 0 + 5}}}
{{{y = 5}}}
So, the vertex is at (0,5)
I'll graph it, too
{{{ graph( 300, 300, -5, 5, -2, 10, x^2 + 5) }}}