How do you find the vertex of a parabola # y=y=[x]^2-5#?

I don't know in which grade you are so there will be two parts in the explanation.

Deduction of the formula:
If you know how to derivate and find max and mins of functions then you have to deduct the formula. If not, jump to the next step.

For doing the deduction i'll use you specific case, it's easy to deduct using #y=ax^2+bx+c#. In a vertex the slope of the function ( it's derivate) HAS to be 0 either if it is a max or a min, you are searching for a vertex so you don't care about it being a max or a min.

The derivate of #f(x)=x^2-5# is #f'(x)=2x# and it has to be equal to zero, then 2x=0 so x=0 and by substitute x=0 in your equation you find y=-5 then the vertex is #(0,-5)#.

In the general case #f'(x)=2ax+b=0# isolating x you get the general formula.

Just using the formula (kids way)
Somehow it's posible to deductthat the vertex of a parabola is placed at #x=-b/(2a)# you don't net to know the deduction ( first part of the solving).
in your equation #y=x^2-5#, cmparing it to #y=ax^2+bx+c# you see that the numer that is in front of #x^2# is a=1 the number in front of x it's b=0 and c=-5.

using the formula #x=-b/(2a)=-0/(2·1)=0# your vertex is placed at x=0

Now you need to know in which y? you have to replace x in y by doing #y=0^2-5=-5#

Now you know that the vertex is placed at (0,-5)