x^2 is never less than zero, so f(x) is never less than -3. The vertex is at (0,-3)
In general, f(x) = ax^2+bx+c has its vertex at x = -b/2a
f(x) = a(x-h)^2 + k
has its vertex at (h,k). In this case, h=0 and k = -3, so the vertex is at (0,-3)
How to find the vertex ?
f(x)= x^2-3
3 answers
If you do not do calculus you complete the square
x^2 - 3 = y
x^2 - 3 + (3/2)^2 = y + 9/4
(x-3/2)^2 = y+9/4
vertex at (3/2 , -9/4)
x^2 - 3 = y
x^2 - 3 + (3/2)^2 = y + 9/4
(x-3/2)^2 = y+9/4
vertex at (3/2 , -9/4)
assumed x with 3