Question

what is the vertex of f(x)= x^2-4

Answer 1

well for this problem there are several methods

1. completing the square...

2. find the line of symmetry

the standard for is

\[y = (x -h)^2 + k\]

which means the parabola has a vertex at (h, k)

so looking at your problem it can be written as

\[y = (x - 0)^2 - 4\]

compare this to the standard form above and you should be able to see
the vertex in the equation.

Answer 2

another way to look at it is

\[y = x^2 \]

has a vertex at (0, 0)

what effect does -4 have on the curve...

and thats where the vertex moves to.