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

Question

campbell_st · Answer

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.

campbell_st · Answer

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.