Question 4114
There are several ways to find the vertex and graph a quadratic function.  One easy way is to realize that in the general case

{{{ y = ax^2 + bx + c }}}

the vertex is at 

{{{ x= -b/2a }}}

Therefore for {{{y = x^2 - 4}}}  the coefficient of x is zero, so the vertex is at x=0.  When x = 0, y = -4, so the vertex is at (0, -4).


The graph opens upward, since the coeffient of x^2 is positive.  It would open downward if the coefficient of x^2 is negative.


The y-intercept is where x=0, which is at the point (0, -4).


The x-interecepts are where y=0, which are at 

{{{0=x^2 - 4}}} or x=2 and at x=-2.