Question 31499
Factorise it: (x-3)(x+1) so the x-intercepts are at x=-1 and x=3


Complete the square: {{{ x^2-2x-3 }}}
2a=2
a=1
--> {{{a^2 = 1}}}


--> {{{ x^2-2x(+1-1)-3 }}}
--> {{{ (x^2-2x+1)-1-3 }}}
--> {{{ (x-1)^2-4 }}}


So, the vertex is at (1,-4) meaning that the line of symmetry in the function is at x=1.


Y-intercept is when x=0, so from {{{ y = x^2-2x-3 }}}, we get 
{{{ y = (0)^2-2(0)-3 }}}
--> y= -3.


A point symmetric to the y-intercept (0,-3) is (2,-3) since both points are equi-distant from the line x=1.


All seen on {{{ graph(300,300,-6,6,-10,10,x^2-2x-3) }}}


As for "give the axis"... what axis?


jon.