Question 1084125
Complete the square to write each function in {{{f(x) = a(x-h)^2 + k}}} form. Determine the vertex and the axis of symmetry of the graph of the function. 
Then plot several points to complete the graph. 

{{{f(x) = x^2 - x - 6}}}

{{{f(x) =(x^2 - x +b^2)-b^2- 6}}}...since {{{a=1}}} and {{{2ab=-1}}} we can determine {{{b}}}

{{{2ab=-1}}}->{{{2*1*b=-1}}}->{{{2b=-1}}}->{{{b=-1/2}}}

{{{f(x) =(x^2 - x +(-1/2)^2)-(-1/2)^2- 6}}}

{{{f(x) =(x -1/2)^2-(1/4)- 6}}}

{{{f(x) =(x -1/2)^2-(1/4)- 24/4}}}

{{{f(x) =(x -1/2)^2-25/4}}}

so, {{{h=1/2}}} and {{{k=-25/4}}}->the vertex is at ({{{1/2}}},{{{-25/4}}})≈({{{0.5}}}, {{{-6.25}}})

The axis of symmetry of a parabola is the vertical line through the vertex.

so, axis of symmetry is {{{x=1/2}}} or {{{x=0.5}}}


table:

{{{x}}}|{{{y}}}
{{{-2}}}|{{{0}}}....{{{f(-2) = (-2)^2 - (-2)- 6=4+2-6=0}}}
{{{-1}}}|{{{-4}}}....{{{f(-1) = (-1)^2 - (-1)- 6=1+1-6=-4}}}
{{{0}}}|{{{-6}}}....{{{f(0) = (0)^2 - (0)- 6=0+0-6=-6}}}
{{{1}}}|{{{-6}}}....{{{f(1) = (1)^2 - (1)- 6=1-1-6=-6}}}
{{{2}}}|{{{-4}}}....{{{f(2) = (2)^2 - (2)- 6=4-2-6=-4}}}

plot points and draw a graph:


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(0.5,-6.25,.12),locate(0.5,-6.35,V(0.5,-6.25)),
circle(-2,0,.12),locate(-2.2,0.6,p(-2,0)),
circle(-1,-4,.12),locate(-1,-4,p(-1,-4)),
circle(0,-6,.12),locate(0.2,-6,p(0,-6)),
circle(1,-6,.12),locate(1,-6,p(1,-6)),
circle(2,-4,.12),locate(2,-4,p(2,-4)),
line(1/2,-10,1/2,10),
 graph( 600, 600, -10, 10, -10, 10, x^2 - x - 6)) }}}