Question 978084
{{{ y=3x^2 -12x-3 }}}

to graph it, you need several points

so, make a table

{{{x}}}|{{{y}}}......choose several values for {{{x}}} and calculate {{{y}}}

{{{-1}}}|{{{12}}}....{{{ y=3*(-1)^2 -12(-1)-3=3*1+12-3=12 }}}

{{{0}}}|{{{-3}}}....{{{ y=3*(0)^2 -12(0)-3=-3 }}}


{{{1}}}|{{{-12}}}....{{{ y=3*(1)^2 -12(1)-3=3*1-12-3=-12 }}}

{{{2}}}|{{{-15}}}....{{{ y=3*(2)^2 -12(2)-3=12-24-3=-12-3=-15 }}}

{{{3}}}|{{{-12}}}....{{{ y=3*(3)^2 -12(3)-3=27-36-3=-12 }}}
plot these points and draw a parabola through:


{{{drawing( 600, 600, -10, 10, -20, 20,
circle(-1,12,.12),circle(1,-12,.12),circle(0,-3,.12),circle(2,-15,.12),
circle(3,-12,.12),locate(2,-15,V(2,-15)),

 graph( 600, 600, -10, 10, -20, 20, 3x^2 -12x-3 )) }}}


compare {{{ y=3x^2 -12x-3 }}} to {{{ y=ax^2 +bx+c }}},  you see that {{{a=3}}} and {{{b=-12}}}

since the {{{x}}} coordinate of the vertex is {{{-b/2a}}}, we have

{{{-b/2a=-(-12)/(2*3)=12/6=2}}}

so,{{{x=2}}}

substitute it in {{{ y=3x^2 -12x-3 }}} and calculate  the {{{y}}} coordinate of the vertex

{{{ y=3*(2)^2 -12(2)-3}}}
{{{y=12-24-3}}}
{{{y=-12-3}}}
{{{y=-15 }}}

so, the vertex is at ({{{2}}},{{{-15}}})