Question 1132947
a) 

{{{y=6x^2+12x-18 }}}

{{{y=(6x^2+12x)-18 }}}

{{{y=6(x^2+2x)-18 }}}

{{{y=6(x^2+2x+b^2)-6b^2-18 }}}...take half of coefficient  of {{{x}}} or {{{b}}}:{{{2/2=1}}}

{{{y=6(x^2+2x+1^2)-6*1^2-18 }}}

{{{y=6(x+1)^2-6-18 }}}

{{{y=6(x+1)^2-24 }}}=>{{{h=-1}}} and {{{k=-24}}}

the vertex occurs at ({{{-1}}},{{{-24}}})


{{{drawing(600, 600, -10, 10, -30, 10,
circle(-1,-24,.12),locate(-1,-24,V(-1,-24)),
 graph(600, 600, -10, 10, -30, 10, 6(x+1)^2-24)) }}}


b) 

{{{y=4x^2-16+10}}}

{{{y=4x^2-6}}}

we cannot complete the square for this equation 

use zero value for {{{x}}} to find minimum where the vertex occurs

{{{y=4*0^2-6}}}

{{{y=-6}}}

=>the vertex occurs at ({{{0}}},{{{-6}}})


{{{drawing(600, 600, -10, 10, -10, 10,
circle(0,-6,.12),locate(0,-6,V(0,-6)),
 graph(600, 600, -10, 10, -10, 10, 4x^2-6)) }}}