Question 1132215
 

 vertex form:{{{y=a(x-h)^2+k}}}

{{{y= 2x^2 - 12x + 16}}} in vertex form will be:


{{{y= 2(x^2 - 6x) + 16}}} 

{{{y= 2(x^2 - 6x+b^2)-2b^2 + 16}}} ...{{{b}}} is equal to {{{6/2=3}}}

{{{y= 2(x^2 - 6x+3^2)-2*3^2 + 16}}} 

{{{y= 2(x -3)^2-2*9 + 16}}} 

{{{y= 2(x -3)^2-18 + 16}}} 

{{{y= 2(x -3)^2-2}}} 

=> vertex is at ({{{3}}},{{{-2}}})


{{{drawing( 600, 600, -10,10, -10, 10,
circle(3,-2,.12),locate(3,-2,V(3,-2)),
 graph( 600, 600, -10,10, -10, 10, 2(x -3)^2-2)) }}}