Question 105674
the vertex. {{{h(x)=(x-2)(x+6)+16}}}

{{{h(x) = x^2 + 6x – 2x – 12 + 16}}}

{{{h(x) = x^2 + 4x + 4}}}

{{{h(x) = (x + 2)^2 + 0 }}}                this is the vertex form, so the vertex is at {{{-2}}},{{{0}}}


*[invoke completing_the_square 1, 4, 4]