Question 1082150
Find ({{{h}}},{{{k}}}) of the parabola {{{x^2 - 2x - 8y - 15 = 0}}}:

 {{{(x^2 - 2x) - 8y -15 = 0}}}..........complete squares

{{{(x^2 - 2x+b^2)-b^2 - 8y -15 = 0}}}

{{{(x^2 - 2x+1^2)-1^2 -15=8y }}}

{{{(x-1)^2-1 -15=8y }}}

{{{(x-1)^2-16=8y }}}

{{{(x-1)^2/8-16/8=y }}}

{{{y=(1/8)(x-1)^2-2 }}}...-> {{{h=1}}} and {{{k=-2}}}

so, vertex is at  ({{{1}}},{{{-2}}})


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(1,-2,.12), locate(1,-2,V(1,-2)),
 graph( 600, 600, -10, 10, -10, 10, (1/8)(x-1)^2-2)) }}}