Question 975089
use the vertex form equation of the parabola: 

{{{y=a(x-h)^2+k}}} where {{{h}}} and {{{k}}} are {{{x}}} and {{{y}}} coordinates of the vertex

so,if  vertex ( {{{2}}},{{{-1}}}), then {{{h=-2}}} and {{{k=-1}}} 

{{{y=a(x-2)^2-1}}}


use point ({{{4}}},{{{ -3}}}), and find {{{a}}}

{{{-3=a(4-2)^2-1}}}

{{{-3+1=a(2)^2}}}

{{{-2=4a}}}

{{{-2/4=a}}}

{{{a=-(1/2)}}}

your equation is: {{{y=-(1/2)(x-2)^2-1}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(2,-1,.12),locate(2,-1,V(2,-1)),
circle(4,-3,.12),locate(4,-3,p(4,-3)),
 graph( 600, 600, -10, 10, -10, 10, -(1/2)(x-2)^2-1)) }}}