Question 1191806
as you can see that is a parabola opening up, has vertex at ({{{3}}},{{{-1}}})
=>{{{h=3}}} and {{{k=-1}}}

so, vertex form equation is

{{{y-k=a(x-h)^2}}}.......substitute {{{h=3}}} and {{{k=-1}}}

{{{y-(-1)=a(x-3)^2}}}

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

{{{y=a(x-3)^2-1}}}

you also see from the graph that parabola passes through the point ({{{2}}},{{{0}}}), so use it to calculate {{{a}}}

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

{{{0=a(-1)^2-1}}}

{{{a=1}}}

and your equation is:


{{{y=(x-3)^2-1}}}


{{{ graph( 600, 600, -10, 10, -10, 10, (x-3)^2-1) }}}