Question 955410
Using the vertex form,
{{{y=a(x-2)^2-1}}}
Then using the point,
{{{-3=a(4-2)^2-1}}}
{{{-3=a(2^2)-1}}}
{{{-3=4a-1}}}
{{{-2=4a}}}
{{{a=-1/2}}}
So,
{{{y=-(x-2)^2/2-1}}}
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{{{drawing(300,300,-2,7,-7,2,grid(1),circle(2,-1,0.2),circle(4,-3,0.2),graph(300,300,-2,7,-7,2,-(x-2)^2/2-1))}}}