Question 343971
In vertex form,
{{{y=a(x-3)^2+1}}}
Using ({{{1}}},{{{3}}}),
{{{3=a(1-3)^2+1}}}
{{{a(-2)^2=2}}}
{{{4a=2}}}
{{{a=1/2}}}
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{{{highlight_green(y=(x-3)^2/2+1)}}}
Now that you have the equation, test all of the points.
I'll do one, you do the remaining four.
A) ({{{-3}}},{{{-1}}}):{{{-1=(-3-3)^2/2+1}}}
{{{-1=(-6)^2/2+1}}}
{{{-1=36/2+1}}}
{{{-1=18+1}}}
{{{-1=19}}}
False, so ({{{-3}}},{{{-1}}}) is not a point on the parabola.
Continue with the remaining points in the same manner.
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