If a point is in the graph of the parabola it needs to satisfy the equation. So we just need to find which values does not satisfy the equation
[tex](-6)^2+3(-6)-6=12,[/tex]then the number A. IS on the graph
[tex](-4)^2+3(-4)-6=-2,[/tex]then the number B. IS on the graph
[tex](2)^2+3(2)-6=4,[/tex]then the number B. IS on the graph, and finally
[tex](3)^2+3(3)-6=12,[/tex]since this is diferent from -6, then (3,-6) is NOT on the graph. Then the answer is D. (3,-6)
The point-slope form is
[tex]y-a=m(x-a)[/tex]we find the slope using the formula
[tex]m=\frac{-1-3}{6-(-2)}=\frac{-4}{8}=-\frac{1}{2}[/tex]Replacing this, and the point (-2,3)
[tex]\begin{gathered} y-(-2)=-\frac{1}{2}(x-3) \\ y+2=-\frac{1}{2}x+\frac{3}{2} \\ 2y+2=-x+3 \\ x+2y=1 \end{gathered}[/tex]The equation of the line is x+2y=1
