Question 1206376
You can start with an expression  {{{a(x+4)(x+2)=y}}}, and find "a" value using the "passes through" given point.


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

{{{a(3)(1)=3}}}

{{{a=1}}}


With that you have {{{y=(x+4)(x+2)}}};
doing the multiplication,
{{{x^2+4x+2x+8}}}

{{{x^2+6x+8}}}-----------------choice  C.