Concept:Since g(x) is quadratic and the composite 4g(f(x)) is also quadratic, f(x) must be linear.Explanation:Let f(x)=ax+b.Given f(0)=−3, so b=−3. Hence f(x)=ax−3.Substitute f(x) into g(x):g(f(x))=3(ax−3)2+2(ax−3)−3.Then 4g(f(x))=4[3(ax−3)2+2(ax−3)−3]=3x2−32x+72.Expand the left side:3(ax−3)2=3(a2x2−6ax+9)=3a2x2−18ax+27.Add 2(ax−3)=2ax−6 and subtract 3, giving 3a2x2−16ax+18.Multiply by 4: 12a2x2−64ax+72=3x2−32x+72.Compare coefficients:For x2: 12a2=3⇒a2=41⇒a=±21.For x: −64a=−32⇒a=21. Hence a=21.Thus f(x)=21x−3.Now compute g(2)=3(2)2+2(2)−3=12+4−3=13.Then f(g(2))=f(13)=21(13)−3=6.5−3=3.5=27.Answer:27 which corresponds to option A.