Concept:A quadratic function with a positive coefficient of x2 has a minimum value at its turning point.Explanation:Given f(x)=3x2+18x+32.Differentiate with respect to x: f′(x)=6x+18.At the turning point, f′(x)=0, so 6x+18=0.Solving gives x=−3.Also, f′′(x)=6>0, so this turning point is a minimum.Substitute x=−3 into f(x): f(−3)=3(−3)2+18(−3)+32.This gives 27−54+32=5.Answer:The least value of the function is 5, so the correct option is C.