jmc

Let $n$ be a positive integer. Setting $x=0,$ $1,$ $2,$ $\dots ,$ $n-1,$ we get $$\begin{array}{rl}f(1)-f(0)& =6\cdot 0+4,\\ f(2)-f(1)& =6\cdot 1+4,\\ f(3)-f(2)& =6\cdot 2+4,\\ & \dots ,\\ f(n)-f(n-1)& =6\cdot (n-1)+4.\end{array}$$Adding all the equations, we get $$f(n)-f(0)=6(0+1+2+\cdots +(n-1))+4n=6\cdot \frac{n(n-1)}{2}+4n=3n^2+n.$$Since this holds for all positive integers $n,$ $$f(x)=3x^2+x+c$$for some constant $c.$ Hence, the leading coefficient of $f(x)$ is $\boxed{3}.$