2. Find the minimum value of the expression $\frac{3 f(1)+6 f(0)-f(-1)}{f(0)-f(-2)}$, if $f(x)=a x^{2}+b x+c$ is an arbitrary quadratic function satisfying the condition $b>2 a$ and taking non-negative values for all real $x$.

(12 points)
Solution. We have $f(1)=a+b+c, \quad f(0)=c, \quad f(-1)=a-b+c, f(-2)=4 a-2 b+c$, $\frac{3 f(1)+6 f(0)-f(-1)}{f(0)-f(-2)}=\frac{3(a+b+c)+6 c-a+b-c}{c-4 a+2 b-c}=\frac{2 a+4 b+8 c}{2 b-4 a}=\frac{a+2 b+4 c}{b-2 a}$.

Since $f(x)=a x^{2}+b x+c \quad-$ is an arbitrary quadratic function that takes non-negative values for all real $x$, then $a>0, \quad D=b^{2}-4 a c \leq 0 \Rightarrow c \geq b^{2} / 4 a$.

Then $\quad \frac{a+2 b+4 c}{b-2 a} \geq \frac{a+2 b+b^{2} / a}{b-2 a}=\frac{a\left(1+2 b / a+b^{2} / a^{2}\right)}{a(b / a-2)}=\frac{t^{2}+2 t+1}{t-2}=\frac{(t+1)^{2}}{t-2}, \quad$ where $\quad t=b / a, \quad t>2$. Consider the function $g(t)=\frac{(t+1)^{2}}{t-2}$ and find its minimum value for $t>2$. $g^{\prime}(t)=\frac{2(t+1)(t-2)-(t+1)^{2}}{(t-2)^{2}}=\frac{(t+1)(2(t-2)-(t+1))}{(t-2)^{2}}=\frac{(t+1)(t-5)}{(t-2)^{2}}$, at $t=5$ the derivative $g^{\prime}(t)$ is 0 and, passing through this point, changes sign from "minus" to "plus", therefore, $t_{\min }=5, \quad g_{\min }=g(5)=12$.

Answer: 12.