1. When $3 x^{2}$ was added to the quadratic trinomial $f(x)$, its minimum value increased by 9, and when $x^{2}$ was subtracted from it, its minimum value decreased by 9. How will the minimum value of $f(x)$ change if $x^{2}$ is added to it?
Answer. It will increase by $\frac{9}{2}$.

Solution. Let $f(x)=a x^{2}+b x+c$. Since the quadratic trinomial takes the minimum value, its leading coefficient is positive, and the minimum value is reached at the vertex of the parabola, i.e., at the point $x_{0}=-\frac{b}{2 a}$. This value is $f\left(x_{0}\right)=a \cdot \frac{b^{2}}{4 a^{2}}-\frac{b^{2}}{2 a}+c=-\frac{b^{2}}{4 a}+c$.

If we add $3 x^{2}$ to $f(x)$, we get the quadratic trinomial $(a+3) x^{2}+b x+c$, for which the minimum value is $-\frac{b^{2}}{4 a+12}+c$. If we subtract $x^{2}$ from $f(x)$, we get the quadratic trinomial $(a-1) x^{2}+b x+c$, for which the minimum value is $-\frac{b^{2}}{4 a-4}+c$. From this, we get the equations

$$
\left\{\begin{array}{l}
-\frac{b^{2}}{4 a+12}+c=-\frac{b^{2}}{4 a}+c+9, \\
-\frac{b^{2}}{4 a-4}+c=-\frac{b^{2}}{4 a}+c-9
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{b^{2}}{4 a}-\frac{b^{2}}{4 a+12}=9, \\
\frac{b^{2}}{4 a-4}-\frac{b^{2}}{4 a}=9
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{3 b^{2}}{4 a(a+3)}=9 \\
\frac{b^{2}}{4 a(a-1)}=9
\end{array}\right.\right.\right.
$$

Dividing one equation of the last system by the other, we get $\frac{3(a-1)}{a+3}=1$, from which $a=3$. Then $b^{2}=216$, and the minimum value of $f(x)$ is $-\frac{216}{12}+c=-18+c$. If we add $x^{2}$ to the quadratic trinomial $f(x)$, we get the function $(a+1) x^{2}+b x+c$, the minimum value of which is $-\frac{b^{2}}{4 a+4}+c=-\frac{216}{16}+c=-\frac{27}{2}+c$, which is $\frac{9}{2}$ more than the minimum of the original function.