1. When $x^{2}$ was added to the quadratic trinomial $f(x)$, its maximum value increased by $\frac{27}{2}$, and when $4 x^{2}$ was subtracted from it, its maximum value decreased by 9. How will the maximum value of $f(x)$ change if $2 x^{2}$ is subtracted from it?
Answer. It will decrease by $\frac{27}{4}$.

Solution. Let $f(x)=a x^{2}+b x+c$. Since the quadratic trinomial takes its maximum value, its leading coefficient is negative, and the maximum 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 $x^{2}$ to $f(x)$, we get the quadratic trinomial $(a+1) x^{2}+b x+c$, for which the maximum value is $-\frac{b^{2}}{4 a+4}+c$. If we subtract $4 x^{2}$ from $f(x)$, we get the quadratic trinomial $(a-4) x^{2}+b x+c$, for which the maximum value is $-\frac{b^{2}}{4 a-16}+c$. From this, we get the equations

$$
\left\{\begin{array} { l } 
{ - \frac { b ^ { 2 } } { 4 a + 4 } + c = - \frac { b ^ { 2 } } { 4 a } + c + \frac { 2 7 } { 2 } , } \\
{ - \frac { b ^ { 2 } } { 4 a - 1 6 } + c = - \frac { b ^ { 2 } } { 4 a } + c - 9 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \frac { b ^ { 2 } } { 4 a } - \frac { b ^ { 2 } } { 4 a + 4 } = \frac { 2 7 } { 2 } , } \\
{ \frac { b ^ { 2 } } { 4 a - 1 6 } - \frac { b ^ { 2 } } { 4 a } = 9 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{b^{2}}{4 a(a+1)}=\frac{27}{2} \\
\frac{b^{2}}{a(a-4)}=9
\end{array}\right.\right.\right.
$$

Dividing one equation of the last system by the other, we get $\frac{a-4}{4(a+1)}=\frac{3}{2}$, from which $a=-2$. Then $b^{2}=108$, and the maximum value of $f(x)$ is $-\frac{108}{-8}+c=\frac{27}{2}+c$. If we subtract $2 x^{2}$ from the quadratic trinomial $f(x)$, we get the function $(a-2) x^{2}+b x+c$, the maximum value of which is $-\frac{b^{2}}{4 a-8}+c=-\frac{108}{-16}+c=\frac{27}{4}+c$, which is $\frac{27}{4}$ less than the maximum of the original function.