1. When $2 x^{2}$ was added to the quadratic trinomial $f(x)$, its maximum value increased by 10, and when $5 x^{2}$ was subtracted from it, its maximum value decreased by $\frac{15}{2}$. How will the maximum value of $f(x)$ change if $3 x^{2}$ is added to it?
Answer. It will increase by $\frac{45}{2}$.

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 $2 x^{2}$ to $f(x)$, we get the quadratic trinomial $(a+2) x^{2}+b x+c$, for which the maximum value is $-\frac{b^{2}}{4 a+8}+c$. If we subtract $5 x^{2}$ from $f(x)$, we get the quadratic trinomial $(a-5) x^{2}+b x+c$, for which the maximum value is $-\frac{b^{2}}{4 a-20}+c$. From this, we get the equations

$$
\left\{\begin{array} { l } 
{ - \frac { b ^ { 2 } } { 4 a + 8 } + c = - \frac { b ^ { 2 } } { 4 a } + c + 1 0 , } \\
{ - \frac { b ^ { 2 } } { 4 a - 2 0 } + c = - \frac { b ^ { 2 } } { 4 a } + c - \frac { 1 5 } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ \frac { b ^ { 2 } } { 4 a } - \frac { b ^ { 2 } } { 4 a + 8 } = 1 0 , } \\
{ \frac { b ^ { 2 } } { 4 a - 2 0 } - \frac { b ^ { 2 } } { 4 a } = \frac { 1 5 } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{2 b^{2}}{4 a(a+2)}=10 \\
\frac{5 b^{2}}{4 a(a-5)}=\frac{15}{2}
\end{array}\right.\right.\right.
$$

By dividing one equation of the last system by the other, we get $\frac{2(a-5)}{5(a+2)}=\frac{20}{15}$, from which $a=-5$. Then $b^{2}=300$, and the maximum value of $f(x)$ is $-\frac{300}{-20}+c=15+c$. If we add $3 x^{2}$ to the quadratic trinomial $f(x)$, we get the function $(a+3) x^{2}+b x+c$, the maximum value of which is $-\frac{b^{2}}{4 a+12}+c=-\frac{300}{-8}+c=\frac{75}{2}+c$, which is $\frac{45}{2}$ more than the maximum of the original function.