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

Solution. Let $f(x)=a x^{2}+b x+c$. Since the quadratic trinomial takes its 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 $x^{2}$ to $f(x)$, we get the quadratic trinomial $(a+1) x^{2}+b x+c$, for which the minimum value is $-\frac{b^{2}}{4 a+4}+c$. 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

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

By dividing one equation of the last system by the other, we get $\frac{a-1}{a+1}=\frac{1}{3}$, from which $a=2$. Then $b^{2}=24$, and the minimum value of $f(x)$ is $-\frac{24}{8}+c=-3+c$. If we add $2 x^{2}$ to the quadratic trinomial $f(x)$, we get the function $(a+2) x^{2}+b x+c$, the minimum value of which is $-\frac{b^{2}}{4 a+8}+c=-\frac{24}{16}+c=-\frac{3}{2}+c$, which is $\frac{3}{2}$ more than the minimum of the original function.