3. Let $m \neq 0$. What is the minimum value of the function

$$
f(x)=\frac{1+m^{2}}{m^{2}} x^{2}-2 \cdot \frac{1+m^{2}}{m} x-1
$$

Write down the answer.
3. From the record of the quadratic function, we understand that the coefficients are $a=\frac{1+m^{2}}{m^{2}}, b=-2 \cdot \frac{1+m^{2}}{m}$, and $c=-1$. We determine that we need to calculate the ordinate of the vertex $q=-\frac{D}{4 a}$. The calculated ordinate is equal to $q=-2-m^{2}$.

The coefficients $a=\frac{1+m^{2}}{m^{2}}, b=-2 \cdot \frac{1+m^{2}}{m}$, and $c=-1 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_cd02ddf507985fa73630g-10.jpg?height=77&width=1642&top_left_y=641&top_left_x=207)

The calculated discriminant $D=4 \cdot \frac{1+m^{2}}{m^{2}} \cdot\left(2+m^{2}\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point

Using the formula for $q=-\frac{D}{4 a}=\frac{4 \cdot \frac{1+m^{2}}{m^{2}} \cdot\left(2+m^{2}\right)}{4 \cdot \frac{1+m^{2}}{m^{2}}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_cd02ddf507985fa73630g-10.jpg?height=51&width=1642&top_left_y=868&top_left_x=207)

Answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 point