Let's examine the variation of the following function as $x$ changes from $-\infty$ to $+\infty$:

$$
y=x^{4}-3 x^{2}+2
$$

Draw the corresponding curve.
The curve is symmetric with respect to the ordinate axis because for every value of $y$, there are two values of $x$ that differ only in sign. Since the function can also be written as

$$
y=\left(x^{2}-1\right)\left(x^{2}-2\right)
$$

we see that $y= \pm \infty$ if $x= \pm \infty$, and if $y=0$, then the values of $x$ are: $\pm 1, \pm \sqrt{2}$.

To determine the minimum of the function, we express $x$ in terms of $y$:

$$
x= \pm \sqrt{\frac{3}{2} \pm \sqrt{\frac{1}{4}+y}}
$$

we see that the minimum of $y$ is $-\frac{1}{4}$, which occurs when $x= \pm \sqrt{\frac{3}{2}}$.

To draw the curve, let's determine a few more values of $y$; the corresponding values of the variables are as follows:

| $x$ | $-\infty$, | $\ldots-2$, | $-\sqrt{2}$, | $-\sqrt{\frac{3}{2}}$, | -1, | $-\frac{2}{3}$, | $-\frac{1}{2}$, | $-\frac{1}{3}$, | 0, | $\frac{1}{3}$, | $\ldots+\infty$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $y$ | $+\infty$, | $\ldots 6$, | 0, | $-\frac{1}{4}$, | 0, | $\frac{70}{81}$, | $1 \frac{5}{16}$, | $1 \frac{55}{81}$, | 2, | $1 \frac{55}{81}$, | $\ldots+\infty$ |

The curve thus decreases from $+\infty$ to $-\frac{1}{4}$, then increases to 2 when $x=0$, and then decreases again to $-\frac{1}{4}$ and increases to $+\infty$. The curve has no asymptotes.

![](https://cdn.mathpix.com/cropped/2024_05_02_661fd3eededc58d375d9g-1.jpg?height=456&width=374&top_left_y=1103&top_left_x=865)

(Jenő Filkorn, Nyitra.)

The problem was also solved by: Bayer B., Czank K., Dányi I., Faith F., Keesz J., Kerekes T., König D., Krausz B., Gy. Krisztián, A. Kürth, Gy. Lukhaub, Gy. Lupsa, Gy. Póka, G. Sasvári, J. Scharff, r. Scheuer, K. Smodics, F. Stromfeld, J. Szöllősy.