Let's represent the function

$$
y^{2}=\frac{4 a^{2} x^{2}+b^{2}\left(x^{2}-1\right)^{2}}{\left(x^{2}+1\right)^{2}}
$$

if $a=3, b=2$ and the unit $=1 \mathrm{~cm}$.
For now, let's abstract from specific numerical data, and just assume that $a > b$.

We can simplify our discussion by making the following observations at the outset:

1. Our curve, due to the two opposite signs of the square root, consists of two branches symmetric with respect to the $x$-axis.
2. The curve is also symmetric with respect to the $y$-axis because only $x^2$ terms appear, and thus the $y$ value at $a + x$ is the same as at $-x$.

Therefore, it is sufficient to focus on positive $x$ and $y$ values. Let's transform the function

$$
y=+\left[\frac{4 a^{2} x^{2}+b^{2}\left(x^{2}-1\right)^{2}}{\left(x^{2}+1\right)^{2}}\right]^{\frac{1}{2}}
$$

as follows:

$$
\begin{gathered}
y=\left[\frac{4 a^{2} x^{2}+b^{2} x^{4}-2 b^{2} x^{2}+b^{2}}{\left(x^{2}+1\right)^{2}}\right]^{\frac{1}{2}}=\left[\frac{4 a^{2} x^{2}-4 b^{2} x^{2}+b^{2}\left(x^{2}+1\right)^{2}}{\left(x^{2}+1\right)^{2}}\right]^{\frac{1}{2}}= \\
=\left[\frac{4\left(a^{2}-b^{2}\right) x^{2}}{\left(x^{2}+1\right)^{2}}+b^{2}\right]^{\frac{1}{2}}=\left[\frac{4\left(a^{2}-b^{2}\right)}{\left(\frac{x^{2}+1}{x}\right)^{2}}+b^{2}\right]^{\frac{1}{2}}
\end{gathered}
$$

But

$$
\left(\frac{x^{2}+1}{x}\right)^{2}=\left(x+\frac{1}{x}\right)^{2}=x^{2}+2+\frac{1}{x^{2}}=4+x^{2}-2+\frac{1}{x^{2}}=4+\left(x-\frac{1}{x}\right)^{2}
$$

and thus our function can be written as:

$$
y=f(x)=\left[\frac{4\left(a^{2}-b^{2}\right)}{4+\left(x-\frac{1}{x}\right)^{2}}+b^{2}\right]^{\frac{1}{2}}
$$

From this form, it is immediately clear that the function depends solely on the denominator, and since the numerator is always positive according to our condition, the function will have a minimum or maximum where the denominator is the largest or smallest, respectively.

a) The denominator is the largest when either $x$ grows beyond all bounds, or $\frac{1}{x}$ grows beyond all bounds, i.e., $x$ becomes zero. In these two cases, $(x=0, \infty)$, the fraction inside the square brackets becomes 0, and thus

$$
y_{\min }=\left[b^{2}\right]^{\frac{1}{2}}=b
$$

Therefore, the curve touches the line $y=b$ from above at $x=0$, and if $x$ grows beyond all bounds $(x \rightarrow \infty)$, the curve approaches the line $y=b$ asymptotically from above (asymptote).

b) The denominator is the smallest when the always positive $\left(x-\frac{1}{x}\right)^{2}=0$, i.e., $x=\frac{1}{x}$, which means $x=1$ (considering only positive values according to our agreement).

Thus, $f(x)$ has a maximum at $x=1$, and

$$
y_{\max }=\left[\left(a^{2}-b^{2}\right)+b^{2}\right]^{\frac{1}{2}}=a
$$

Therefore, the curve touches the line $y=a$ from below at $x=1$.

Thus, as $x$ varies from 0 to 1, $y$ increases from $b$ to $a$, and as $x$ increases from 1 to beyond all bounds, $y$ decreases from $a$ to $b$ in such a way that

$$
f(x)=f\left(\frac{1}{x}\right)
$$

which we can verify directly by substitution:

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

Now, transitioning to the numerical data $a=3, b=2$,

$$
y=f(x)=\left[\frac{20}{4+\left(x-\frac{1}{x}\right)^{2}}+4\right]^{\frac{1}{2}}
$$

and calculating the values of $f(x)$ at some points, we obtain the following table of values:

| $\mathrm{x}$ | 0 | 0.1 | 0.2 | 0.4 | 0.5 | 0.6 | 0.8 | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | $\infty$ | 10 | 5 | 2.5 | 2 | 1.67 | 1.25 |  |
| $\mathrm{y}$ | 2 | 2.05 | 2.18 | 2.53 | 2.68 | 2.81 | 2.96 | 3 |

The complete curve is obtained by reflecting it across the $x$ and $y$ axes. (Figure 1)

![](https://cdn.mathpix.com/cropped/2024_05_02_2da8724094a81a5fc358g-2.jpg?height=526&width=1193&top_left_y=339&top_left_x=455)

A common error in many diagrams is that the curve does not touch the lines $y= \pm 2$ and $y= \pm 3$ at the extreme values.