Consider the following function:

$$
y=\frac{2 x^{2}-5 a x+8}{x-2 a}
$$

$1^{0}$. For which values of $a$ will this function be strictly increasing?

$2^{0}$. For which values of $a$ will this function have a maximum and a minimum?

$3^{0}$. The values of $a$ that separate the conditions in $1^{0}$. and $2^{0}$. are certain $a_{1}$ and $a_{2}$. What happens to the function for these values of $a$?

$4^{0}$. Imagine all the curves $C$ that correspond to the function for all possible values of $a$, plotted on the same coordinate system. What geometric locus is described by the intersection points of the asymptotes of these $C$ curves? Show that all $C$ curves pass through two fixed points!
$1^{0}$. A function is constantly increasing if its derivative is positive for all values of $x$.

$$
y^{\prime}=\frac{(4 x-5 a)(x-2 a)-\left(2 x^{2}-5 a x+8\right)}{(x-2 a)^{2}}=\frac{2}{(x-2 a)^{2}}\left[(x-2 a)^{2}+a^{2}-4\right]
$$

The denominator of $y^{\prime}$ is positive; therefore, its sign depends only on the numerator, which is the quadratic function

$$
f(x) \equiv(x-2 a)^{2}+a^{2}-4
$$

As we can see, $f(x)$ is positive for all values of $x$ if $a^{2}-4>0$, that is,

$$
\text { if } a+2
$$

In these cases, the equation $f(x)=0$ has no real roots. The function $y$ is constantly increasing. There is a discontinuity at $x=2 a$ where $y= \pm \infty$.

$2^{0}$. If $-2<a<+2$, then the quadratic function $f(x)$ changes sign at the points where $f(x)=0$. If these points are $x^{\prime}$ and $x^{\prime \prime}$, with $x^{\prime}<x^{\prime \prime}$, then $f(x)$ changes from positive to negative values at $x^{\prime}$, so $y$ has a maximum there.

At $x^{\prime \prime}$, $f(x)$ changes from negative to positive values, so $y$ has a minimum there.

$3^{0}$. From the above, we see that $a_{1}=-2$ and $a_{2}=+2$ are the values that separate the $a$ values in case $1^{\circ}$ from those in case $2^{\circ}$. If $a_{1}=-2$ and $x+4 \neq 0$, then

$$
y=\frac{2 x^{2}+10 x+8}{x+4}=\frac{2(x+4)(x+1)}{x+4}=2(x+1)
$$

If $a=+2$ and $x-4 \neq 0$, then $y=\frac{2 x^{2}-10 x+8}{x-4}=\frac{2(x-4)(x-1)}{x-4}=2(x-1)^{2}$

$4^{0}$. It is obvious that the function $y$ corresponds to a hyperbola. If we perform the division,

$$
y=2 x-a+\frac{8-2 a^{2}}{x-2 a}
$$

From this, we can see that the lines

$$
x=2 a \quad \text { and } \quad y=2 x-a
$$

are the asymptotes of the curve $C$. The coordinates of the intersection point of these two lines, which is the center of the curve, are:

$$
x=2 a, \quad y=3 a
$$

By eliminating $a$, we get the relationship between the coordinates of the center:

$$
y=\frac{3 x}{2}
$$

This is the equation of a line that passes through the origin. This line describes the center of the hyperbola $C$ as $a$ takes on all possible values.

$5^{0}$. The equation of the curve $C$, after removing the denominator, can be written as

$$
2 x^{2}-x y+8=a(5 x-2 y)
$$

If $5 x-2 y=0$, then for all values of $a$,

$$
2 x^{2}-\frac{5}{2} x \cdot x+8 \equiv 8-\frac{x^{2}}{2}=0, \quad \text { or } \quad x= \pm 4 \quad \text { and thus } \quad y= \pm \frac{5}{2} \cdot 4= \pm 10
$$

Therefore, all curves pass through the

$$
P_{1}(4,10), \quad P_{2}(-4,-10)
$$

solid points.

Freud Géza (Berzsenyi Dániel g. VI. o. Bp. V.)

Note. If, for example, $a=-2$, then we are dealing with a pair of lines $x=-4, y=2 x+2$. The line $x=-4$ passes through the point $(x=-4, y=-10)$, and the line $y=2 x+2$ passes through the point $(x=4, y=10)$.[^0]


[^0]:    ${ }^{1}$ Indeed, $f(x)$ is a quadratic function of $x$ where the coefficient of $x^{2}$ is positive.

    ${ }^{2} y=\frac{2 x^{2}-5 a x+8}{x-2 a}$ can be written as $2 x^{2}-x y-5 a x+2 a y+8=0$. If, for example, $a=-2$, then $2 x^{2}+(10-y) x+8-4 y \equiv(x+4)(2 x+2-y)=0$. The two lines of the pair are: $x+4=0$ and $y=2 x+2$.