Let's examine $y$ as a function of $x$, given that the following relationship holds:

$$
y=2 x-1-\sqrt{y^{2}-2 x y+3 x-2} .
$$

Plot the function.
In the given expression, $y$ appears in multiple places (implicitly defining the dependence of $y$ on $x$). To express $y$ (giving its explicit dependence on $x$), transform the relationship:

$$
\sqrt{y^{2}-2 x y+3 x-2}=2 x-y-1
$$

The left side is non-negative, so the right side must also be non-negative, i.e.,

$$
2 x-y-1 \geqq 0, \quad \text { which implies } \quad y \leqq 2 x-1
$$

From (1), squaring both sides and then collecting the terms involving $y$ on the left side:

$$
\begin{gathered}
2 x y-2 y=4 x^{2}-7 x+3, \quad \text { hence } \\
2 y(x-1)=4(x-1)(x-3 / 4)=(x-1)(4 x-3) \\
\text { finally }(x-1)(2 y-4 x+3)=0
\end{gathered}
$$

Accordingly, if $x-1 \neq 0$, i.e., $x \neq 1$, then

$$
2 y-4 x+3=0, \quad y=2 x-1.5
$$

If, however, $x=1$, then (4) holds for all $y$ allowed by (2). Thus, at $x=1$, the function value is not uniquely determined: it can take infinitely many values. Since (5) always satisfies the (2) requirement, our function only differs from the function $y=2 x-1.5$ in that it is undefined at $x=1$. Therefore, its graph is the line $y=2 x-1.5$, excluding the point $(x=1, y=0.5)$ (this is indicated by the empty circle on the graph).

![](https://cdn.mathpix.com/cropped/2024_05_02_ec7e6d56fc980168da31g-1.jpg?height=585&width=466&top_left_y=1109&top_left_x=816)

Huber Tibor (Budapest, Kossuth L. mech. eng. tech. IV. grade)

Remarks. 1. The product form (4) can also be obtained by reducing equation (3) to 0 and solving for $x$:

$$
4 x^{2}-(7+2 y) x+(2 y+3)=0
$$

the discriminant:

$$
\begin{gathered}
49+28 y+4 y^{2}-32 y-48=4 y^{2}-4 y+1=(2 y-1)^{2}, \quad \text { thus } \\
x=\frac{1}{8}[7+2 y \pm(2 y-1)], \quad x_{1}=1, \quad x_{2}=0.5 y+0.75
\end{gathered}
$$

2. Several solutions add the half-line of the line $x=1$ that satisfies the condition $y \leqq 1$ from (2) to the graph of the function. This corresponds to accepting that the function assigns infinitely many values to $x=1$, or, in other words, if we are looking for the geometric figure whose points satisfy the given equation, or if geometry is the primary focus and the numerical relationships serve as auxiliary tools. It is precisely in this prioritization of roles that coordinate geometry and function graphing differ.