Translating a reciprocal

Consider the rational function $y = \dfrac{1}{x - 1} + 2$ . It is the reciprocal $y = \dfrac{1}{x}$ translated by $1$ in the $x$ -direction and $2$ in the $y$ -direction: replace $x$ with $x - 1$ and add $2$ to the whole.

Everything moves together

Item	$y = \dfrac{1}{x}$	$y = \dfrac{1}{x-1} + 2$
Vertical asymptote	$x = 0$	$x = 1$
Horizontal asymptote	$y = 0$	$y = 2$
Centre of symmetry	$(0, 0)$	$(1, 2)$
Domain	$x \neq 0$	$x \neq 1$
Range	$y \neq 0$	$y \neq 2$

The forbidden values are exactly the asymptotes. The original reciprocal had point symmetry about the origin, so after the translation the image of the origin takes over that role.

Monotonicity

The derivative is $y' = -\dfrac{1}{(x - 1)^2}$ , negative everywhere on the domain, so the function decreases on $x < 1$ and decreases on $x > 1$ . As with $\dfrac{1}{x}$ , the two branches cannot be compared across the asymptote.

The same shape in disguise

This shape hides inside expressions that look quite different. Take $y = \dfrac{2x - 1}{x - 1}$ and divide the numerator by the denominator.

\frac{2x - 1}{x - 1} = \frac{2(x - 1) + 1}{x - 1} = 2 + \frac{1}{x - 1}

That is precisely our function. Any ratio of linear expressions $y = \dfrac{ax + b}{cx + d}$ can be reduced by this division to a translated, vertically scaled reciprocal. That is why the graphs of such functions all look like the same hyperbola in different places.

Checking the point symmetry

Point	Value	Departure from $2$
$x = 1 + t$	$2 + \dfrac{1}{t}$	$+\dfrac{1}{t}$
$x = 1 - t$	$2 - \dfrac{1}{t}$	$-\dfrac{1}{t}$

The departures from the centre's height are equal and opposite, so the two points sit diametrically across $(1, 2)$ : the graph is symmetric about that point.

The graph draws the horizontal asymptote as a line, and the large dot marks the centre; near the vertical asymptote the graph shoots up steeply.