# How do you graph f(x)=2/(x-1) using holes, vertical and horizontal asymptotes, x and y intercepts?

Jul 29, 2018
• $x = 1$ the vertical asymptote of $f \left(x\right)$
• $y = 0$ the horizontal asymptote of $f \left(x\right)$
• $\left(0 , 2\right)$ the $y$-intercept
• $f \left(x\right)$ has no $x$-intercept

#### Explanation:

$f \left(0\right)$ gives the value of the $y$-intercept:

$f \left(0\right) = \frac{2}{- 1} = - 2$

$f \left(x\right)$ has no intersection with the $x$ axis; solving $f \left(x\right) = 0$ for $x$ yields no real result:

$f \left(x\right) = \frac{2}{x - 1} = 0$
$2 \ne 0 \cdot \left(x - 1\right)$

The graph of $f \left(x\right) = \frac{2}{x - 1}$ closely relates to that of $f \left(x\right) = \frac{1}{x}$.
$y = \frac{1}{x}$ has the following features:

• $x = 0$ a vertical asymptote
• $y = 0$ a horizontal asymptote
• $y = 1 / x$ contains two separate branches and is decreasing on both intervals.

Translating the graph of $y = \frac{1}{x}$ to the right by $1$ unit (and stretching the result vertically by a factor of two) gives the graph of $y = f \left(x\right) = \frac{2}{x - 1}$. The two asymptotes are translated in the same fashion:

• $x = 0 + 1$ the vertical asymptote of $f \left(x\right)$
• $y = 0 \times 2 = 0$ the horizontal asymptote of $f \left(x\right)$

The branched feature from $y = \frac{1}{x}$ is translated and shall also present in $y = f \left(x\right)$. Mark all the three graphical features [horizontal/vertical asymptote, intercept(s)] on the Cartesian plane and chart the two branches:

graph{2/(1-x) [-10, 10, -5, 5]}