# How do you graph y=(6x-1)/(3x-1) using asymptotes, intercepts, end behavior?

Nov 30, 2016

graph{(6x-1)/(3x-1) [-10, 10, -5, 5]}

#### Explanation:

The domain of

$y \left(x\right) = \frac{6 x - 1}{3 x - 1}$

is all $\mathbb{R}$ except the point $x = \frac{1}{3}$ where the denominator vanishes.

Look at the behavior of the function at the limits of the intervals of definition:

${\lim}_{x \to - \infty} y \left(x\right) = 2$

${\lim}_{x \to + \infty} y \left(x\right) = 2$

The limits are finite, so $y \left(x\right)$ has an horizontal asymptote $y = 2$ on both sides.

${\lim}_{x \to {\left(\frac{1}{3}\right)}^{-}} y \left(x\right) = - \infty$

${\lim}_{x \to {\left(\frac{1}{3}\right)}^{+}} y \left(x\right) = + \infty$

There is a vertical asymptote at $x = \frac{1}{3}$

The intercepts with the axis can be found as:

$y \left(x = 0\right) = 1$

$0 = \frac{6 x - 1}{3 x - 1}$ at $x = \frac{1}{6}$

The derivative is:

$y ' \left(x\right) = \frac{6 \left(3 x - 1\right) - 3 \left(6 x - 1\right)}{{\left(3 x - 1\right)}^{2}} = - \frac{3}{3 x - 1} ^ 2$

and is always negative, so the function is strictly decreasing in the whole domain.

Summing it up, as $x$ goes from $- \infty$ to $+ \infty$, $y \left(x\right)$ starts from below the horizontal line $y = 2$, decreases constantly crossing the $y$ axis at $y = 1$ and the $x$ axis at $x = \frac{1}{6}$, then goes negative without bound until the vertical line $x = \frac{1}{3}$.
To the right of this line, $y \left(x\right)$ starts from $+ \infty$ and decreases approaching to the horizontal line $y = 2$ as $x$ grows.