# How do you graph f(x)=(2x)/(3x-1) using holes, vertical and horizontal asymptotes, x and y intercepts?

Feb 24, 2018

$x = \frac{1}{3}$ is the vertical asymptote.

$y = \frac{2}{3}$ is the horizontal asymptote.

$x = 0$ is the x-intercept.

$y = 0$ is the y intercept.

#### Explanation:

The vertical asymptote is found when $f \left(x\right)$ tends to infinity. $f \left(x\right)$ normally tends to infinity when the denominator tends to $0$.

So here:

$3 x - 1 = 0$

$3 x = 1$

$x = \frac{1}{3}$ is the vertical asymptote.

For the horizontal asymptote, we use the degrees of the numerator and the denominator. Say $m$ is the former and $n$ the latter. If:

$m > n$, then there is no horizontal asymptote, only a slant.
$m = n$, the horizontal asymptote is at the quotient of the leading coefficient of the numerator and denominator
$m <$$n$, the asymptote is at $y = 0$.

Here, $m = 1$ and $n = 1$. So $m = n$.

We must divide the leading coefficients of the numerator $\left(2\right)$ and the denominator ($3$).

$y = \frac{2}{3}$ is the horizontal asymptote.

The x-intercept is found when $f \left(x\right) = 0$. Here,

$\frac{2 x}{3 x - 1} = 0$

$2 x = 0$

$x = 0$ is the x-intercept.

The y-intercept is the answer to $f \left(0\right)$. Inputting:

$\frac{2 \cdot 0}{3 \cdot 0 + 1}$

$\frac{0}{1}$

$y = 0$ is the y intercept.