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

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

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Answer 1 · 2018-02-24T03:32:27

$x=1/3$ is the vertical asymptote.

$y=2/3$ is the horizontal asymptote.

$x=0$ is the x-intercept.

$y=0$ is the y intercept.

Explanation:

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

So here:

$3x-1=0$

$3x=1$

$x=1/3$ 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 $(2)$ and the denominator ( $3$ ).

$y=2/3$ is the horizontal asymptote.

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

$(2x)/(3x-1)=0$

$2x=0$

$x=0$ is the x-intercept.

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

$(2*0)/(3*0+1)$

$0/1$

$y=0$ is the y intercept.

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

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Explanation:

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