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How do you find the asymptotes for $f (x) = \frac{x}{3 x (x - 1)}$ $f(x) = x / (3x(x-1))$ ?

1 Answer

Jan 4, 2016

Vertical: $x = 1$ $x=1$
Horizontal: $y = 0$ $y=0$

Explanation:

First, notice that the $x$ $x$ terms will cancel, leaving the function as

$f (x) = \frac{1}{3 (x - 1)}, \times \times x \neq 0$ $f(x)=1/(3(x-1)),color(white)(xxxx)x!=0$

However, canceling the $x$ $x$ terms leaves a hole, or a removable discontinuity, at $x = 0$ $x=0$ .

Vertical asymptotes:

Vertical asymptotes will occur when the denominator equals $0$ $0$ .

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

Solved, this gives

$x = 1$ $x=1$

Thus the vertical asymptote occurs at $x = 1$ $x=1$ .

Even though $3 x$ $3x$ is in the denominator in the original function, its cancellation makes it just a hole and not also a vertical asymptote.

Horizontal asymptotes:

Since the degree of the denominator is larger than the degree of the numerator, the horizontal asymptote is the line $y = 0$ $y=0$ .

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

Don't be fooled—there is a hole at $x = 0$ $x=0$ , despite appearances.

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