How do you identify all asymptotes or holes and intercepts for $f (x) = \frac{1}{{(x - 2)}^{2}}$ $f(x)=1/(x-2)^2$ ?

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How do you identify all asymptotes or holes and intercepts for $f (x) = \frac{1}{{(x - 2)}^{2}}$ $f(x)=1/(x-2)^2$ ?

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Answer 1 · 2017-12-16T11:07:12

The $y$ $y$ -intercept is $(0, \frac{1}{4})$ $(0,1/4)$ ; $y = 0$ $y=0$ is the horizontal asymptote; $x = 2$ $x=2$ is the vertical asymptote.

Explanation:

When $x = 0$ $x=0$ we have $f (0) = \frac{1}{{(0 - 2)}^{2}} = \frac{1}{4}$ $f(0)=1/(0-2)^2=1/4$ , so the $y$ $y$ -intercept is $(0, \frac{1}{4})$ $(0,1/4)$ .

The equation $f (x) = 0$ $f(x)=0$ has no solutions so there are no $x$ $x$ -intercepts.

Since the degree of the numerator is less than the degree of the denominator we know that $y = 0$ $y=0$ is the horizontal asymptote. (Or you could justify it by saying $lim_(x\to oo)f(x)=0$ if you've gotten to that concept yet.)

There are no holes because there are no common factors in the numerator and denominator.

$x=2$ is the vertical asymptote because it is the zero of the denominator. (Or you could use limits to justify that as well.)

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How do you identify all asymptotes or holes and intercepts for $f (x) = \frac{1}{{(x - 2)}^{2}}$ $f(x)=1/(x-2)^2$ ?

1 Answer

Explanation:

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How do you identify all asymptotes or holes and intercepts for f(x)=1/(x-2)^2f(x)=1(x−2)2f(x)=1/(x-2)^2?

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How do you identify all asymptotes or holes and intercepts for $f (x) = \frac{1}{{(x - 2)}^{2}}$ $f(x)=1/(x-2)^2$ ?