What are the asymptotes of: x(x-7)/(x^3-49x)

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embizze | High School Teacher | (Level 1) Educator Emeritus

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There are three types of asymptotes; horizontal,vertical, and slant.

(1) Vertical asymptotes occur in rational functions when the rational function, written with no common factors in the numerator and denominator, has a factor in the denominator that is zero.

Factoring the expression we get:

`(x(x-7))/(x^3-49x)=(x(x-7))/(x(x+7)(x-7))`

Note the common factor of x-7 -- the function will not be defined at x=7, but there will not be a vertical asymptote there. Most graphing utilities will not even show the "hole" that occurs when x=7.

Also, there is a common factor of x in both numerator and denominator. Again, there will not be a vertical asymptote at x=0.

There will be a vertical asymptote at x=-7.

(2) Horizontal asymptotes occur in rational functions if:

(a) The degree of the numerator is less than the degree of the denominator. Then the horizontal asymptote is y=0.

(b) If the degree of the denominator is equal to the degree of the numerator then the horizontal asymptote occurs at `y=a/b` where a is the leading coefficient of the numerator and b is the leading coefficient of the denominator.

(c) If the degree of the numerator is greater than the degree of the denominator there is no horizontal asymptote, but there may be a slant asymptote.

Here the degree of the numerator is smaller than the degree of the denominator so the horizontal asymptote is y=0.

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There is a vertical asymptote at x=-7. (The function is undefined when x=0 or x=7.)

There is a horizontal asymptote at y=0.

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The graph:

** Note that the function is not defined at x=0 or x=7, despite the apparent value in the graph. There are "holes" in the graph at those points.

Sources:
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Zaca | Student, Undergraduate | (Level 1) Salutatorian

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The two types of asymptotes are vertical and horizontal.

Vertical asymptotes happen when there is an invalid `x` value (most likely causing the denominator to be zero).

Horizontal asymptotes happen when a certain `y` value cannot be reached.

First, let's simplify the function:
`(x(x-7))/(x^3-49)`

Factor out an x from the denominator and factor the difference of squares:
`(x(x-7))/(x(x+7)(x-7))`

To find vertical asymptotes, we need to find out what values would make the denominator zero.

By setting `x` , `(x+7)` , and `(x-7)`  to zero, we find that these x values are `0, -7, 7`.

To find horizontal asymptotes, let's simplify the function again.

We can cancel out `x` and `(x-7)` , resulting in:
`1/(x+7)`

This is just a translation of the inverse function `1/x` , a function which has a horizontal asymptote of zero.

So the horizontal asymptote is `y=0`

Therefore, the vertical asymptotes of this function are:
`x=0`
`x=7`
` `  `x=-7`
And the horizontal asymptote is:
`y=0`

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