# what are the vertical asymptotes of: x(x-7)/(x^3-49x)

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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When the graph of a function f(x) is drawn, vertical asymptotes are points where the value of f(x) goes to `+-oo` .

Here is the graph of `f(x) = (x(x-7))/(x^3-49x)`

The vertical asymptotes lie at the values of x for which the denominator of the function is equal to 0.

f(x) = (x(x-7))/(x^3-49x)

= (x(x-7))/(x*(x^2 - 49))

= (x(x-7))/(x*(x - 7)(x+7))

= 1/(x + 7)

The denominator of the function is 0 at x = -7

The vertical asymptote of the function lies at x = -7.

llltkl | College Teacher | (Level 3) Valedictorian

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Vertical asymptotes are vertical lines which correspond to the zeroes of the denominator of a rational function.

To find the vertical asymptotes of:` (x(x-7))/(x^3-49x)`

At first check the zeroes of the denominator.

`x^3-49x=0`

`rArr x(x^2-49)=0`

`rArr x(x-7)(x+7)=0`

Hence,` x=0,7,-7`

The vertical asymptote is at x=-7 but not at x=0 and x=7. The reason is that at x=0 and x=7, the numerator is also zero, and thus, there will be a hole at x=0 and x=7.

Therefore, the vertical asymptote is at x=-7 .

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