Show all work to identify the asymptotes and zero of the function f(x) = 3x/x^2-9

Obviously, −3 and 3 would be set to equal [tex]x[/tex], knowing that they are the roots of 9. Or, you can choose to factor. The divisor of [tex]{x}^{2} - 9[/tex] is a product of two binomials:

[tex][x - 3][x + 3][/tex]

Then, knowing that you set both factors equal to zero, −3 and 3 would be your zeros, which are also two non-removable discontinuities, or vertical asymptotes because you cannot have zero in the denominator.

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ahirohit963 ahirohit963 · Answer 2 · 2021-11-13T07:26:20+01:00

The vertical asymptotes are +3 and -3 and the horizontal asymptote is 0 and this can be determined by putting the denominator equal to zero.

Given :

Function - [tex]f(x) = \dfrac{3x }{x^2 -9}[/tex]

The following steps can be used to determine the asymptotes and zero of the given function.

Step 1 - Determine the vertical asymptotes of the given function.

[tex]x^2-9=0[/tex]

(x-3)(x+3) = 0

x = +3 , -3

Step 2 - Determine the horizontal asymptotes of the given function.

[tex]y = \lim_{n \to \infty} \dfrac{3x}{x^2 -9}[/tex]

y = 0

The vertical asymptotes are +3 and -3 and the horizontal asymptote is 0.

For more information, refer to the link given below:

https://brainly.com/question/4084552