Step-by-step explanation:
There is a vertical asymptote x=3 and x=-1 so our rational function will contain
(x-3)(x+1).
Since we have a horinzintal asymptote, our numerator will be either a constant function, or linear function.
we have a zero at approximately, at 1 so our numerator will include
[tex] \frac{x - 1}[/tex]
So our rational function is
[tex] \frac{x - 1}{(x - 3)(x + 1)} [/tex]
There is a vertical asymptote x=3 and x=-1 so our rational function will contain option C [tex]f(x) = \dfrac{x-1}{(x-3)(x+1)}[/tex].
The line y = a is a horizontal asymptote if the function f(x) tends to 'a' from the upside of that line y = a, or from downside of that line.
Since we have a horizontal asymptote, our numerator will be either a constant function or a linear function.
We have given a zero at approximately, 1 so our numerator will include (x-1).
So our rational function is
[tex]f(x) = \dfrac{x-1}{(x-3)(x+1)}[/tex]
Learn more about horizontal asymptotes here:
https://brainly.com/question/2513623
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