# How do you find any asymptotes of #g(x)=(2x-3)/(x^2-6x+9)#?

The vertical asymptote is

There is no oblique asymptote

graph{(2x-3)/(x-3)^2 [-10, 10, -5, 5]}

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To find the asymptotes of the function g(x) = (2x-3)/(x^2-6x+9), we need to consider the behavior of the function as x approaches positive or negative infinity.

First, we check for vertical asymptotes by finding the values of x that make the denominator equal to zero. In this case, the denominator x^2-6x+9 can be factored as (x-3)^2. Therefore, the function has a vertical asymptote at x = 3.

Next, we examine the horizontal asymptote. To determine this, we compare the degrees of the numerator and denominator. Since the degree of the numerator (1) is less than the degree of the denominator (2), the horizontal asymptote is y = 0.

In summary, the function g(x) = (2x-3)/(x^2-6x+9) has a vertical asymptote at x = 3 and a horizontal asymptote at y = 0.

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