How do you find the horizontal asymptote for #(x-3)/(x-2)#?

To find the horizontal asymptote for $\frac{x - 3}{x - 2}$:

1. Check the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$.
2. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote.
3. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

In this case, both the numerator and the denominator have the same degree (which is 1), so we proceed to step 2.

Divide the leading coefficient of the numerator by the leading coefficient of the denominator:

$\frac{1}{1} = 1$

Therefore, the horizontal asymptote for $\frac{x - 3}{x - 2}$ is $y = 1$.

To find the horizontal asymptote of the function $\frac{x-3}{x-2}$, you need to examine the behavior of the function as $x$ approaches positive or negative infinity.

Since the degree of the numerator and the denominator is the same (both are 1), the horizontal asymptote can be found by dividing the leading coefficients of the numerator and denominator.

In this case, the horizontal asymptote is given by the ratio of the leading coefficients, which is:

$\frac{1}{1} = 1$

Therefore, the horizontal asymptote of $\frac{x-3}{x-2}$ is $y = 1$.