Find the asymptotes of the function: 3+ (2x/x-3). Show the steps.

Sure, here are the steps to find the asymptotes of the function: 3+ (2x/x-3).

1. First, we need to identify the vertical and horizontal asymptotes.

2. Vertical asymptotes are found by setting the denominator equal to zero and solving for x. So, we set x-3 = 0. Solving for x gives us x = 3. Therefore, x = 3 is the vertical asymptote.

3. Horizontal asymptotes are found by looking at the degree (the highest exponent) of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the x-axis (y = 0) is the horizontal asymptote. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

4. In this case, the degree of the numerator (2x, which is degree 1) is equal to the degree of the denominator (x-3, which is also degree 1). Therefore, the horizontal asymptote is the ratio of the leading coefficients, which are 2 and 1. So, the horizontal asymptote is y = 2/1 = 2.

5. However, the function has a constant term, 3, added to it. This shifts the entire function up by 3 units. Therefore, the actual horizontal asymptote is y = 2 + 3 = 5.

So, the asymptotes of the function 3+ (2x/x-3) are x = 3 (vertical) and y = 5 (horizontal).