How do you find vertical, horizontal and oblique asymptotes for #y=3/(x-2)+1#?

Question

Elijah Allen · Accepted Answer

The vertical asymptote is #x=2#
The horizontal asymptote is #y=1#
There is no oblique asymptote.

As we cannot divide by #0#, the vertical asymptote is #x=2# And #lim_(x->+-oo)y=1# So #y=1#is a horizontal asymptote. There is no oblique asymptote as the degree of the numerator #=# the degree of the denominator graph{(y-1-3/(x-2))(y-1)=0 [-14.24, 14.24, -7.12, 7.12]}

Serenity Johnson · Answer

undefined To find the vertical asymptote(s), set the denominator of the rational function equal to zero and solve for x.

For the function y = 3/(x - 2) + 1:

x - 2 = 0
x = 2

So, the vertical asymptote is x = 2.

To find the horizontal asymptote, examine the behavior of the function as x approaches positive or negative infinity.

As x approaches positive infinity, y = 3/(x - 2) + 1 approaches 0, because the term with x in the denominator becomes negligible compared to x as x becomes very large. Therefore, the horizontal asymptote is y = 1.

For the oblique asymptote, if the degree of the numerator is one greater than the degree of the denominator, there is an oblique (slant) asymptote. To find it, perform polynomial long division of the numerator by the denominator.

In this case, the degree of the numerator (0) is not greater than the degree of the denominator (1), so there is no oblique asymptote.

In summary:
- Vertical asymptote: x = 2
- Horizontal asymptote: y = 1