How do you identify all asymptotes or holes and intercepts for #f(x)=(2x-3)/(3x+1)#?

Question

Theodore Amidon · Accepted Answer

There is an asymptote at x =#-1/3#

1 divided by 0.10=10. Or how many 0.1s can you get in 1 1 divided by 0.01 =100.... The smaller the divisor gets, the bigger the answer

In the limit as the divisor approaches zero the result approaches infinity and that is where we have an asymptote on the graph

So as 3x+1 approaches zero the function approaches infinity

Asymptote where 3x+1=0 or x =#-1/3#

Christian Antunez · Answer

undefined To identify asymptotes, holes, and intercepts for $ f(x) = \frac{2x - 3}{3x + 1} $:

1. Vertical asymptote: Set the denominator equal to zero and solve for $ x $. The vertical asymptote is at $ x = -\frac{1}{3} $.

2. Horizontal asymptote: If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote.

3. To find any holes, factor both the numerator and denominator, then simplify to see if there are any common factors that cancel out.

4. To find $ y $-intercept, plug in $ x = 0 $ and solve for $ y $.

5. To find $ x $-intercept, plug in $ y = 0 $ and solve for $ x $.