What are all the asymptote of #(x^2+3x-4)/ (x+2)#?

Question

Christian Antunez · Accepted Answer

Vertical asymptotes is #x=-2# and obliqe asymptote is given by #y=x#

To find all the asymptotes for function #y=(x^2+3x−4)/(x+2)#, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or #x+2=0# i.e. #x=-2#, which is the only vertical asymptote..

As the highest degree of numerator is #2# and of denominator is #1# and is higher by one degree, we have only slant / oblique asymptote is given by #y=x^2/x=x# i.e. #y=x# (Had the degree been equal, we would have horizontal asymptote).

Hence, while vertical asymptotes is #x=-2# and obliqe asymptote is given by #y=x#

graph{x+x/(x+2) [-20, 20, -10, 10]}

James Allen · Answer

undefined The rational function $ \frac{x^2 + 3x - 4}{x + 2} $ has two types of asymptotes: vertical and horizontal.

Vertical asymptote:
- Vertical asymptote occurs where the denominator of the rational function becomes zero.
- In this case, the vertical asymptote occurs at $ x = -2 $.

Horizontal asymptote:
- To find the horizontal asymptote, compare the degrees of the numerator and denominator of the rational function.
- If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $ y = 0 $.
- If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients of both.
- In this case, since the degree of the numerator (2) is equal to the degree of the denominator (1), divide the leading coefficients: $ \frac{1}{1} = 1 $.
- Therefore, the horizontal asymptote is $ y = 1 $.