How do you find vertical, horizontal and oblique asymptotes for #(x^2+3x-4)/x#?

1 Answer
Sep 15, 2016

Answer:

vertical asymptote at x = 0
oblique asymptote y = x +3

Explanation:

The denominator of the function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #x=0rArrx=0" is the asymptote"#

Horizontal asymptotes occur when the degree of the numerator# <=# degree of the denominator. This is not the case here (numerator-degree 2 , denominator-degree 1 ) Hence there is no horizontal asymptote.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. Hence there is an oblique asymptote.

divide numerator by x.

#x^2/x+(3x)/x-4/x=x+3-4/x#

as #xto+-oo,x+3-4/xtox+3-0#

#rArry=x+3" is the asymptote"#
graph{(x^2+3x-4)/x [-20, 20, -10, 10]}