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

1 Answer
Jul 12, 2016

Answer:

vertical asymptote x = 0
slant asymptote y = x + 3

Explanation:

The denominator cannot be zero as this is undefined. Setting the denominator equal to zero and solving will give us the value that x cannot be and if the numerator is non-zero for this value of x then it is a vertical asymptote.

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

When the degree of the numerator > degree of denominator as is the case here (numerator-degree 2, denominator-degree 1) Then we have a slant asymptote but no horizontal asymptote.

divide function gives #f(x)=(x^2)/x+(3x)/x-4/x=x+3-4/x#

as #xto+-oo,f(x)tox+3-0#

#rArry=x+3" is the asymptote"#
graph{(x^2+3x-4)/x [-10, 10, -5, 5]}