How do you find vertical, horizontal and oblique asymptotes for # (x² - 3x - 7)/(x+3)#?

1 Answer
Jun 26, 2018

Answer:

#"vertical asymptote at "x=-3#
#"oblique asymptote "y=x-6#

Explanation:

#"let "f(x)=(x^2-3x+7)/(x+3)#

The denominator of f(x) cannot be zero as this would make f(x) 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+3=0rArrx=-3" is the asymptote"#

Since the degree of the numerator is greater than the degree of the denominator there is an oblique asymptote but no horizontal asymptote.

#"dividing numerator by denominator gives"#

#f(x)=x-6+18/(x+3)#

#"as "xto+-oo,f(x)tox-6+0#

#y=x-6" is the asymptote"#
graph{(x^2-3x-7)/(x+3) [-10, 10, -5, 5]}