How do you find vertical, horizontal and oblique asymptotes for # (x^2 + 5x + 6)/(x+3)#?
1 Answer
Aug 4, 2018
Explanation:
#"factor the numerator"#
#=((x+2)cancel((x+3)))/cancel((x+3))=x+2#
#"the removal of the factor "(x+3)" indicates a removable"#
#"discontinuity (hole) at "x=-3#
#"The simplified version is a linear equation with "#
#"no asymptotes"#
graph{(x^2+5x+6)/(x+3) [-10, 10, -5, 5]}
