How do you find vertical, horizontal and oblique asymptotes for #(3x^2 + 2x - 1 )/ (x + 1)#?

2 Answers
Jul 3, 2017

Answer:

Straight line ( slant asymptote), #y=3x-1#

Explanation:

#(3x^2+2x-1)/(x+1) #, Vertical asymptote is # x+1= 0 or x= -1# .But

#(3x^2+2x-1)/(x+1) =( (3x-1)cancel( (x+1)))/cancel((x+1))#, this

disconinuity is removable.

The degree of numerator is #1# more than dgree of denominator , so horizontal asymptote is absent.

By long division we have slant asymptote as #y=3x-1#

The graph will show a straight line (slant asymptote) #y=3x-1# only

graph{(3x^2+2x-1)/(x+1) [-10, 10, -5, 5]} [Ans]

Jul 3, 2017

Answer:

# y=3x-1#

Explanation:

#"factorising the numerator"#

#((3x-1)cancel((x+1)))/cancel((x+1))=3x-1#

#"the removal of the factor " (x+1)#

#"from the numerator/denominator indicates a hole at x = - 1"#

#"the graph of " (3x^2+2x-1)/(x+1)" simplifies to the line"#

#y=3x-1 " which has no hole"#
graph{(3x^2+2x-1)/(x+1) [-10, 10, -5, 5]}