How do you find the asymptotes for #f(x)= (2x^2 +x -1 )/( x-1)#?

1 Answer
Apr 23, 2017

Answer:

#"vertical asymptote at " x=1#
#"oblique asymptote " y=2x+3#

Explanation:

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-1=0rArrx=1" is the asymptote"#

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

#"using the divisor as a factor in the numerator"#

#color(red)(2x)(x-1)color(magenta)(+2x)+x-1#

#=color(red)(2x)(x-1)color(red)(+3)(x-1)color(magenta)(+3)-1#

#=color(red)(2x)(x-1)color(red)(+3)(x-1)+2#

#rArrf(x)=2x+3+2/(x-1)#

as #xto+-oo,f(x)to2x+3#

#rArry=2x+3" is the asymptote"#
graph{(2x^2+x-1)/(x-1) [-24.98, 24.98, -12.48, 12.5]}