Question #ef8d7

1 Answer
Aug 21, 2017

Answer:

#"vertical asymptote at "x=2#
#"slant asymptote "y=x+2#

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-2=0rArrx=2" is the asymptote"#

Horizontal asymptotes occur when the degree of the numerator #<=# the degree of the denominator. This is not the case here hence there are no horizontal asymptotes.

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here hence there is a slant asymptote.

#"dividing out gives"#

#x(x-2)+2(x-2)+5#

#rArrf(x)=(x^2+1)/(x-2)=x+2+5/(x-2)#

#"as " xto+-oo,f(x)tox+2#

#rArry=x+2" is the asymptote"#
graph{(x^2+1)/(x-2) [-40, 40, -20, 20]}