How do you identify all asymptotes or holes and intercepts for #f(x)=(x^2+1)/((x-1)(2x-4))#?

1 Answer
Jul 26, 2017

Answer:

#"vertical asymptotes at "x=1" and "x=2#
#"horizontal asymptote at "y=1/2#

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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#"solve "(x-1)(2x-4)=0#

#rArrx=1" and "x=2" are the asymptotes"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

divide terms on the numerator/denominator by the highest power of x, that is #x^2#

#f(x)=(x^2/x^2+1/x^2)/((2x^2)/x^2-(6x)/x^2+4/x^2)=(1+1/x^2)/(2-6/x+4/x^2)#

as #xto+-oo,f(x)to(1+0)/(2-0+0)#

#rArry=1/2" is the asymptote"#

Holes occur when there are common factors on the numerator/denominator. This is not the case here hence there are no holes.
graph{(x^2+1)/(2x^2-6x+4) [-20, 20, -10, 10]}