How do you identify all asymptotes or holes for #f(x)=(x^2-x)/(2x^2+4x-6)#?

1 Answer
Sep 9, 2016

Answer:

hole at #(1,1/8)#
vertical asymptote at x = 3
horizontal asymptote at #y=1/2#

Explanation:

Holes occur when a duplicate factor is cancelled in the numerator/denominator of f(x).
The first step is therefore to factorise and simplify f(x).

#f(x)=(xcancel((x-1)))/(2(x+3)cancel((x-1)))=x/(2(x+3))#

set #x-1=0rArrx=1#

substitute x = 1 into 'simplified' version of f(x)

#rArrf(1)=1/(2(1+3))=1/8#

#rArr"there is a hole at"(1,1/8)#

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: #2(x+3)=0rarrx=-3" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

#f(x)=(x/x)/((2x)/x+6/x)=1/(2+6/x)#

as #xto+-oo,f(x)to1/(2+0)#

#rArry=1/2" is the asymptote"#
graph{(x^2-x)/(2x^2+4x-6) [-10, 10, -5, 5]}