How do you identify all asymptotes or holes for #f(x)=(x^2+3x-4)/(2x^2+10x+8)#?

1 Answer
Oct 30, 2016

Answer:

vertical asymptote at x = - 1
horizontal asymptote at #y=1/2#
hole at x = -4

Explanation:

The first step is to factorise the numerator/denominator of f(x).

#f(x)=(x^2+3x-4)/(2x^2+10x+8)=(cancel((x+4))(x-1))/(2cancel((x+4))(x+1))=(x-1)/(2x+2)#

excluded value is x ≠ -4. This means that the original function has a hole at x = -4, while the simplified version does not.

The denominator of simplified 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 : #2x+2=0rArrx=-1" 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-1/x)/((2x)/x+2/x)=(1-1/x)/(2+2/x)#

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

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