How do you find the vertical asymptotes and holes of #f(x)=(x+2)/(x^2+3x-4)#?

1 Answer
Nov 17, 2016

Answer:

vertical asymptotes at x = - 4 and x = 1
horizontal asymptote at y = 0

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^2+3x-4=0rArr(x+4)(x-1)=0#

#rArrx=-4" and " x=1" are the asymptotes"#

Horizontal asymptotes occur as

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

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

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

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

#rArry=0" is the asymptote"#

Holes occur when there are duplicate factors on the numerator/denominator. This is not the case here hence f(x) has no holes.
graph{(x+2)/(x^2+3x-4) [-10, 10, -5, 5]}