How do you find the asymptotes for #y=(x+3)/((x-4)(x+3))#?

1 Answer
Jul 12, 2018

Answer:

#"vertical asymptote at "x=4#
#"horizontal asymptote at "y=0#

Explanation:

#"simplify by cancelling"#

#y=cancel(x+3)/((x-4)cancel((x+3)))=1/(x-4)#

#"the removal of "(x+3)" from numerator/denominator"#
#"indicates a hole at "x=-3#

#"the graph of the simplification "1/(x-4)" is the same as"#

#"the graph of "y=(x+3)/((x-4)(x+3))" without the hole"#

The denominator of y cannot be zero as this would make y 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-4=0rArrx=4" is the asymptote"#

#"Horizontal asymptotes occur as"#

#lim_(xto+-oo),ytoc" ( a constant)"#

#"divide terms on numerator/denominator by "x#

#y=(1/x)/(x/x-4/x)=(1/x)/(1-4/x)#

#"as "xto+-oo,yto0/(1-0)#

#y=0" is the asymptote"#
graph{1/(x-4) [-10, 10, -5, 5]}