How do you graph #y=(x+2)/(x+3)# using asymptotes, intercepts, end behavior?

1 Answer
Dec 20, 2016

Answer:

graph{(x+2)/(x+3) [-10, 10, -5, 5]}

#y#-intercept is #2/3#

#x#-intercept is #-2#

vertical asymptote at #x=-3#

#y<1# when #x> -3#

#y>1# when #x< -3#

Explanation:

asymptote:

#n/0# = undefined

#therefore# if #x+3 = 0#, #y# is undefined.

this means that it is not on the graph, and so is shown as the asymptote.

when #x+3 = 0#, #x = 0-3#

#x = -3#

intercepts:

#y#:

the #y#-intercept is when #x = 0#

#y = (x+2)/(x+3)#

#y = 2/3#

#y#-intercept is #2/3#

#x#:

the #x#-intercept is when #y=0#

#(x+2)/(x+3)=0#

the numerator has to be #0#, since #0/n = 0#

this means that #x+2=0#

when #x+2 = 0#, #x = 0-2#

#x#-intercept is #-2#

end behaviour:

#y<1#

#x+2# is always less than #x+3#. with both positive and negative numbers.

the fraction #(x+2)/(x+3)# cannot be simplified to #(>=1)/1# if #x> -3#

however, it is vice versa for negative numbers, since smaller negative numbers have a higher absolute value (distance from #0#),

e.g. #-3/-2 = |3|/|2| = 3/2#

this means that #(x+2)/(x+3)# cannot be simplified to #(<=1)/1# if #x< -3#