How do you graph #y=8/(x^2-x-6)# using asymptotes, intercepts, end behavior?

1 Answer
Oct 15, 2017

Answer:

Vertical asymptote: x=3,-2
Horizontal asymptote: #y=0#
#x#- intercept: none
#y#-intercept: #-4/3#
End behaviour:
As #x rarr -∞,y rarr ∞#
As #x rarr ∞, y rarr ∞#

Explanation:

Denote the function as #(n(x))/[d(x)]#

To find the vertical asymptote,
Find #d(x)=0#
#rArr x^2-x-6=0#
#(x-3)(x+2)=0#

#:.# The vertical asymptotes are at #x=3,-2#

To find the #x#-intercept, plug in #0# for #y# and solve for #x#.
#0=(8)/(x^2-x-6)#

#:.# There are no #x#-intercepts.

To find the #y#-intercept, plug in #0# for #x# and solve for #y#.
#y=(8)/(0^2-0-6#
#y=-4/3#

#:.# The #y#-intercept is at #-4/3#.

To find the horizontal asymptote,
Compare the leading degrees of the numerator and denominator.
For #n(x)#, the leading degree is #0#, since #8*x^0# would give #8#. Denote this as #color(pink)m#.

For #d(x)#, the leading degree is #2#. Denote this as #color(brown)n#.

If #color(pink)n < color(brown)m#, then the horizontal asymptote is #y=0#.

To find the end behaviour of the graph, plot it in a graphic calculator.
As #x rarr -∞,y rarr ∞#
As #x rarr ∞, y rarr ∞#