How do you graph #y=3/(x+8)-10# using asymptotes, intercepts, end behavior?

1 Answer
Binayaka C.
Jul 7, 2018

Vertical asymptote is at #x=-8# ,Horizontal asymptote is at # y=-10#, x intercept at #(-7.7,0)#, y intercept at #(0, -9.625)#, end behavior: #y-> -10 # as #x -> -oo andy-> -10 # as #x -> oo#

Explanation:

#y =3/(x+8) - 10 or y = (3-10 x-80)/(x+8)# or

#y = (-10 x-77)/(x+8)#

Vertical asymptote occur when denominator is zero.

# x+8=0 :. x= -8; lim(x->8^-) y -> -oo #

#lim (x->8^+) y - > oo #

Vertical asymptote is at #x=-8#

Horizontal asymptote: #lim (x->-oo) ; y =-10/1=-10 #

Horizontal asymptote is at # y=-10#

x intercept: Putting #y=0# in the equation we get,

#-10 x -77=0 :. x = -7.7# or at #(-7.7,0)#

y intercept: Putting #x=0# in the equation we get,

#y= -77/8= -9.625# or at #(0, -9.625)#

End behavior: #y-> -10 # as #x -> -oo# and

#y-> -10 # as #x -> oo#

graph{3/(x+8)-10 [-90, 90, -45, 45]} [Ans]

Related questions
See all questions in Limits - End Behavior and Asymptotes
Impact of this question
1839 views around the world
You can reuse this answer
Creative Commons License