Question

50PTS! Describe the vertical asymptotes and holes for the graph of y = x-5/x^2 -1

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Ok

first of all, for q(x)/p(x)
if the degree of q(x) is less than the degree of p(x),then the horizontal assemtote is 0

then simplify
any factors you factored out is now a hole, remember them

to find the vertical assemtotes of a function, set the SIMPLIFIED denomenator equal to 0 and solve

so

y=(x-5)/(x^2-1)
q(x)<p(x)
horizontal assemtote is y=0

no factors to simplify so no holes

set denomenator to 0 to find vertical assemtote
x^2-1=0
(x-1)(x+1)=0
x-1=0
x=1

x+1=0
x=-1

the horizontal assemtotes are x=1 and -1