# How do I find the horizontal asymptote in this question?

My current thought process is to find the equation of f(x), use the coefficients of the x^2 expressions in the numerator and demoninator to find the horizontal asymptote.

f(x) = (x^2-5x-6)/(x^2-1)

is the current state of my assumed equation.

Which would make HA = 1

But this is wrong because the lim as x-> -1 would not be 4.

I'm currently thrown off by that bit. Can you help? Thank you in advance. Relevance

HA occur if the functon approaches a y value as x approaches infinity

...  nobdy gives a rats **** what happens at x -> -1 or even x -> 1000  for the asymptote.

VA  at x = 1 ... what does that mean?  as x --> 1 .. the y --> +/- infinity.

HA ===  as x --> +/- infinity what does y approach??

removable discontinuity (called a 'hole' by many books / teachers) means that both the numerator and denom. have a factor of (x + 1) ..  which cancels creating the removable discontinuity.

f(x)  =  [P(x)]/[Q(x)]  factors as  f(x)  =  [(x+1)(ax + b)] / [(x+1)(cx + d)]  ...  reduces to  f(x) = (ax + b) / (cx + d)  for x not = -1

as x --> -1   ....   (-a + b)/(-c+d) = 4

,,,  your analysis is correct .. the HA will occur at  a/c

...  need more?

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• Let f(x)=(ax^2+bx+c)/[(x-1)(cx+d)]

limit f(x)=4 & f(x) has a removable discontinuity at x=-1

x->-1

=>

a-b+c=0-----(1)

-2(-c+d)=0----(2)

f(6)=0

=>

36a+6b+c=0-----(3)

From (1) & (3), get

b=-5a

c=-6a

=>

f(x)=(x^2-5x-6)/[-6(x^2-1)]

Check:

x=1 is the vertical asymptote.

f(-1)=(1+5-6)/[-6(1-1)]=0/0=> there is a removable

continuity at x=-1.

f(6)=(36-30-6)/[-6(36-1)]=0

f(x) is valid.

Let y=mx+b be the slant aysmptote.

m=limit {(x^2-5x-6)/[6x(x^2-1)]+b/x}

.....x->+/-inf.

m->{(x^2)(1-5/x-6/x^2)/[6(x^3)(1-x^2)]->0 as x->+/-inf.

b=limit {(x^2-5x-6)/[-6(x^2-1)]

...x->+/-inf.

b->{(x^2)(1-5/x-6/x^2)/[-6x^2(1-1/x^2)]}->-1/6, as x->+/-inf.

Thus, the horizontal asymptote is

y=-1/6

• Correct...thought I was going mad!

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• Cancelling the removable dicontinuity, we get:

f(x) = (x - 6)/(x - 1)

To compensate for the limiting constraint we have:

f(x) = a(x - 6)/(x - 1)

Now, as x --> -1 we have:

a(-7)/(-2) = 4

=> a = 8/7

Therefore, f(x) = (8/7)(x - 6)/(x - 1)

As x -->  ∞, (x - 6)/(x - 1) --> 1

Then, (8/7)(x - 6)/(x - 1) --> 8/7 => horizontal asymptote

:)>

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• Hint: The denom can be written in the form k(x²-1).

I'm sure you can take it from there now!

• Good suggestion. Makes a change from the usual condemnation and critical comments.

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