. asked in Science & MathematicsMathematics · 1 month ago

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. 

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  • david
    Lv 7
    1 month ago
    Favorite Answer

    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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  • 1 month ago

    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

  • 1 month ago

    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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  • rotchm
    Lv 7
    1 month ago

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

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

    • Wayne DeguMan
      Lv 7
      1 month agoReport

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

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  • Multiply the numerator by 4 and see what happens.

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