CALCULUS HELP !

posted by tony

f(x)=(-3x^3-x^2-9x-8)/(6x^2+4x+3. Find the equation of the non-vertical asymptote. y=
Does f(x) intersect its non-vertical asymptote ?
What is the smallest value of x at which f(c) intersects its non-vertical asymptote ?

please show all the work. I did this and got the wrong answer. I want to know where I made a mistake. Thank you

  1. Reiny

    By long division I got

    (-3x^3-x^2-9x-8)/(6x^2+4x+3) = (-1/2)x+1/6 + ( (-49/6)x - 17/2 )/(6x^2 + 4x + 3

    As x becomes ± large, the last term ----> 0

    and the asymptote is
    y = (-1/2)x + 1/6

    The denominator can never be zero, solving 6x^2 + 4x + 3 = 0 produces complex roots.
    Thus the function is continuous, and can never cross its asymptote.

  2. Anonymous

    The value of c we find solving the equation
    (-49/6)x-17/2=0

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