CALCULUS HELP !
posted by tony .
f(x)=(3x^3x^29x8)/(6x^2+4x+3. Find the equation of the nonvertical asymptote. y=
Does f(x) intersect its nonvertical asymptote ?
What is the smallest value of x at which f(c) intersects its nonvertical 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

CALCULUS HELP ! 
Reiny
By long division I got
(3x^3x^29x8)/(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. 
CALCULUS HELP ! 
Anonymous
The value of c we find solving the equation
(49/6)x17/2=0
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