Calculus
posted by HELP!!! DESPERATE D; .
f(x)=(9x3–5x2+4x–2)/(–7x2–9x+7)
What is the smallest value of x at which f(x) intersects its nonvertical asymptote?

First we calculate the nonvertical (slant) asymptote as the leading term in the quotient represented by f(x), that is:
(9x3–5x2+4x–2)/(–7x2–9x+7)
=9x/7 + .....
So g(x)=9x/7 is the nonvertical asymptote.
To solve f(x)=g(x), we can set
h(x)=f(x)g(x)=0 and solve for x.
Thus,
h(x)=(9x3–5x2+4x–2)/(–7x2–9x+7)9x/7
we need to take the common denominator of 49*x^2+63*x49 and add the two terms together to get:
h(x)=((46*x^235*x14))/(49*x^2+63*x49)
h(x) will vanish if and only if the numerator vanishes, which is the condition:
n(x)=(46*x^235*x14)=0
You only need to solve the quadratic equation n(x)=0 and select the smaller root.
Note: you need to check that the denominator of h(x) does not vanish at the same point!
(46*x^235*x14)
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