# Calculus

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f(x)=(-9x3–5x2+4x–2)/(–7x2–9x+7)

What is the smallest value of x at which f(x) intersects its non-vertical asymptote?

• Calculus -

First we calculate the non-vertical (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 non-vertical 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*x-49 and add the two terms together to get:
h(x)=(-(46*x^2-35*x-14))/(49*x^2+63*x-49)

h(x) will vanish if and only if the numerator vanishes, which is the condition:
n(x)=-(46*x^2-35*x-14)=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^2-35*x-14)

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