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August 30, 2014

August 30, 2014

Posted by **HELP!!! DESPERATE D;** on Friday, June 1, 2012 at 4:56am.

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

- Calculus -
**MathMate**, Friday, June 1, 2012 at 8:06amFirst 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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