How do you find the oblique asymptote?

Question

Please find the slany asymptote, if any, of the graph of the raional function....

f(x) = (x2-2x+6) / (x+6)

Answer 1

For an equation such as that, the oblique asymptote can usually be found by putting the function in this form:

f(x) = a/q(x) + r(x)

The oblique asymptote will be y = r(x). I'll demonstrate.

f(x) = (x^2 - 2x + 6) / (x + 6)

Use synthetic long division to solve this; (x + 6) into (x^2 - 6x + 6). Since long division is difficult to show on here, I'll do this another method.

Split -2x into +6x and -8x.

f(x) = (x^2 + 6x - 8x + 6) / (x + 6)

Factor the first two terms.

f(x) = ( x(x + 6) - 8x + 6 ) / (x + 6)

Split -6 into -48 and +42.

f(x) = ( x(x + 6) - 8x - 48 + 42 ) / (x + 6)

Factor.

f(x) = ( x(x + 6) - 8(x + 6) + 42 ) / (x + 6)

Factor (x + 6) from the first two terms.

f(x) = [ (x + 6)(x - 8) + 42 ] / (x + 6)

Put each term over the denominator.

f(x) = (x + 6)(x - 8)/(x + 6) + 42/(x + 6)

We get a cancellation.

f(x) = (x - 8) + 42/(x + 6)

Swap the terms.

f(x) = 42/(x + 6) + (x - 8)

Our oblique asymptote is the line y = x - 8

Answer 2

Divide (x2-2x+6) by (x+6) getting x-8 with a remainder.
Ignore the remainder so you have y = x-8 as your slant asymptote.