How do you identify the oblique asymptote of $f(x)= (9-x^3)/(3x^2)$ ?

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Answer 1 · 2018-04-16T10:52:15

$color(blue)(y=-1/3x)$

Oblique asymptotes occur if the degree of the numerator is greater than the degree of the denominator:

To find the oblique asymptote we divide the numerator by the denominator. We only have to divide until we have the equation of a line, $y=mx+b$ :

We can divide this using polynomial division, or, since we only need to obtain the equation of the line we could do the following.

Write the numerator as:

$-x^3+0x^2+0x+9$

Now divide by $3x^2$

$(-x^3)/(3x^2)+(0x^2)/(3x^2)+(0x)/(3x^2)+9/(3x^2)$

$-1/3x+0+0+3/x^2$

$3/x^2$ is just the remainder and we don't need this. We therefore have the line equation:

$y=-1/3x$

This is the oblique asymptote.

The graph confirms the findings:

enter image source here

How do you identify the oblique asymptote of f(x)= (9-x^3)/(3x^2)f(x)= (9-x^3)/(3x^2)?