Question

How do I find the slant asymptote of f(x) = x + (32/x^2)

Answer 1

\[f(x) = x + \frac{ 32 }{ x ^{2} }\]

Answer 2

what is an asymptote?

Answer 3

A vertical asymptote (VA) is an x value where as we approach that value the function goes either toward infinity or negative infinity. The VA is the x value that would make the denominator equal 0.

A Horizontal asymptote (HA) is a Y value such that when X reaches infinity the function comes very close (but not equal to) the Y value. The HA is the same value for the limit as X approaches infinity (or negative infinity).

I haven't yet graphed slant asymptotes so I'm not sure yet exactly what the slant asymptote is.

Answer 4

To find the slant asymptote the numerator needs to be divided by the denominator using long division. So the original function can be re-written as:

\[\frac{ x(x ^{2})+32 }{ x ^{2} }\]

Using long division the slant asymptote = x

Answer 5

Slant asymptote can be found if the degree in the numerator is greater than in the denominator. If you have a function that has these characteristics, then you do long division or synthetic division (whatever works best for you) to divide. After dividing, you'll get a quotient, that quotient is you slant asymptote.
y= quitient is the slant asymetote

Answer 6

@Zale101 \[f' = 1 - \frac{ 64 }{ x^3 }\]

Answer 7

The critical points are x = 4 and x = 0, but , I don't understand why. Apparently the equation can be re-written as \[\frac{ (x-4)(x^2 + 4x + 16) }{ x^3 }\]
This looks like the difference of cubes formula but there is no X in the numerator of f", so I'm not sure why it can be rewritten like that. I do understand that it is equivalent to \[\frac{ 1^3 - 4^3 }{ x^3 }\] But I don't get why x = 4 is a zero

Answer 8

I made a mistake in what I wrote earlier \[1 - \frac{ 64 }{ x^3 } = \frac{ (1)(x^3) - 64}{ x^3 } = \frac{ (x^3) - 4^3}{ x^3 } \]
therefore the difference of cubes formula can be used

Answer 9

Your slant asymptote is x, just as you said...

Answer 10

man, that was long huh