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

1. anonymous

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

2. nincompoop

what is an asymptote?

3. anonymous

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.

4. anonymous

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

5. Zale101

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

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. Zale101

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

10. nincompoop

man, that was long huh