anonymous
  • anonymous
How do I find the slant asymptote of f(x) = x + (32/x^2)
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[f(x) = x + \frac{ 32 }{ x ^{2} }\]
nincompoop
  • nincompoop
what is an asymptote?
anonymous
  • 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.

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anonymous
  • 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
Zale101
  • 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
anonymous
  • anonymous
@Zale101 \[f' = 1 - \frac{ 64 }{ x^3 }\]
anonymous
  • 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
anonymous
  • 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
Zale101
  • Zale101
Your slant asymptote is x, just as you said...
nincompoop
  • nincompoop
man, that was long huh

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