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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x) = x + \frac{ 32 }{ x ^{2} }\]

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0what is an asymptote?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To find the slant asymptote the numerator needs to be divided by the denominator using long division. So the original function can be rewritten as: \[\frac{ x(x ^{2})+32 }{ x ^{2} }\] Using long division the slant asymptote = x

Zale101
 one year ago
Best ResponseYou've already chosen the best response.1Slant 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
 one year ago
Best ResponseYou've already chosen the best response.0@Zale101 \[f' = 1  \frac{ 64 }{ x^3 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The critical points are x = 4 and x = 0, but , I don't understand why. Apparently the equation can be rewritten as \[\frac{ (x4)(x^2 + 4x + 16) }{ x^3 }\] 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1Your slant asymptote is x, just as you said...

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.0man, that was long huh
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