anonymous
  • anonymous
Leading coefficient test?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
I have a question that wants me to use the leading coefficient test to determine the end behavior of f(x) = x^2(x + 2)
amistre64
  • amistre64
expand the product to a bunch of terms that are added together
amistre64
  • amistre64
then forget the stuff after the first term; the end behaviours of the function act in the same manner as the leading term

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anonymous
  • anonymous
What do you mean?
amistre64
  • amistre64
|dw:1360618376605:dw| the end behaviours of both sketches is the same .... they both behave like x^2
anonymous
  • anonymous
I mean what did you mean by expand the product to abunch of terms added together ?
amistre64
  • amistre64
what does: x^2(x + 2) look like after you multiply the x^2 into the (x+2) ??
anonymous
  • anonymous
2x^23?
anonymous
  • anonymous
I mean 2x^3?
amistre64
  • amistre64
close: x^2(x+2) = x^3 + 2x^2 now, end behaviour is what the graph acts like for values of x that go way off to the left or right zero. would you agree that the 2x^2 term contributes very little to the equation for say x=1000000000000 ?? if so, then the end bahviours (the behaviour of large values of x) gets dominated by the first term
anonymous
  • anonymous
I'm not sure, I wouldnt think so?
amistre64
  • amistre64
the leading term of a polynomial always takes control for larger values of x. so we can know what the end behaviour of a graph of any given polynomial is by just comparing it to the standard graph of its leading term. How do the ends of x^3 behave?
anonymous
  • anonymous
I have no idea, I don't get any this.
amistre64
  • amistre64
you need to be able to know a few basic graph shapes :/ |dw:1360619317867:dw|
amistre64
  • amistre64
since the leading term of the expanded product is x^3; the ends will act in the same manner is x^3 .... another thing to keep in mind is that all even powers act alike; and all odd powers act aloike
anonymous
  • anonymous
Thanks for trying to explain this to me, I think Im just gonna have to find me a tutor. Problems like this is foreign language to me, very hard to under stand.

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