Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

gjhfdfg

  • one year ago

Leading coefficient test?

  • This Question is Closed
  1. gjhfdfg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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)

  2. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    expand the product to a bunch of terms that are added together

  3. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    then forget the stuff after the first term; the end behaviours of the function act in the same manner as the leading term

  4. gjhfdfg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What do you mean?

  5. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1360618376605:dw| the end behaviours of both sketches is the same .... they both behave like x^2

  6. gjhfdfg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I mean what did you mean by expand the product to abunch of terms added together ?

  7. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    what does: x^2(x + 2) look like after you multiply the x^2 into the (x+2) ??

  8. gjhfdfg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    2x^23?

  9. gjhfdfg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I mean 2x^3?

  10. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  11. gjhfdfg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm not sure, I wouldnt think so?

  12. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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?

  13. gjhfdfg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have no idea, I don't get any this.

  14. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you need to be able to know a few basic graph shapes :/ |dw:1360619317867:dw|

  15. amistre64
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  16. gjhfdfg
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  17. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.