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anonymous

  • 4 years ago

An equation of the line tangent to the graph of y=x^3+3x^2+2 at its point of inflection is?

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  1. amistre64
    • 4 years ago
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    how do we find inflection points?

  2. anonymous
    • 4 years ago
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    second deriv

  3. amistre64
    • 4 years ago
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    good, and whats our 2nd derivative?

  4. anonymous
    • 4 years ago
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    6x+6

  5. amistre64
    • 4 years ago
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    and that is zero for what value of x?

  6. anonymous
    • 4 years ago
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    -6

  7. amistre64
    • 4 years ago
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    -1 ... but yes what is the value of the first derivative at x=-1 to determine slope at x=-1

  8. anonymous
    • 4 years ago
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    i don't know...

  9. amistre64
    • 4 years ago
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    whats our 1st derivative?

  10. anonymous
    • 4 years ago
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    3x^2+6x

  11. amistre64
    • 4 years ago
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    right, and when x=-1 we get? -3 right?

  12. anonymous
    • 4 years ago
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    Wait, second derivative is zero is not sufficient for point of inflection. One also needs the lowest-order non-zero derivative to be of odd order.

  13. amistre64
    • 4 years ago
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    it is good for this case since its a cubic ...

  14. anonymous
    • 4 years ago
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    Yes, but still you have show it right? Otherwise teacher will deduct some marks ?!

  15. amistre64
    • 4 years ago
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    slope at x=-1 is -3; we need to know the value of y at -1 to determine a point for this:

  16. amistre64
    • 4 years ago
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    i dunno what a teacher does .... :)

  17. anonymous
    • 4 years ago
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    please continue :)

  18. anonymous
    • 4 years ago
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    They are there to deduct marks :P

  19. anonymous
    • 4 years ago
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    there is a point of inflection at -1

  20. amistre64
    • 4 years ago
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    what is the value of y when x=-1? y = x^3 +3x^2 + 2, when x=-1

  21. anonymous
    • 4 years ago
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    4

  22. amistre64
    • 4 years ago
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    4? (-1,4) so, the equation of the tangent at the inflection is putting all the parts together: tanY = -3x+3(-1)+4

  23. anonymous
    • 4 years ago
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    thank youuuuuuuuuuuu!

  24. amistre64
    • 4 years ago
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    youre welcome

  25. anonymous
    • 4 years ago
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    can you look at my other question as well?

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