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anonymous

  • 5 years ago

Determine the equation of the line tangent to the graph of the following function at x = 0. g(x) = ex(−4 + 3x + 3x2)

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  1. anonymous
    • 5 years ago
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    the eqn is y+4-6x^2-3x

  2. anonymous
    • 5 years ago
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    y+4-6x^2-3x = 0 ***

  3. anonymous
    • 5 years ago
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    how did you figure that one man?

  4. anonymous
    • 5 years ago
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    first you the y-value when x=0(so y=-4) and then when you find the equation of the tangent at any piont (aka, the slope) you use the slope, and your x and y value and sub them into the y=mx+b eqn to get it or you can use slope=(-4-y)/(o-x)

  5. anonymous
    • 5 years ago
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    only one problem. we're looking for the equation of a straight line. so it couldn't have any squared values in it

  6. anonymous
    • 5 years ago
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    true, okay the answer is y=3x+4

  7. anonymous
    • 5 years ago
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    that was considered the wrong answer when i submitted

  8. anonymous
    • 5 years ago
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    what grade are you in?

  9. anonymous
    • 5 years ago
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    im a senior in high school

  10. anonymous
    • 5 years ago
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    okay, and what's the ex in front of the eqn?

  11. anonymous
    • 5 years ago
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    g(x) = e^x(−4 + 3x + 3x^2)

  12. anonymous
    • 5 years ago
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    e to the power of x?

  13. anonymous
    • 5 years ago
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    yezzir

  14. anonymous
    • 5 years ago
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    whoa thats wierd, i didnt even consider that

  15. anonymous
    • 5 years ago
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    just get the derivative using product rule, then sub 0 in for x and find the slope at that point, and then use y=mx+b, no big.

  16. anonymous
    • 5 years ago
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    a'ight could you check my answer for me after i work it?

  17. anonymous
    • 5 years ago
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    k go, young grasshopper

  18. anonymous
    • 5 years ago
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    is there answers in the back of your text book or something?

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spraguer (Moderator)
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