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pottersheep

  • one year ago

Please help. Calculus! Find the point of tangency and the equation of the tangent line passing through the point (-1.-23) to the graph of the function f(x) = 3 + 2x + 3x^2

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  1. pottersheep
    • one year ago
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    The answer says:q (-4,43) y = -22x - 45 AND y = 14x - 9 at q(2,19)

  2. satellite73
    • one year ago
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    the derivative is \(f'(x)=2-6x\) the slope of the tangent line through \((x,2+2x+3x^2)\) and \((-1,-23)\) is \[\frac{2+2x+2x^2+23}{x+1}\]

  3. satellite73
    • one year ago
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    set \[\frac{2+2x+2x^2+23}{x+1}=2+6x\]and solve for \(x\)

  4. satellite73
    • one year ago
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    oops, i meant the derivative is \(f'(x)=2+6x\)

  5. pottersheep
    • one year ago
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    Thank you for your time. Sorry, why is it (2+2x+2x2+23)/x+1? Where did the x+1 come from?

  6. satellite73
    • one year ago
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    slope is \(\frac{y_2-y_1}{x_2-x_1}\)

  7. satellite73
    • one year ago
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    here you have the point \((-1,-23)\) so the denominator is \(x+1\)

  8. pottersheep
    • one year ago
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    What exactly does the point P(-1,-23) mean?

  9. satellite73
    • one year ago
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    it says that \((-1,-23)\) is a point on the tangent line, so it is given to you

  10. pottersheep
    • one year ago
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    What does it give me? Is it because a tangent is a straight line so I can use that point and the slope formula?

  11. pottersheep
    • one year ago
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    Ah, I thought it was where the tangent met f(x) = 3 + 2x + 3x^2....but that didnt make sense to me

  12. satellite73
    • one year ago
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    a generic point on the curve looks lik e \[(x,3+2x+3x^2)\] you also have the point \((-1,-23)\)

  13. satellite73
    • one year ago
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    the slope of the line through those two points is \[\frac{3+2x+2x^2+23}{x+1}\]

  14. satellite73
    • one year ago
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    and you know that the line with this slope is tangent to the curve at some point the slope my therefore be equal to the derivative at that point, which is \(2+6x^2\)

  15. pottersheep
    • one year ago
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    I get it now!!!

  16. pottersheep
    • one year ago
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    Thank you so much :)

  17. satellite73
    • one year ago
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    ok good yw

  18. pottersheep
    • one year ago
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    You are such a good teacher =D

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