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pottersheep Group Title

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

  • one year ago
  • one year ago

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

    • one year ago
  2. satellite73 Group Title
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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 \[\frac{2+2x+2x^2+23}{x+1}\]

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

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

    • one year ago
  5. pottersheep Group Title
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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?

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

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

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

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

    • one year ago
  10. pottersheep Group Title
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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?

    • one year ago
  11. pottersheep Group Title
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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

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

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

    • one year ago
  14. satellite73 Group Title
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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 \(2+6x^2\)

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

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

    • one year ago
  17. satellite73 Group Title
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    ok good yw

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

    • one year ago
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