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pottersheep

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

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

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

  5. pottersheep
    • 2 years 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
    • 2 years ago
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    slope is \(\frac{y_2-y_1}{x_2-x_1}\)

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

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

  9. satellite73
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    I get it now!!!

  16. pottersheep
    • 2 years ago
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    Thank you so much :)

  17. satellite73
    • 2 years ago
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    ok good yw

  18. pottersheep
    • 2 years ago
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    You are such a good teacher =D

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