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burhan101

Determine the coordinates of the points on f(x)=x³+2x² where the tangent lines are perpendicular to y-x+2=0. How do I approach this question ?

  • 10 months ago
  • 10 months ago

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  1. burhan101
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    m=-1

    • 10 months ago
  2. burhan101
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    :O why'd you delete it

    • 10 months ago
  3. genius12
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    So basically you have to find the coordinates of points on f(x) such that the derivative, or the slope of the tangent, is the negative reciprocal of the slope of the given line y = x - 2. This implies that the tangents at those coordinates on f(x) will be perpendicular to the line y = x - 2. So we know that the derivative must be the negative reciprocal of the slope of y = x - 2 at those coordinates so f'(x) must be -1 at those coordinates. Let's find f'(x): \[\bf f'(x)=3x^2+4x\] We know that the slope of the tangent at those coordinates of f(x) or the derivative is -1. So we equate f'(x) with -1 and solve for x:\[\bf 3x^2+4x=-1 \rightarrow 3x^2+4x+1=0 \rightarrow (3x+1)(x+1)=0 \implies x = -\frac{ 1 }{ 3 },-1\]Plug these x-values back in to f(x) to get the y-values and that will give you the coordinates. @burhan101

    • 10 months ago
  4. burhan101
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    @genius12 thank you for the explanation ! very clear :)

    • 10 months ago
  5. burhan101
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    How would I determine the equations for the tangent for f(x) ?

    • 10 months ago
  6. genius12
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    Why do you need the equation of the tangent lines at the coordinates? @burhan101

    • 10 months ago
  7. burhan101
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    it's part b of the question

    • 10 months ago
  8. genius12
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    You have the coordinates right?

    • 10 months ago
  9. genius12
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    @burhan101

    • 10 months ago
  10. burhan101
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    Yes \[(\frac{ -1 }{ x },\frac{ 5 }{ 27 })\] and (-1,1)

    • 10 months ago
  11. genius12
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    whats the x there for.

    • 10 months ago
  12. genius12
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    @burhan101

    • 10 months ago
  13. burhan101
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    typo it was supposed to be a 3

    • 10 months ago
  14. genius12
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    oh ok.

    • 10 months ago
  15. genius12
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    Anyway, to find the equation of the tangent line at the coordinates, we need a point and a slope. We already know that since the tangent lines at these coordinates are perpendicular to y = x - 2, the slope for both tangents will be -1. We already have the points each tangent goes through. First one goes through (-1/3, 5/27) and second one goes through (-1, 1). The equation of tangent line will be int he form:\[\bf y= mx+b\]Where m is the slope, which is -1, and we need b. Since we know a point for each tangent, plug in the x-value and y-value of the coordinates and solve for b. This will then give you the slope-intercept equation of the tangents at the coordinates. @burhan101

    • 10 months ago
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