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anonymous

  • 5 years ago

f(x)=(3x^2-2x-1) find equation of the line that is tangent and parallel to this line 8x-2y+12=0

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  1. bahrom7893
    • 5 years ago
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    wait to which line? f(x)=(3x^2-2x-1) or 8x-2y+12=0?

  2. anonymous
    • 5 years ago
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    8x-2y+12=0 cant figure it out because i need the two points (x and Y) my book is not clear

  3. anonymous
    • 5 years ago
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    howdy. First we can find the slope of the line. I'll rewrite it in y=mx+b form to make it apparent: y = 4x+6 -> slope is 4

  4. anonymous
    • 5 years ago
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    Then we need to find where the derivative (slope of the tangent) is equal to 4

  5. anonymous
    • 5 years ago
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    The derivative is: f'(x)=6x-2 4=6x-2 -> x = 1

  6. anonymous
    • 5 years ago
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    This means that the tangent line to f(x) at x = 1 has a slope of 4 and is thus parallel to the given line. Now we just need to find f(1) to write the equation of a line: y(1) = 3(1^2)-2(1)-1=0

  7. anonymous
    • 5 years ago
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    Thus the line is: y = 4(x-1) + 0 = 4x-4

  8. anonymous
    • 5 years ago
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    thank you very helpful

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