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anonymous

  • 4 years ago

Let f(x)=x^2+x+13. What is the value of x for which the tangent line to the graph of y=f(x) is parallel to the x-axis?

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  1. anonymous
    • 4 years ago
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    i have no idea how to do this

  2. Xishem
    • 4 years ago
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    \[f(x)=x^2+x+13\]\[f'(x)=2x+1\] Any line is parallel to the x-axis when its slope is 0, so find x for when the tangent line's slope is equal to 0: 0 = 2x+1 -1=2x x = -1/2

  3. anonymous
    • 4 years ago
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    y = -2

  4. anonymous
    • 4 years ago
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    @above? how is the line vertical and tangent to that graph at the same time?

  5. anonymous
    • 4 years ago
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    oh i see, maybe, sorry

  6. Xishem
    • 4 years ago
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    The line is horizontal, because the x-axis is horizontal. Two horizontal lines are parallel.

  7. anonymous
    • 4 years ago
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    yes i didnt see that it said x-value, i put the actual line that was tangent and parallel

  8. anonymous
    • 4 years ago
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    and i messed that part up too!

  9. anonymous
    • 4 years ago
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    the equation for the tangent line would be y = 51/4

  10. Xishem
    • 4 years ago
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    The equation for the tangent line would be: \[y=\frac{-1}{2}x+13\]

  11. anonymous
    • 4 years ago
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    no, plug the axis of symmetry (-1/2) in for x to get 12 3/4

  12. anonymous
    • 4 years ago
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    and yours isnt even horizontal

  13. Xishem
    • 4 years ago
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    I apologize. The equation for the tangent line would be: \[y=13\] \[(\frac{-1}{2},\frac{51}{4}) \]is actually the vertex point for the original equation

  14. anonymous
    • 4 years ago
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    y = 13? That would touch it at two points though. You got the vertex, now you just need a horizontal line with the y-coordinate of the vertex so that it touches the vertex.

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