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anonymous

  • 4 years ago

This what I have so far...

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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    f'(x)=4x x=2 f'(2)=4(2)=8 which is m x=2 f(2)=2(2)^2+3=11 which is y (8,11) y-11=8(x-2) y=8x-5 what do i do next to find y1 and y2?

  3. anonymous
    • 4 years ago
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    you need the equation for the line tangent to the graph. did you find that?

  4. anonymous
    • 4 years ago
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    huh?

  5. anonymous
    • 4 years ago
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    you found it right?

  6. anonymous
    • 4 years ago
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    the tangent line at (2,11) is \[y=8x-5 \]

  7. anonymous
    • 4 years ago
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    so if you want the y value at some other point, replace x by the x value given

  8. anonymous
    • 4 years ago
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    16-5=11

  9. anonymous
    • 4 years ago
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    yes, and you knew that one to begin with right?

  10. anonymous
    • 4 years ago
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    yes

  11. anonymous
    • 4 years ago
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    since that is the point you used in the first place to find the line now replace x by -1 to find another point on the line

  12. anonymous
    • 4 years ago
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    -8-5=-13

  13. anonymous
    • 4 years ago
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    i guess. was the other x value -1?

  14. anonymous
    • 4 years ago
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    yes

  15. anonymous
    • 4 years ago
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    if so then you are right.

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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