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anonymous

  • 5 years ago

Find an equation for a line that is tangent to f(x)=x^2-2x-3

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  1. myininaya
    • 5 years ago
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    for any point? or do you want a formula for any point?

  2. anonymous
    • 5 years ago
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    it doesn't specify, it just says to find an equation.

  3. myininaya
    • 5 years ago
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    ok say we want to find the tangent line at x=a

  4. myininaya
    • 5 years ago
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    f'(x)=2x-2

  5. myininaya
    • 5 years ago
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    so the slope at x=a is f'(a)=2a-2

  6. myininaya
    • 5 years ago
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    a line has the form y=mx+b

  7. myininaya
    • 5 years ago
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    we know the slope m is 2a-2

  8. myininaya
    • 5 years ago
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    so now we have y=(2a-2)x+b we now need to find the y-intercept

  9. myininaya
    • 5 years ago
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    we know a point on this line (a,f(a))

  10. myininaya
    • 5 years ago
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    f(a)=a^2-2(a)-3

  11. myininaya
    • 5 years ago
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    a^2-2a-3=(2a-2)a+b

  12. myininaya
    • 5 years ago
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    so we now solve this for b to find the y intercept

  13. myininaya
    • 5 years ago
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    so we have y=(2a-2)x-a^2-3

  14. myininaya
    • 5 years ago
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    this is the general equation for the tangent line at (a,f(a))

  15. anonymous
    • 5 years ago
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    you think i can just pick a point off the original quadratic function and use those points, such as the point (3,0)?

  16. myininaya
    • 5 years ago
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    It does say to find *an* so maybe it doesn't matter, but make sure the point is on the curve ( I mean verify that it is) Does f(3)=0?

  17. anonymous
    • 5 years ago
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    since a point was not specified in the original problem and yes, f(3)=0.

  18. myininaya
    • 5 years ago
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    ok you can use this point as long as we aren't looking for anything general.

  19. anonymous
    • 5 years ago
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    ok thanks so much, i don't need to find the x and y intercepts right?

  20. myininaya
    • 5 years ago
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    no. you can just use the general tangent line that I came up with. lol

  21. myininaya
    • 5 years ago
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    but your teacher will probably want you to show work

  22. anonymous
    • 5 years ago
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    ok thanks so much for your help :)

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