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micolperu317 Group Title

Find an equation for the line tangent to the graph of the given function at the indicated point. f(x) = x2-3 at (3,6)

  • 2 years ago
  • 2 years ago

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  1. sputnik Group Title
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    The first thing you need to do is find out how fast the function f(x) is changing at the given point. And to do that you need the derivative f '(x). Would you like help finding the derivative?

    • 2 years ago
  2. micolperu317 Group Title
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    Yes, please

    • 2 years ago
  3. sputnik Group Title
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    You can look at the parts of f(x) individually. Umm, is the function f(x) = 2x - 3 or f(x) = x^2 - 3?

    • 2 years ago
  4. micolperu317 Group Title
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    \[f(x)= x ^{2}-3 at (3,6)\]

    • 2 years ago
  5. sputnik Group Title
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    Nice. So to find the derivative of the x^2 part, you need to use the power rule. The two comes down and becomes the coefficient, and that is raised to 2-1 = 1. Which is just x so I'm going to leave it out. The derivative of the - 3 is zero, because the derivative is the rate of change and the rate of change of anything constant is 0. The whole derivative is the sum of the parts: f ' (x) = 2x + 0 Which is just f ' (x) = 2x.

    • 2 years ago
  6. sputnik Group Title
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    Next you need to find out what the derivative is when x = 3. f ' (3) = 2 * 3 f ' (3) = 6 So that is going to be the slope m in your y = mx + b for the tangent. Hopefully this makes sense, I've never tutored anyone before in my life. ;D

    • 2 years ago
  7. micolperu317 Group Title
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    It's telling me the answer should be y=2x-12

    • 2 years ago
  8. micolperu317 Group Title
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    Thank you, your actually helping me see things I haven't noticed. I'm sorry, this course is really hard for me. I've been doing my homework for like 4 hours :(

    • 2 years ago
  9. sputnik Group Title
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    No prob. I wish I got the *$(#) problem right though. ;/

    • 2 years ago
  10. sputnik Group Title
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    No, it shouldn't be that line y = 2x - 12. I just plotted them both on my graphing calculator and they don't even intersect.

    • 2 years ago
  11. sputnik Group Title
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    And I plotted y = 6x - 12 and it looks right. Should we go on working through the problem and see if we get - 12 for the y intercept?

    • 2 years ago
  12. micolperu317 Group Title
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    It is 6x-12, I'm sorry I must not being seeing right. I'm getting tired of learning lol. How do you plot it and get an equation?

    • 2 years ago
  13. sputnik Group Title
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    Once you have figured out the slope from the derivative, it is really easy to get the equation for the tangent line. You know it has the form y = mx + b. And you have an m from solving the derivative. And you have an x and y, points it passes through. The only variable you don't have is b. 6 = 6 * 3 + b And you solve for b. 6 = 18 + b - 12 = b And you put that back into your y = mx + b for the tangent line: y = 6x - 12.

    • 2 years ago
  14. micolperu317 Group Title
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    Thanks! woW, WHAT A RELIEF.

    • 2 years ago
  15. micolperu317 Group Title
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    Since you seem to know about this stuff, do you know how to differentiate? This problem has been breaking my head for hours: f(x)= \[f(x)= (x ^{2}-4x+2)(5x ^{3}-x ^{2}+5)\]

    • 2 years ago
  16. sputnik Group Title
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    The whole thing f(x) is a product of two parts. So you should use the product rule. f ' (x) = (the derivative of the 1st part) * (the 2nd part) + (the derivative of 2nd part) * (the first part). The 1st part = x^2 - 4x + 2 The 2nd part = 5x^3 - x^2 + 5 You find the derivatives for the 1st and 2nd parts by using the power rule like for the other problem. Are you cool with how to do that or would you like help?

    • 2 years ago
  17. micolperu317 Group Title
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    Yes i know I have to do that, but I keep missing up and I don't get the answer. Help me please :)

    • 2 years ago
  18. micolperu317 Group Title
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    messing up*

    • 2 years ago
  19. sputnik Group Title
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    OK. I'll the 1st and let you try the 2nd? So for the first part, the first term is x ^ 2. The coefficient in front of the x^2 is 1. So when I bring the exponent down and multiply it by the coefficient, it is 2 x 1 = 2. The new exponent is the previous power, 2, minus 1. So 2 - 1 = 1. For the next part, -4x, the derivative is just -4. And for the constant at the end, +2, constants don't change so the derivative of that is 0. To get the entire expression for the derivative of the 1st part, add up those parts. Derivative 1st part = 2x - 4 + 0. Hopefully that makes sense.

    • 2 years ago
  20. micolperu317 Group Title
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    Is the second one this: \[15x ^{4}-2x+0\]

    • 2 years ago
  21. micolperu317 Group Title
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    well without the x for the #2,

    • 2 years ago
  22. sputnik Group Title
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    For the first term, you brought the 5*3 right. But then you added to the exponent, 3, instead of subtracting from it. It should look like 15x^2 - 2x + 0. Lemme play with the equation editor thingbob... \[15x^2 - 2x + 0\]

    • 2 years ago
  23. micolperu317 Group Title
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    Is that what I do now? (2x+4)(15x^2+2x) + (15x^2+2x)(2x+4)

    • 2 years ago
  24. sputnik Group Title
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    f ' (x) = (deriviatve 1st part)*(2nd part) + (derivative 2nd part)*(1st part) 1st part = x^2 - 4x + 2 2nd part = 5x^3 - x^2 + 5 Derivative 1st part = 2x - 4 Drivative 2nd part = 15x^2 - 2x So: f '(x) = \[f'(x) = (2x - 4)(5x^3 - x^2 + 5) + (15x^2 - 2x)( x^2 - 4x + 2)\] Nasty to simplify but it is all algebra from here. :D

    • 2 years ago
  25. sputnik Group Title
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    I think the algebra works out to: Expand... \[10x^4 - 2x^3 + 5x - 20x^3 + 4x^2 - 20 + 15x^4 - 60x^3 + 30 - 2x^3 + 8x^2 - 4x\] Combine like terms of the same power: \[ 25x^4 - 84x^3 + 12x^2 + x - 20\] Hopefully all the algebra came out in the wash.

    • 2 years ago
  26. micolperu317 Group Title
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    I GOT IT! lol sorry i was trying to solve it, but it took me forever! :)

    • 2 years ago
  27. micolperu317 Group Title
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    Your really good at this!!!

    • 2 years ago
  28. sputnik Group Title
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    Thanks. You'll be too, it just takes some patience and getting used to. Then calc is fun.

    • 2 years ago
  29. micolperu317 Group Title
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    LOL I wish it was fun, I've been at it for days. I just dioscover this website today

    • 2 years ago
  30. sputnik Group Title
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    Me too. It looks really fun but I don't know anyone here. :D I have to go soon but it has been real nice talking with you!

    • 2 years ago
  31. sputnik Group Title
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    They look like they really know their stuff here.

    • 2 years ago
  32. micolperu317 Group Title
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    You do too! Good Night :)

    • 2 years ago
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