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anonymous

  • one year ago

Find an equation of the tangent line to the graph f(x)=2-7x^2 at (2,-26)

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  1. amistre64
    • one year ago
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    what are 2 things we need to know to define a line?

  2. anonymous
    • one year ago
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    two points?

  3. amistre64
    • one year ago
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    well, 2 points are useful yes, but we only have one point given to us ... what else could we use to define a line with? a point and ....

  4. anonymous
    • one year ago
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    slope?

  5. amistre64
    • one year ago
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    correct, so a point and a slope can be used to define a line. its called the point slope form of the equation. now, how do we determine the slope of a curve at a given point?

  6. anonymous
    • one year ago
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    the change in y over the change in x? so the point slope equation?

  7. amistre64
    • one year ago
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    hmm, im assuming you know what a derivative is for this to work.

  8. anonymous
    • one year ago
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    yeah but I don't really get derivatives

  9. anonymous
    • one year ago
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    I know there's 4 types

  10. amistre64
    • one year ago
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    you should have a simple rule you can apply constant rule, and a power rule

  11. amistre64
    • one year ago
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    \[\frac{d}{dx}k=0\] \[\frac{d}{dx}x^n=n~x^{n-1}\]

  12. anonymous
    • one year ago
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    I'm just having a hard time applying the equation to these rules

  13. amistre64
    • one year ago
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    9 - x^2 ^ ^ ^ this is a power of x ^ and this is a constant

  14. amistre64
    • one year ago
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    what is the derivative of 9? what is the derivative of -x^2?

  15. amistre64
    • one year ago
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    e3rr, -7x^2 ... forgot the equation we were dealing with :/

  16. anonymous
    • one year ago
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    so for 2 it would be 0 and -7x^2 would be -7x^-6?

  17. amistre64
    • one year ago
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    \[\frac{d}{dx}(2-7x^2)\] \[\frac{d}{dx}(29)-\frac{d}{dx}(7x^2)\] \[\frac{d}{dx}(2)-7\frac{d}{dx}(x^2)\] \[0-7(2x^{2-1})\]

  18. anonymous
    • one year ago
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    how did you get 29?

  19. amistre64
    • one year ago
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    typo, slip of the fingers.

  20. anonymous
    • one year ago
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    can you please explain how you got the answer step by step?

  21. amistre64
    • one year ago
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    i jsut did ... except for 29 which is spose to be a 2 instead use the constant rule for the constant, and the power rule for the power of x there is also a 'pull out' rule which lets us pull out that 7 to the outside.

  22. amistre64
    • one year ago
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    your -7x^2 derivative is off ... try it again

  23. amistre64
    • one year ago
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    the format for the tangent line is just gonna be: y - f(a) = f'(a) (x-a) so as long as we can determine f'(a) we are set

  24. anonymous
    • one year ago
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    sorry my book shows different explanations so I'm a little confused right now

  25. amistre64
    • one year ago
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    tell me your thoughts then, what does your book explain?

  26. anonymous
    • one year ago
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    it basically shows me all these rules but all these variables are confusing me

  27. anonymous
    • one year ago
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    but the examples only show me if I have a point and slope

  28. amistre64
    • one year ago
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    what is your rule for powers of x?

  29. amistre64
    • one year ago
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    can you take a picture of it?

  30. anonymous
    • one year ago
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    For any real number k if y=x^k then d/dx(x^h)= k*x^k-1

  31. amistre64
    • one year ago
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    let k=2 x^2 become 2x^(2-1) or simply 2x

  32. amistre64
    • one year ago
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    so our derivative is -7x^2 derives to -7(2x), at x=2 that is ... -7(4)

  33. anonymous
    • one year ago
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    so the 2 becomes x?

  34. amistre64
    • one year ago
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    out point (2,-26) defines the value of x as 2

  35. amistre64
    • one year ago
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    *our point ...

  36. amistre64
    • one year ago
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    2x, when x=2 is just ... 2(2) right?

  37. anonymous
    • one year ago
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    how did x^2 become 2x?

  38. amistre64
    • one year ago
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    by the derivative rule you posted. let k=2

  39. amistre64
    • one year ago
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    x^k derives to k*x^(k-1)

  40. anonymous
    • one year ago
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    Ohhhh I get it now

  41. amistre64
    • one year ago
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    by first principles of limit \[\lim_{h\to0}\frac{(x+h)^2-x^2}{(x+h)-x}\] \[\lim_{h\to0}\frac{x^2+h^2+2xh-x^2}{h}\] \[\lim_{h\to0}\frac{h^2+2xh}{h}\] \[\lim_{h\to0}\frac{h(h+2x)}{h}\] \[\lim_{h\to0}(h+2x)=2x\]

  42. anonymous
    • one year ago
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    but I'm still pretty confused

  43. amistre64
    • one year ago
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    the 'rules' are really a table of outcomes that can be applied so that you dont have to use the long limiting process each and every time.

  44. anonymous
    • one year ago
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    so now that we know the derivative of -7x^2 to be -7(2x) which is -7(2*2)= -28 what do we do next?

  45. amistre64
    • one year ago
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    well, thats our slope of our tangent line. the rest is from algebra class.

  46. amistre64
    • one year ago
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    construct the line using your slope and the stated point

  47. anonymous
    • one year ago
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    y=mx+b

  48. anonymous
    • one year ago
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    or point slope equation y-y1=m(x-x1)

  49. amistre64
    • one year ago
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    y - f(a) = f'(a) (x-a) or rewritten as y = f'(a) (x-a) +f(a)

  50. amistre64
    • one year ago
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    point slope equation

  51. anonymous
    • one year ago
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    y-(-26)=-28(x-2)

  52. amistre64
    • one year ago
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    thatll do it

  53. anonymous
    • one year ago
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    y=-28x+30

  54. anonymous
    • one year ago
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    could I get a two out of it to be 2(-14x+15)? or is okay like that?

  55. amistre64
    • one year ago
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    usually we dont factor it, keep it in y=mx+b form is conventional

  56. anonymous
    • one year ago
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    so that's my final answer? thank you! is there any way I reward you on here? it's my first time using this site

  57. amistre64
    • one year ago
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    there should be a 'best response' button to everything ive posted. clicking it gives out a medal. but saying thank you and learning a little bit is reward enough for me to be honest.

  58. anonymous
    • one year ago
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    okay thank you!!

  59. amistre64
    • one year ago
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    youre welcome :) and good luck

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