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anonymous

  • 2 years ago

PLEASE HELP :( !! y=13e^(3x^2)-6x at x=3 in the form y=mx +b has

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  1. anonymous
    • 2 years ago
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    f'(x)= 78xe^(3x^2)-6

  2. anonymous
    • 2 years ago
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    slope= 234e^27-6 y=13e^27-18

  3. zepdrix
    • 2 years ago
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    So you're looking for an equation of the line tangent to the curve f(x) at x=3?

  4. anonymous
    • 2 years ago
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    omg this is so simple

  5. anonymous
    • 2 years ago
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    wat is america coming to

  6. anonymous
    • 2 years ago
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    yes, Determine the equation of the line tangent to the graph of

  7. anonymous
    • 2 years ago
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    yeah but I'm not sure why I'm getting this wrong.

  8. zepdrix
    • 2 years ago
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    @akotto4897 wtf is wrong with you? Why would you call someone elses problem `so simple`. It's like someone in 5th grade calling 3rd grade math SUPER EASY! Just because you understand it doesn't mean that everyone else is on your level...

  9. zepdrix
    • 2 years ago
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    Ok that's enough of my rant :) lemme check your work a sec :3

  10. anonymous
    • 2 years ago
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    lol, i appreciate that. but yes, let's begin.

  11. zepdrix
    • 2 years ago
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    Mmm ok so your slope looks good. So what coordinate pair are we using to find our y-intercept?\[\Large f(3)\quad=\quad 13e^{27}-18\]Mmm ok ok I see you have the written down already :3 \[\Large y_{\tan}\quad=\quad mx+b\]So if we plug in our coordinate pair, we get something like this, yes?\[\Large 13e^{27}-18\quad=\quad (234e^{27}-6)\cdot 3+b\]

  12. anonymous
    • 2 years ago
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    woah so the slope is good and the y too.

  13. anonymous
    • 2 years ago
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    oh i see there's an x there.

  14. anonymous
    • 2 years ago
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    so would it be 247e^27-18?

  15. zepdrix
    • 2 years ago
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    for your y_tan? or for b?

  16. anonymous
    • 2 years ago
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    y tan

  17. zepdrix
    • 2 years ago
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    Hmm after you find your b value, you should plug it back into the \[\Large\bf y_{tan}\quad=\quad \color{royalblue}{m}x+\color{royalblue}{b}\] For our final answer, we DON'T want the coordinate pair plugged in. We're only trying to fill in these blue pieces.

  18. anonymous
    • 2 years ago
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    so how do i get to the answer.

  19. anonymous
    • 2 years ago
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    at first i thought it was 247e^27 -6x

  20. zepdrix
    • 2 years ago
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    grr sorry website crashed on me >:c

  21. anonymous
    • 2 years ago
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    its fine

  22. anonymous
    • 2 years ago
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    i just don't understand what to do at the point slope formula.

  23. anonymous
    • 2 years ago
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    i keep getting the answer wrong.

  24. zepdrix
    • 2 years ago
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    \[\Large\bf y_{\tan}\quad=\quad \color{royalblue}{m}x+\color{royalblue}{b}\]Plugging in our slope,\[\Large\bf y_{\tan}\quad=\quad \color{royalblue}{(234e^{27}-6)}x+\color{royalblue}{b}\]Then we plug in our coordinate pair to find b,\[\Large\bf \color{#DD4747 }{13e^{27}-18}\quad=\quad \color{royalblue}{(234e^{27}-6)}\cdot\color{#DD4747 }{3}+\color{royalblue}{b}\]

  25. zepdrix
    • 2 years ago
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    So we have some -18's cancelling out. And I think we get a b value of,\[\Large\bf \color{royalblue}{b\quad=\quad -689e^{27}}\]Something like that?

  26. anonymous
    • 2 years ago
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    isn't it y-(13e^27-18)=234e^27-6x+18

  27. zepdrix
    • 2 years ago
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    y-?

  28. anonymous
    • 2 years ago
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    right, y-y1=m(x-x1)

  29. anonymous
    • 2 years ago
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    so in this case, slope= 234e^27-6 y=13e^27-18

  30. anonymous
    • 2 years ago
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    and x1= 3

  31. zepdrix
    • 2 years ago
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    No that's `point-slope` form of a line. y-y1 = m( x-x1) We don't want to use that. We were told to use the `slope-intercept` form of a line. y=mx+b

  32. anonymous
    • 2 years ago
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    can we do it the way i had it please?

  33. anonymous
    • 2 years ago
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    yeah but i thought i could use that to find the equation of a tangent line

  34. anonymous
    • 2 years ago
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    I've been using that for other problems and it worked.

  35. zepdrix
    • 2 years ago
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    Yes we can use that, so let's see if we can get it set up correctly. Lemme see if I can match what you wrote a sec :)

  36. anonymous
    • 2 years ago
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    ok

  37. zepdrix
    • 2 years ago
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    \[\Large\bf y-y_1\quad=\quad \color{royalblue}{m}(x-x_1)\]\[\Large\bf y-y_1\quad=\quad \color{royalblue}{(234e^{27}-6)}(x-x_1)\]Then plugging in our point:\[\Large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad \color{royalblue}{(234e^{27}-6)}(x-\color{#DD4747 }{3})\]

  38. anonymous
    • 2 years ago
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    yes correct! :)

  39. anonymous
    • 2 years ago
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    after that I'm lost lol

  40. zepdrix
    • 2 years ago
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    Ok so now we just need to get in into the form y=mx+b. So we need to multiply out the brackets, then we gotta isolate the y term.

  41. zepdrix
    • 2 years ago
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    \[\large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad 234e^{27}x-6x -702e^{27}+18\]I think it expands like that, yes? Give it a try :)

  42. anonymous
    • 2 years ago
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    yup, but i seem to be doing that wrong unfortunately.

  43. anonymous
    • 2 years ago
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    wait don't we just multiply out the -6 and -6*-3 only?

  44. anonymous
    • 2 years ago
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    and leave the

  45. zepdrix
    • 2 years ago
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    Yes, true. We probably want don't want the x's expanded like that,\[\large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad (234e^{27}-6)x -702e^{27}+18\]

  46. zepdrix
    • 2 years ago
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    \[\Large 234e^{27}\cdot-3\]

  47. anonymous
    • 2 years ago
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    why don't we multiply 234e^27 with x?

  48. anonymous
    • 2 years ago
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    and then -3

  49. zepdrix
    • 2 years ago
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    \[\large\bf (234e^{27}-6)(x-3)\quad=\\ \large\bf\quad (234e^{27}-6)x-(234e^{27}-6)3\]And we only want to multiply out the part with the 3.

  50. anonymous
    • 2 years ago
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    im so confused. i do't understand why we did this problem differently than all the others when it comes to this part.

  51. anonymous
    • 2 years ago
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    so for (234e^27−6)(x−3) i thought we do foil

  52. zepdrix
    • 2 years ago
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    Yes, foiling should give you this,\[\large\bf 234e^{27}x-6x -702e^{27}+18\]

  53. zepdrix
    • 2 years ago
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    Are you not getting that when you foil..?

  54. anonymous
    • 2 years ago
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    yes i have that! :)

  55. zepdrix
    • 2 years ago
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    We don't want multiple x terms, so we'll factor an x out of each of the first two terms.\[\Large\bf =\quad \large\bf (234e^{27}-6)x -702e^{27}+18\]

  56. anonymous
    • 2 years ago
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    now what? :(

  57. anonymous
    • 2 years ago
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    where does my foiling go?

  58. anonymous
    • 2 years ago
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    great so we factor?

  59. zepdrix
    • 2 years ago
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    We had this,\[\Large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad \color{royalblue}{(234e^{27}-6)}(x-\color{#DD4747 }{3})\]It foiled to give us this,\[\large\bf y-(13e^{27}-18)\quad=\quad 234e^{27}x-6x -702e^{27}+18\]We factored out an x to give us this,\[\large\bf y-(13e^{27}-18)\quad=\quad (234e^{27}-6)x -702e^{27}+18\]

  60. zepdrix
    • 2 years ago
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    We finish by solving for y.

  61. anonymous
    • 2 years ago
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    right

  62. anonymous
    • 2 years ago
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    x(-468+12)

  63. zepdrix
    • 2 years ago
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    \[\large\bf y=\\\large\bf (234e^{27}-6)x -702e^{27}+18+13e^{27}-18\]

  64. zepdrix
    • 2 years ago
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    what..? :(

  65. zepdrix
    • 2 years ago
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    the -6 and the 18 are not like terms, we can't combine those.

  66. zepdrix
    • 2 years ago
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    I don't know why you insisted on using point-slope form for this one. It seems way more complicated :c

  67. anonymous
    • 2 years ago
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    but it's the only way for me… i have to stick to this or else i'll be completely lost in the exam

  68. zepdrix
    • 2 years ago
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    This one only ended up being complicated because our slope was `2 terms`. So it ended up being a ton of multiplication.

  69. anonymous
    • 2 years ago
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    damn im so confused.

  70. anonymous
    • 2 years ago
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    =[

  71. anonymous
    • 2 years ago
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    what would be he answer btw?

  72. zepdrix
    • 2 years ago
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    \[\large\bf y-(13e^{27}-18)\quad=\quad (234e^{27}-6)x -702e^{27}+18\]From here: you add (13e^{27}-18) to each side,\[\large\bf y=\\ \large\bf(234e^{27}-6)x -702e^{27}+18\color{orangered}{+(13e^{27}-18)}\]Does that step make sense? +_+ That's how we isolate the y term.

  73. zepdrix
    • 2 years ago
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    From there we can drop the brackets since nothing in being applied to them, \[\large\bf y=\\ \large\bf(234e^{27}-6)x -702e^{27}+18+13e^{27}-18\]It looks like the 18's will cancel out,\[\large\bf y=\\ \large\bf(234e^{27}-6)x -702e^{27}+\cancel{18}+13e^{27}-\cancel{18}\]

  74. zepdrix
    • 2 years ago
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    And the exponential terms ( not the one attached to the x ) are like terms, so we'll combine them,\[\large\bf y=\\ \large\bf(234e^{27}-6)x \color{orangered}{-702e^{27}+13e^{27}}\] \[\large\bf y=(234e^{27}-6)x \color{orangered}{-689e^{27}}\]And I think that's our final answer, assuming I didn't make any boo boos along the way.

  75. anonymous
    • 2 years ago
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    wait which is the mx and b?

  76. anonymous
    • 2 years ago
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    oh it is correct….

  77. zepdrix
    • 2 years ago
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    \[\large\bf y=\color{royalblue}{(234e^{27}-6)}x \color{orangered}{-689e^{27}}\]\[\large\bf y=\color{royalblue}{m}x +\color{orangered}{b}\]

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