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

1. anonymous

f'(x)= 78xe^(3x^2)-6

2. anonymous

slope= 234e^27-6 y=13e^27-18

3. zepdrix

So you're looking for an equation of the line tangent to the curve f(x) at x=3?

4. anonymous

omg this is so simple

5. anonymous

wat is america coming to

6. anonymous

yes, Determine the equation of the line tangent to the graph of

7. anonymous

yeah but I'm not sure why I'm getting this wrong.

8. zepdrix

@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

Ok that's enough of my rant :) lemme check your work a sec :3

10. anonymous

lol, i appreciate that. but yes, let's begin.

11. zepdrix

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

woah so the slope is good and the y too.

13. anonymous

oh i see there's an x there.

14. anonymous

so would it be 247e^27-18?

15. zepdrix

for your y_tan? or for b?

16. anonymous

y tan

17. zepdrix

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

so how do i get to the answer.

19. anonymous

at first i thought it was 247e^27 -6x

20. zepdrix

grr sorry website crashed on me >:c

21. anonymous

its fine

22. anonymous

i just don't understand what to do at the point slope formula.

23. anonymous

i keep getting the answer wrong.

24. zepdrix

$\Large\bf y_{\tan}\quad=\quad \color{royalblue}{m}x+\color{royalblue}{b}$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

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

isn't it y-(13e^27-18)=234e^27-6x+18

27. zepdrix

y-?

28. anonymous

right, y-y1=m(x-x1)

29. anonymous

so in this case, slope= 234e^27-6 y=13e^27-18

30. anonymous

and x1= 3

31. zepdrix

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

33. anonymous

yeah but i thought i could use that to find the equation of a tangent line

34. anonymous

I've been using that for other problems and it worked.

35. zepdrix

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

ok

37. zepdrix

$\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

yes correct! :)

39. anonymous

after that I'm lost lol

40. zepdrix

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

$\large\bf y-\color{#DD4747 }{(13e^{27}-18)}\quad=\quad 234e^{27}x-6x -702e^{27}+18$I think it expands like that, yes? Give it a try :)

42. anonymous

yup, but i seem to be doing that wrong unfortunately.

43. anonymous

wait don't we just multiply out the -6 and -6*-3 only?

44. anonymous

and leave the

45. zepdrix

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

$\Large 234e^{27}\cdot-3$

47. anonymous

why don't we multiply 234e^27 with x?

48. anonymous

and then -3

49. zepdrix

$\large\bf (234e^{27}-6)(x-3)\quad=\\ \large\bf\quad (234e^{27}-6)x-(234e^{27}-6)3$And we only want to multiply out the part with the 3.

50. anonymous

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

so for (234e^27−6)(x−3) i thought we do foil

52. zepdrix

Yes, foiling should give you this,$\large\bf 234e^{27}x-6x -702e^{27}+18$

53. zepdrix

Are you not getting that when you foil..?

54. anonymous

yes i have that! :)

55. zepdrix

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

now what? :(

57. anonymous

where does my foiling go?

58. anonymous

great so we factor?

59. zepdrix

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

We finish by solving for y.

61. anonymous

right

62. anonymous

x(-468+12)

63. zepdrix

$\large\bf y=\\\large\bf (234e^{27}-6)x -702e^{27}+18+13e^{27}-18$

64. zepdrix

what..? :(

65. zepdrix

the -6 and the 18 are not like terms, we can't combine those.

66. zepdrix

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

67. anonymous

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

This one only ended up being complicated because our slope was 2 terms. So it ended up being a ton of multiplication.

69. anonymous

damn im so confused.

70. anonymous

=[

71. anonymous

what would be he answer btw?

72. zepdrix

$\large\bf y-(13e^{27}-18)\quad=\quad (234e^{27}-6)x -702e^{27}+18$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

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

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

wait which is the mx and b?

76. anonymous

oh it is correct….

77. zepdrix

$\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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