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

1. mathcalculus

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

2. mathcalculus

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. akotto4897

omg this is so simple

5. akotto4897

wat is america coming to

6. mathcalculus

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

7. mathcalculus

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. mathcalculus

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. mathcalculus

woah so the slope is good and the y too.

13. mathcalculus

oh i see there's an x there.

14. mathcalculus

so would it be 247e^27-18?

15. zepdrix

for your y_tan? or for b?

16. mathcalculus

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. mathcalculus

so how do i get to the answer.

19. mathcalculus

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

20. zepdrix

grr sorry website crashed on me >:c

21. mathcalculus

its fine

22. mathcalculus

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

23. mathcalculus

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. mathcalculus

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

27. zepdrix

y-?

28. mathcalculus

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

29. mathcalculus

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

30. mathcalculus

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. mathcalculus

33. mathcalculus

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

34. mathcalculus

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. mathcalculus

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. mathcalculus

yes correct! :)

39. mathcalculus

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. mathcalculus

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

43. mathcalculus

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

44. mathcalculus

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. mathcalculus

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

48. mathcalculus

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. mathcalculus

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

51. mathcalculus

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. mathcalculus

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. mathcalculus

now what? :(

57. mathcalculus

where does my foiling go?

58. mathcalculus

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. mathcalculus

right

62. mathcalculus

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. mathcalculus

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. mathcalculus

damn im so confused.

70. mathcalculus

=[

71. mathcalculus

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. mathcalculus

wait which is the mx and b?

76. mathcalculus

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}$