## anonymous one year ago Find the equation of the tangent line to f(x) = 9 − x^2 through the point (5, 0) that touches f (x) at a point below the x-axis.

1. dan815

first you need to find the slope at this point

2. anonymous

-2x ?

3. dan815

that is the function of all the slopes wrt to x

4. dan815

f(x) is telling you all the heights at every x f'(x) is teelling you all the slopes at every x now at what x do we need to know our f'(x)

5. anonymous

the point where it touches f(x) which is below the x axis?

6. dan815

take a look at your point

7. dan815

(5, 0)

8. dan815

whats the x value of the point

9. anonymous

5

10. dan815

right so we need to know f'(5) = ?

11. anonymous

But that point doesn't even lie on the curve...

12. dan815

we just want the slope there at x=5

13. anonymous

so that would be -2(5) = -10, right?

14. dan815

oh wait 5,0 isnt on this curve really?

15. dan815

u are right lol xD

16. anonymous

Yeah cause try plugging it in the function

17. anonymous

XD

18. anonymous

I think we should first find a point that apparently lies somewhere under the x-axis

19. dan815

k lets graph and see what exactly is going on here

20. dan815

|dw:1435217659239:dw|

21. dan815

|dw:1435217751709:dw|

22. dan815

okay so eventually this has to happen, they want us to find this x,y

23. anonymous

Yep!

24. anonymous

So are we solving for the secant line?

25. anonymous

between (5,0) and (a, 9-a^2)

26. dan815

you already got that the slope was F'(x)=-2x

27. dan815

right

28. anonymous

Right

29. dan815

now write the slope with those 2 points

30. ganeshie8

Any line through \((5,0)\) is of form \[y=m(x-5)\tag{1}\] we want that line to "just touch" below parabola \[y=9-x^2\tag{2}\] simply set both equations equal to each other and use the discriminant to find the slope, \(m\)

31. dan815

and equate it to the F'(x)

32. dan815

|dw:1435217973229:dw|

33. anonymous

Yeah lol i was typing the same xD

34. anonymous

We get a quadratic equation when we solve for a... that means we'll get 2 values for a...

35. anonymous

a^2 - 10a + 9 = 0

36. anonymous

thats a 9 :P

37. dan815

yes

38. ganeshie8

Yes, two tangent lines can be drawn to a parabola form an external point

39. anonymous

lol xD ok so a=1 and x=9 ? Is that right?

40. anonymous

i mean a* = 9

41. ganeshie8

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42. ganeshie8

a=9 is right, but how did u figure ?

43. anonymous

but neither of the 2 points i found for 'a' lie under the x-axis

44. anonymous

I solved the quadratic a^2 -10a +9

45. ganeshie8

okay i see you simply discarded a=1 without any reason

46. anonymous

nope i wrote that as well lol ^^ check again :P

47. anonymous

What happens next?

48. ganeshie8

Okay just making sure :) plugin x=a=9 into the equation, you do get a point under the x axis

49. anonymous

I do? Ohh thats right! xD

50. anonymous

So that means I get 2 tangent lines?

51. ganeshie8

plugin x=a=1, you get y = 9-x^2 = 9-1 = 8 that means the tangent line touches the parabola at (1, 8) which is above x axis so discard this

52. anonymous

Ohh nice!!

53. anonymous

I love discarding stuff in math xD haha

54. ganeshie8

lol we're very much alike :P

55. dan815

i am not following exactly what u are doing here ganeshie y=m(x-5) y=9-x^2

56. dan815

are u then setting m=-2x to solve it

57. dan815

arent there multiple solutions to inf solutions to m that intersect with this poiint on the parabola, how are you limiting to the case where its just touching

58. ganeshie8

nope set both equations equal to each other and use discriminant : m(x-5) = 9-x^2 x^2 + mx - (5m+9) = 0 For this quadratic to have a single solution, the discriminant must be 0 : m^2 + 4(5m+9) = 0

59. ganeshie8

solving that gives the slope of two tangent lines passing thru (5, 0)

60. dan815

oh i see! what you mean i didnt understand what discriminant meant lol

61. dan815

nice trick

62. anonymous

neither did I o.o XD lol

63. Astrophysics

Lol that had me confused to but I was too scared to ask, nice one.

64. ganeshie8

thats okay, stick to dan's method as it is more calculussy

65. Astrophysics

I feel like an idiot now xD

66. anonymous

No I totally understood everything we did!! I just didn't get what you were explaining to dan at the end xP

67. anonymous

@Astrophysics pshhhh in that case i'm more of an idiot that everyone! XD lol

68. dan815

basically what he did was, well normally to a line passing through that point (5,0) to your parabola there will be 2 intersections everywhere except for 1 place

69. ganeshie8

reminds me of 3 idiots movie

70. anonymous

Ahhh i see

71. dan815

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72. anonymous

Lmao XD truuueee :D so much humour the night before my exam :D lol

73. anonymous

hmmm i see it now

74. ganeshie8

just to conclude : x=a = 9 gives slope=-2x=-2*9 = -18 so the equation of desired tangent line is y = -18(x-5)

75. dan815

right, and u know how from the quadratic equation the 2 differnet solutions come from that +/- part in squareroot

76. dan815

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77. anonymous

Yep! Got it all :)

78. dan815

k good

79. anonymous

Thanks so much!!! :)

80. dan815

yeah sure =]

81. Astrophysics

Thank you teachers!

82. anonymous

Yeshh :D

83. Astrophysics

Are all these questions part of a practice exam?

84. anonymous

Yeah, they're some sample/practice questions... or whatever they call it XD lol

85. anonymous

@Astrophysics :D sorry about the late response, wasn't paying attention lol

86. Astrophysics

Lol no worries was helping someone anyways, haha and good luck, calc exams have always been long >.<

87. anonymous

thank you! :) Yeah it's gonna be 2.5 hours 8:30 in the morning >.<

88. Astrophysics

Eek, haha get some sleep as well! Later :P

89. anonymous

I'd better xD lol otherwise its a big fat 0 xD haha