## anonymous one year ago Someone Please Help, Giving A Medal A.Find the slope of the tangent line to the graph of f at the given point. A.Find the slope-intercept equation of the tangent line to the graph of f at the given point.

1. anonymous

2. welshfella

find the derivative f(x) = 4x + 2 can you find f'(x)?

3. anonymous

-1/2+x/4

4. anonymous

no the derivative of that is 4

5. welshfella

yes thats right so the slope is - at every point on the graph as its a straight line

6. welshfella

* slope is 4

7. welshfella

the second function is:- f(x) = x^(1/2) can you find the derivative of this one?

8. anonymous

x^-1/2/2

9. welshfella

|dw:1435264790652:dw|

10. welshfella

so to find the slope at (16,4) plug x = 16 into the the above value for f'(x)

11. welshfella

then use the general form y-y1 = m(x-x1) and plug in slope m and the point (x1,y1) will be (16,4)

12. anonymous

for the first step is it 2

13. welshfella

1 / 2*x(1/2) = 1 / 2* (16)^1/2 = 1/ 2*4 = 1/8

14. welshfella

so m = 1/8

15. anonymous

im lost. I don't understand this at all.

16. welshfella

m stands for the slope in the formula the slope is given by the derivative of sqrt x - and is 1 / 2* x (1/2) o r 1 / 2 sqrtx here x = 16 ( point is (16/4) so m = 1 / 2 sqrt16 = 1 / 2*4 = 1/8

17. anonymous

ok but its two parts to the problem that's the answer to both?

18. anonymous

@Preetha

19. anonymous

@misssunshinexxoxo

20. anonymous

@peachpi

21. anonymous

@xapproachesinfinity

22. anonymous

@dan815

23. anonymous

@Compassionate

24. anonymous

@uri

25. anonymous

@surjithayer

26. anonymous

@LynFran

27. LynFran

ok what exactly u dont understand here?

28. anonymous

i dont understand the problem or the question this whole thing confuses me.

29. LynFran

ok lets go through this step by step from the beginning ok

30. anonymous

ok

31. LynFran

first of all the slope of a tangent line is the gradient of that line ok

32. anonymous

what is a gradient

33. LynFran

now to find the slope of the tangent line is y/x...|dw:1435279410553:dw|

34. LynFran

The Gradient (also called Slope) of a straight line shows how steep a straight line is.

35. anonymous

the question has two parts is 4 the answer for both

36. LynFran

|dw:1435279812417:dw|

37. LynFran

4 is just the slope of the tangent line we have to find the slope intercept equation at the given point ok to complete this question ok

38. anonymous

ok whats the next step

39. LynFran

|dw:1435280154077:dw|

40. LynFran

what equation u get??

41. anonymous

y=4x+2

42. LynFran

correct

43. anonymous

ok so thats the second part to the first question?

44. LynFran

we just did 2nd question and yes that was this part; .Find the slope-intercept equation of the tangent line to the graph of f at the given point.

45. anonymous

ok so 1.4 and 2. y=4x+2

46. LynFran

yes Find the slope of the tangent line to the graph of f at the given point. is 4

47. LynFran

for that question

48. anonymous

ok got it

49. LynFran

Did u do the other question?

50. anonymous

can you help me with that one too?

51. LynFran

ok

52. LynFran

|dw:1435281016248:dw|

53. LynFran

|dw:1435281095053:dw|

54. LynFran

now to take the derivative of it we get|dw:1435281258598:dw|

55. LynFran

|dw:1435281332523:dw|

56. LynFran

|dw:1435281411275:dw|

57. anonymous

ok im with you, so whats next

58. LynFran

well we have to plug in the value of x into the derivative we just found

59. LynFran

(16,4) and x=16 there ok so.. plug it into the derivative and let me know what u get

60. LynFran

|dw:1435282159975:dw|

61. anonymous

ok so when i plug in 16 and solve to find the derivative i get 0

62. LynFran

no u cant get 0 lets work it together

63. anonymous

ok

64. LynFran

|dw:1435282473866:dw|

65. LynFran

so 1/8 is this part Find the slope of the tangent line to the graph of f at the given point.

66. anonymous

ok

67. LynFran

so can u find this part now .Find the slope-intercept equation of the tangent line to the graph of f at the given point. |dw:1435282844490:dw|

68. anonymous

4-y1=1/8(16-x1)

69. LynFran

no we leave the y and x term alone and uses the y1 and x1

70. LynFran

|dw:1435283138396:dw|

71. anonymous

ok y-4=1/8(x-16)

72. LynFran

correct now the next step is to multiply out the bracket

73. anonymous

y-4=1/8x -2

74. LynFran

good next step is to bring the -4 over the equal sign ...remember when we bring a # over the equal sign , the sign of the # changes

75. anonymous

y=1/8x+2

76. LynFran

yes and u have officially completed the question ok now i would like to share this site with u it was very useful to me so https://www.symbolab.com/solver/calculus-calculator/%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft(%5Csqrt%7Bx%7D%5Cright)/?origin=button

77. anonymous

ok thank You so much for your help.

78. LynFran

ur welcome