## anonymous 4 years ago find the equations of three tangent lines to the graph of y = 4x^3 -40x^2 + 125x - 126 which pass through point (1, -1) (but not on the curve)

1. anonymous

$y = 4x^3 -40x^2 + 125x -126$

2. allank

Cool question. I'd start by calculating the first derivative of y which will give me the slope of the lines we are looking for.

3. allank

Can you take it from here?

4. anonymous

yea

5. allank

Great. Good luck. :)

6. anonymous

thanks

7. allank

You're welcome :)

8. anonymous

i get $y' = 12x^2 - 80x +125$ what now

9. anonymous

im supposed to get 3 slopes

10. anonymous

factor right?

11. anonymous

$y = (2x-5)(6x-25)$

12. anonymous

y prime i mean

13. anonymous

slopes so far are 5/2, 25/6, and...?

14. allank

Just a sec iop360...

15. anonymous

actually is that even right?

16. anonymous

ok

17. allank

Now, what we want to do is first get the positions where the tangents touch the curve...

18. allank

Think about it...how would we do that?

19. allank

Hint: Simultaneous equations.

20. anonymous

so use m = m with one m representing the first derivative i just calculated and the other m representing rise/run

21. anonymous

equate the two and solve for x?

22. allank

Sorry for taking forever to reply...

23. anonymous

$\frac{ (4x^3 - 40x^2 +125x -126) - (-1) }{ x-1 }= 12x^2 -80x + 125$

24. anonymous

its alright

25. allank

Okay, we have the gradient: m = 12x^2-80x+125 and we have a point (1,-1). I'm thinking we can use the general form of an equation of a line: y-y1 = m(x-x1) where y1 and x1 are -1 and 1 respectively, and m is the gradient. I'm not sure whether that will work. Lemme try....

26. anonymous

we would have to find m first

27. allank

True. Trying something different...

28. anonymous

i think i found the slopes, tell me what you have too slope = 0, 5/2, 4

29. allank

Well, I have used a different method, where I first determine the positions where the tangents touch the curve. Want to finish off and compare answers?

30. anonymous

ok

31. anonymous

you mean now find the line equations?

32. allank

Yep.

33. anonymous

k

34. anonymous

y = -1 y = (5/2)x - (7/2) y = 4x - 5

35. anonymous

doesnt seem right when i graph it...

36. anonymous

hopefully u have a better one

37. allank

Well, I've graphed yours and It seems okay. :) My equation is giving me imaginary solutions -_-

38. anonymous

u graphed mine?

39. allank

Yep.

40. anonymous

it doesnt look tangent to the curve

41. allank

Plotting on a calculator?

42. anonymous

yes

43. anonymous

they just all intersect together and dont touch the curve

44. allank

Darn. Okay, maybe I'm just sleepy to notice that on a computer grapher...so wanna see my method? Though it may not lead to an answer...

45. anonymous

sure

46. anonymous

perhaps i should bump this

47. allank

Okay, so we have the slope m = 12x^2-80x+125. Remember that the value of m depends on the x value on the curve.... Knowing the point (1,-1) , the equation of the tangent(s) will be y+1 = m(x-1) Plugging in m: y = m(x-1)-1 y = (12x^2-80x+125)(x-1) -1 y = 12x^3 - 80x^2 +125x - 12x^2 + 80x -125 - 1 y = 12x^3 - 92x^2 + 205x - 126 (equation 1) So, we have an equation that depends on the very same x value on the curve as the original equation: y = 4x^3 -40x^2 + 125x - 126 (equation 2) Two nice simultaneous equations that will give the x values on the curve where the tangents touch the curve: y = 12x^3 - 92x^2 + 205x - 126 (equation 1) y = 4x^3 -40x^2 + 125x - 126 (equation 2)

48. allank

I hope I've done the math right. Subtracting the second from the first eliminates y and the 126. y = 12x^3 - 92x^2 + 205x - 126 (equation 1) y = 4x^3 -40x^2 + 125x - 126 (equation 2) 0 = 8x^3 +52x^2 + 330x And solving for x gives x=0 and two other imaginary solutions. *sigh*

49. anonymous

i see

50. allank

Do bump @iop360. I'll sleep on it and in case you haven't gotten a valid response by the time I'm awake, I will post the solution.

51. anonymous

alright

52. anonymous

ill leave this question open

53. anonymous

thanks for helping

54. anonymous

hey, i got the question right! if you want to know how, the values for x i found (0, (5/2), and 4) were actually the x coordinates of the tangents so i plugged these values in the original equation to get their respective y values after that i used each (x,y) coordinates of the three tangents with the given point in the question (1,-1) in the rise/run formula to get the slope for each line

55. anonymous

plug in the found slopes in point-slope formula and it should be right

56. allank

Nice. Good job! :)