## iop360 Group Title 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) one year ago one year ago

1. iop360 Group Title

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

2. allank Group Title

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 Group Title

Can you take it from here?

4. iop360 Group Title

yea

5. allank Group Title

Great. Good luck. :)

6. iop360 Group Title

thanks

7. allank Group Title

You're welcome :)

8. iop360 Group Title

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

9. iop360 Group Title

im supposed to get 3 slopes

10. iop360 Group Title

factor right?

11. iop360 Group Title

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

12. iop360 Group Title

y prime i mean

13. iop360 Group Title

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

14. allank Group Title

Just a sec iop360...

15. iop360 Group Title

actually is that even right?

16. iop360 Group Title

ok

17. allank Group Title

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

18. allank Group Title

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

19. allank Group Title

Hint: Simultaneous equations.

20. iop360 Group Title

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

21. iop360 Group Title

equate the two and solve for x?

22. allank Group Title

Sorry for taking forever to reply...

23. iop360 Group Title

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

24. iop360 Group Title

its alright

25. allank Group Title

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. iop360 Group Title

we would have to find m first

27. allank Group Title

True. Trying something different...

28. iop360 Group Title

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

29. allank Group Title

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. iop360 Group Title

ok

31. iop360 Group Title

you mean now find the line equations?

32. allank Group Title

Yep.

33. iop360 Group Title

k

34. iop360 Group Title

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

35. iop360 Group Title

doesnt seem right when i graph it...

36. iop360 Group Title

hopefully u have a better one

37. allank Group Title

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

38. iop360 Group Title

u graphed mine?

39. allank Group Title

Yep.

40. iop360 Group Title

it doesnt look tangent to the curve

41. allank Group Title

Plotting on a calculator?

42. iop360 Group Title

yes

43. iop360 Group Title

they just all intersect together and dont touch the curve

44. allank Group Title

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. iop360 Group Title

sure

46. iop360 Group Title

perhaps i should bump this

47. allank Group Title

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 Group Title

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. iop360 Group Title

i see

50. allank Group Title

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. iop360 Group Title

alright

52. iop360 Group Title

ill leave this question open

53. iop360 Group Title

thanks for helping

54. iop360 Group Title

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. iop360 Group Title

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

56. allank Group Title

Nice. Good job! :)