AonZ 2 years ago Help please :) Find the values of a and b if the parabola y=a(x+b)^2 −8 has tangent y = 2x at the point P(4, 8),

1. AonZ

i have gotten gradient to be 2a(x+b)

2. myininaya

ok great so far y' at (x,y)=(4,8) is 2a(4+b)=8a+2b And we are also given at that point we have y'=2x But that point is (4,8) so at that point we have y'=8 So this means we have 8=8a+2b Did you get this far?

3. myininaya

We should be looking for another liner equation

4. myininaya

Any ideas?

5. AonZ

not really

6. AonZ

i got 8=8a +2ab

7. myininaya

Well you also know (4,8) is on the parabola. See if you can use that.

8. myininaya

this should give you two equations with the only unknowns a and b

9. myininaya

You know how to solve a system of equations, correct?

10. AonZ

err yea

11. myininaya

Did you think to do 8=a(4+b)^2-8

12. AonZ

i got 8=8a +2ab not 8=8a +2b btw

13. myininaya

wait how did you get the first equation?

14. myininaya

oh wait i messed up when i wrote it

15. myininaya

2a(4+b)=8a+2b I didn't distribute there correctly 8a+2ab would be correct

16. myininaya

so yeah we have 8a+2ab=8

17. AonZ

2nd equation is 16a + 8ba + ab^2 - 16 right?

18. myininaya

all of that = 0 yes

19. AonZ

then we just solve right?

20. myininaya

yes

21. AonZ

16a + 8ba + ab^2 - 16 = 0 and 8a+2ab-8= 0 err how?

22. myininaya

Did you try solving the linear equation either for a or b and replace the a or b in the first equation you ahve there

23. AonZ

im pretty sure we made a mistake somewhere

24. AonZ
25. AonZ

the answers in the back of my book says 1/16 = a and 12= b

26. myininaya

27. AonZ

oh typed it wrong before

28. AonZ

oh wait...

29. AonZ

can we start from the begining again?

30. AonZ

the derivative is 8a+2ab that means the gradient is 8a + 2ab How did it become 8= 8a + 2ab

31. myininaya

y'=2x at (4,8)

32. myininaya

we also know that y'=2a(x+b) at (4,8) so at (4,8) we know 2x=2a(x+b)

33. myininaya

x=4 then

34. myininaya

so we have 2*4=2a(4+b)

35. myininaya

8=8a+2ab

36. AonZ

ok alright

37. myininaya

38. AonZ

hmm

39. myininaya

so if a=1/16, b=12 and we have that y'=2x at (4,8) and y'=2a(x+b) at (4,8) then we have 2*4=2*1/16(4+12) correct? is this true?

40. myininaya

2*4=8 2/16*(4+12)=1/8*(16)=2 2 does not equal 8

41. AonZ

err yea

42. myininaya

The equation you put is correct right? y=a(x+b)^2-8

43. myininaya

the point is (4,8) where we have the tangent line y=2x

44. AonZ

yes right question

45. myininaya

i see nothing wrong with our resulting equations

46. AonZ

hmm yea

47. AonZ

but i dont think we could sub (4,8) into 16a + 8ba + ab^2 - 16=0

48. myininaya

we already did to get that equation

49. myininaya

remember y=a(x+b)^2-8 (4,8) is on that parabola

50. myininaya

8=a(4+b)^2-8

51. myininaya

a(4+b)^2-16=0 a(16+8b+b^2)-16=0 ab^2+8ab+16a-16=0

52. myininaya

So we have ab^2+8ab+16a-16=0 4=4a+ab

53. myininaya

Solve that bottom equation for either a or b then plug into the first equation.

54. AonZ

a = 1, b=0

55. myininaya

By the way I just divided our earlier linear equation by 2 to get 4=4a+ab 4=a(4+b) 4/(4+b)=a Assuming b does not equal -4. Then we have 0=4/(4+b)*b^2+8*(4/(4+b))*b+16*(4/(4+b))-16

56. myininaya

Multiply both sides by 4+b 0=4b^2+32b+64-16(4+b)

57. AonZ
58. myininaya

yep so i did make a mistake there the slope of 2x is 2 I put 8. haha So we had 2=8a+2ab was suppose to be that one linear equation not 8=8a+2ab