## A community for students. Sign up today

Here's the question you clicked on:

## anonymous 4 years ago I AM CERTAIN THAT NO ONE WILL ANSWER THIS QUESTION... Suppose that l be a line through the points (0,3)and (5,-2) and tangent to the graph of the function $F(x)=\frac{ x^2-p }{ 3x+1 }$ where ‘p’ is real number. What is the equation of a line parallel to l and tangent to the graph of F. A) 4x+4y+7 = 0 B) 9x+9y+37 = 0 C) x+y-3 = 0 D) 4x+6y-37 = 0

• This Question is Closed
1. anonymous

I will post the answer exactly after one hour.

2. mathslover

Hint: Never challenge OpenStudy

3. anonymous

m= -2-3/5-0 = -1 the equation will be y-3= -1(x-0) x+y-3 =0

4. TheViper

Right @mathslover :)

5. hartnn

so its not D

6. anonymous

@hartnn that means???

7. hartnn

that means answer can be a,b or c

8. anonymous

yep.... but the answer hasn't yet finished... @hartnn

9. anonymous

I will the slope of the line through the two points

10. hartnn

p=-7 B)9x+9y+37=0

11. hartnn

solve y=3-x=F(x)=.... u get two roots of x, equate them because the line is tangent and can have only one intersection point. from this i got p=-7. now answer can only be either A or B so solving both of them with F(x), gave only B) option with one intersection point, A) doesn't even intersect.

12. anonymous

PLZ DONT POST THE SOLUTION

13. anonymous

I am here to post the answer.

14. anonymous

ok 5min

15. anonymous

ok

16. hartnn

whats wrong with my answer? and method ??

17. anonymous

w/c is your ans?

18. hartnn

B)

19. hartnn

and p=-7

20. anonymous

correct but the approach not.

21. hartnn

didn't u get p=-7, the same way as i got ?

22. anonymous

yape

23. hartnn

then ?

24. anonymous

you use the options to answer the question, isn't it?

25. anonymous

Time is up @sauravshakya

26. anonymous

OK.......

27. anonymous

Here is the answer

28. anonymous

First of all what you have to do is finding l so as to get p $l:\frac{ y-3 }{ x-0 }=\frac{ -2-3 }{ 5-0 }$ $l:y=-x+3$thus $F(x)=\frac{ x^2-p }{ 3x+1} =-x+3=y$simplifying the above expression $4x^{2}-8x-p-3=0$Now since l is tangent to F(x), the quadratic equation above has only one solution. That means $b^{2}-4ac=0$ $(-8)^{2}-4(4)(-p-3)=0$thus $(-8)^{2}-4(4)(-p-3)=0$$p=-7$Now we get F(x) to be $F(x)=\frac{ x^2+7 }{ 3x+1 }$Therefore lets solve F’(x)=-1 so as to get the coordinate of the unknown line. $F'(x)=\frac{ 2x(3x+1)-3(x^{2}+7) }{ (3x+1)^2 }=-1$ $3x^{2}+2x-21=-9x^{2}-6x-1$ $12x^{2}+8x-20=0$thus $x=1 or x=\frac{ -5 }{ 3 }$and $F(1)=2 and F(\frac{ -5 }{ 3 })=\frac{ -22 }{ 9 }$Thus we got this two points (1,2) and (-5/3,-22/9). The two possible lines therefore are $\frac{ y-2 }{ x-1 }=-1{\rightarrow}y=-x+3$w/c is the given line l. And $\frac{ y-(-\frac{ 22 }{9 }) }{ x-(-\frac{ 5 }{ 3 })}=-1{\rightarrow}9y+9x+37=0$w/c is the answer.

29. hartnn

why f'(x) = -1 ?

30. anonymous

b/c the slope at that point is -1

31. anonymous

Because the line l and the required line are parallel

32. anonymous

So, both of their slope is -1

33. hartnn

okk.

34. hartnn

it wasn't that difficult.

35. anonymous

yes but need smart approach

36. anonymous

YEP....... Actually I did a algebra mistake, as usual

37. hartnn

lol! checking options(when given) is the smartest approach!! :P

38. anonymous

agreed with @hartnn

39. anonymous

that saves time in exam

40. anonymous

For exam only

41. hartnn

atleast i got the answer...

42. anonymous

yape

43. anonymous

ok thanks for your help....... closing

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy