## anonymous 5 years ago the equation of the line tangent to the curve y=kx+8/k+x at x= -2 is y=x+4. what is the value of k?

1. anonymous

check your question , possibly a typo

2. anonymous

no, thats it word for word

3. anonymous

y= (k+1)x +(8/k) ( eqn of the "curve" , even though its a line )

4. anonymous

the only way to have a tangent to a line , is for the tangent to be the line

5. anonymous

so (k+1)x +(8/k) = x+4

6. anonymous

now we are told they are a tangent at x=-2, so we could sub x=-2 into the above to find k , but we could also sub any value of x we like , these lines have to be the same for all x

7. anonymous

but ill just sub x=-2 -2(k+1) + 8/k = 2 -2k(k+1) +8 = 2k -2k^2 -2k -2k +8 =0 -2k^2 -4k+8 =0 -2(k^2 +2k -4) =0

8. anonymous

k^2 +2k -4 =0

9. anonymous

how did you get that function?

10. anonymous

(k+1)^2 = 5 k +1 = +-sqrt(5) k = -1+-sqrt(5)

11. anonymous

12. anonymous

13. anonymous

i doubt that you dont even know what the equation of the line tangent is

14. anonymous

this is a question from an ap test so there are no typos

15. anonymous

lol was the curve $y=kx + \frac{8}{k+x}$ y =

16. anonymous

if it was................ :|

17. anonymous

use brackets when you are writting fractions on the net :|

18. anonymous

put the kx on top also

19. anonymous

sorry? didnt know

20. anonymous

lol

21. anonymous

y=(kx+8)/k+x

22. anonymous

With the given info, one finds the tangent vector of the curve to be <1, -1> and the point of tangency to be (-2, 2). You can find k by plugging in (-2, 2).