When 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?

A. -3
B. -1
C. 1
D. 3
E. 4

Question

When 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?

A. -3
B. -1
C. 1
D. 3 
E. 4

anonymous · Answer

$$y=\frac{kx+8}{k+x}\to f^{\prime}(x) = \frac{(k)(k+x)-(1)(kx+8)}{(k+x)^2}$$
So when $$f^{\prime}(-2)=1=\frac{(k)(k+(-2))-(1)(k(-2)+8)}{(k+(-2))^2}$$
solve for k.

anonymous · Answer

f'(-2)=(k^2-8)/(k-2)^2 
what am i supposed to do next?

tkhunny · Answer

At x = -2, the value of the derivative is +1.  Why do I know that?

anonymous · Answer

Um...I have no idea...

tkhunny · Answer

It's hiding in the problem statement: " x=-2 is y=x+4"  What's the slope of that line that was given as reference?  There it is!

anonymous · Answer

can you simplify the derivative dy/dx = ???   what u got?

anonymous · Answer

i got k=3?

anonymous · Answer

yep...

tkhunny · Answer

Did you circle back around and prove it in the original equation?  That will build your confidence.

anonymous · Answer

I realized that the derivative has to equal to the slope of the tangent line which is 1.

tkhunny · Answer

Victory Prize!  Go challenge the next one, now.

anonymous · Answer

thank you!