Question Detail
Question
The tangent to \( y = \frac{kx+8}{k+x} \) at \( x = -2 \) is \( y = x + 4 \). Find \( k \).
Correct Answer
Option C is the correct answer.
Detailed Explanation
Slope of tangent is 1, so ( y'(-2) = 1 ). Using quotient rule: ( y' = \frac{k(k+x) - (kx+8)(1)}{(k+x)^2} = \frac{k^2 - 8}{(k+x)^2} ). Set ( y'(-2) = \frac{k^2 - 8}{(k-2)^2} = 1 \implies k^2 - 8 = k^2 - 4k + 4 \implies 4k = 12 \implies k = 3 ).
Hint
The derivative of the function at the given point equals the slope of the tangent line.