Determine the parameter n such that the line y = x+n is

Determine the parameter $n$ such that the line $y = x+n$ is tangent to the curve $y = \dfrac{x}{x+4}$.

Step 1: Derivative

By the quotient rule, \[y' = \frac{(x+4)\cdot1 - x\cdot1}{(x+4)^{2}} = \frac{4}{(x+4)^{2}}. \tag{1}\]

Step 2: Slope condition

The line $y=x+n$ has slope $1$. At the point of tangency $x_{0}$, the slope of the curve must equal $1$: \[\frac{4}{(x_{0}+4)^{2}} = 1 \;\Longrightarrow\; (x_{0}+4)^{2}=4 \;\Longrightarrow\; x_{0}+4=\pm2 \;\Longrightarrow\; x_{0} = -2 \ \text{ or } \ x_{0}=-6. \tag{2}\]

Step 3: Contact condition

At $x_{0}$ the point lies on both the curve and the line: \[\frac{x_{0}}{x_{0}+4} = x_{0}+n \;\Longrightarrow\; n = \frac{x_{0}}{x_{0}+4}-x_{0}. \tag{3}\]

Case $x_{0}=-2$: $n = \dfrac{-2}{2} - (-2) = -1+2 = 1.$

Case $x_{0}=-6$: $n = \dfrac{-6}{-2} - (-6) = 3+6 = 9.$

Step 4: Verification

\[n=1,\ (-2,-1)\text{: curve } \tfrac{-2}{2}=-1,\ \text{line } -2+1=-1,\ \text{slope } \tfrac{4}{4}=1\ \checkmark\]

\[n=9,\ (-6,3)\text{: curve } \tfrac{-6}{-2}=3,\ \text{line } -6+9=3,\ \text{slope } \tfrac{4}{4}=1\ \checkmark\]