Find the equations of the tangent and the normal

Find the equations of the tangent and the normal to the curve $y = x^{2}+2x$ at an arbitrary point of the curve.

Step 1: The point and the slope

Let $x_{0}$ be arbitrary; the point of the curve is $P = \bigl(x_{0},\,x_{0}^{2}+2x_{0}\bigr)$, and

\[ y' = 2x+2 \;\Longrightarrow\; k = y'\bigl(x_{0}\bigr) = 2x_{0}+2. \tag{1} \]

Step 2: The tangent

\[ y-\bigl(x_{0}^{2}+2x_{0}\bigr) = \bigl(2x_{0}+2\bigr)\bigl(x-x_{0}\bigr), \] and expanding, the terms $2x_{0}$ cancel and $-2x_{0}^{2}+x_{0}^{2} = -x_{0}^{2}$ remains: \[y = \bigl(2x_{0}+2\bigr)x-x_{0}^{2}. \tag{2}\]

Step 3: The normal

For $k\neq0$, that is $x_{0}\neq-1$, the normal has slope $-\dfrac1k$: $y-\bigl(x_{0}^{2}+2x_{0}\bigr) = -\frac{1}{2\bigl(x_{0}+1\bigr)}\bigl(x-x_{0}\bigr)$.

\[ y = -\frac{x}{2\bigl(x_{0}+1\bigr)}+\frac{x_{0}}{2\bigl(x_{0}+1\bigr)}+x_{0}^{2}+2x_{0}. \tag{3} \]

Step 4: The exceptional point

At $x_{0} = -1$ the vertex of the parabola, $k = 0$: the tangent is the horizontal line $y = -1$ and the normal the vertical line $x = -1$, which has no slope and so falls outside formula (3).

Step 5: Verification

\[\text{Tangent at } x = x_{0}:\ \bigl(2x_{0}+2\bigr)x_{0}-x_{0}^{2} = x_{0}^{2}+2x_{0} \;\;\checkmark\]

\[\text{Normal at } x = x_{0}:\ -\frac{x_{0}}{2\bigl(x_{0}+1\bigr)}+\frac{x_{0}}{2\bigl(x_{0}+1\bigr)}+x_{0}^{2}+2x_{0} = x_{0}^{2}+2x_{0} \;\;\checkmark\]

\[\text{Slopes: } k\cdot\left(-\frac1k\right) = -1 \;\;\checkmark\]