We'll find the tangent plane for the
bowl
\[
\begin{align}
z &\depends (x, y)
\\
z &= x^2 + y^2
\end{align}
\]
centered at the point \( p = (1, -1) \). We start by taking partials.
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= 2 x
&
\operatorname{\overset{\mathit{y}}{\slope}} z &= 2 y
\end{align}
\]
Localizing at the point \( p = (1, -1) \), we find constants:
\[
\begin{align}
&x = 1,
&
&y = -1,
&
&z = 2,
\\
&\operatorname{\overset{\mathit{x}}{\slope}} z = 2,
&
&\operatorname{\overset{\mathit{y}}{\slope}} z = -2.
&
&
\end{align}
\]
This is everything we need to write down an equation for the tangent plane!
\[
\begin{align}
z_T - 2 &= 2 \mult (x_T - 1) - 2 \mult (y_T + 1)
\end{align}
\]