\(f(-1) = -6 \text{,}\) dus \(A (-1 , -6) \text{.}\)

\(f(x) = 2 x^{3} - 3 x^{2} - 5 x - 6\) geeft \(f'(x) = 6 x^{2} - 6 x - 5 \text{.}\)

Stel \(l{:}\,y = a x + b\) met \(a = f'(-1) = 7 \text{.}\)

\(\begin{rcases}y = 7 x + b \\ \text{door } A (-1 , -6)\end{rcases} \begin{matrix}7 ⋅ -1 + b = -6 \\ -7 + b = -6 \\ b = 1\end{matrix}\)
Dus \(l{:}\,y = 7 x + 1 \text{.}\)

\(f(-1) = --1^{3}+\sqrt{-3 \cdot -1+22} = 6 \text{,}\) dus \(A (-1 , 6)\)

\(f(x) = -x^{3}+\sqrt{-3 x+22}\) geeft
\(f'(x) = -3 x^{2}+\frac{-3}{2 \sqrt{-3 x+22}}\)

\(l{:}\,y = a x + b\) met \(a = f'(-1) = -3 \cdot -1^{2}+\frac{-3}{2 \sqrt{-3 \cdot -1+22}} = {-33 \over 10}\)

\(\begin{rcases}y = {-33 \over 10} x + b \\ \text{door } A (-1 , 6)\end{rcases} \begin{matrix}{-33 \over 10} -1 + b = 6 \\ {33 \over 10} + b = 6 \\ b = {27 \over 10}\end{matrix}\)
Dus \(l{:}\,y = {-33 \over 10} x + {27 \over 10} \text{.}\)