\(f(-3) = 4 \text{,}\) dus \(A (-3 , 4) \text{.}\)

\(f(x) = x^{3} + 5 x^{2} + 4 x - 2\) geeft \(f'(x) = 3 x^{2} + 10 x + 4 \text{.}\)

Stel \(l{:}\,y = a x + b\) met \(a = f'(-3) = 1 \text{.}\)

\(\begin{rcases}y = x + b \\ \text{door } A (-3 , 4)\end{rcases} \begin{matrix}1 ⋅ -3 + b = 4 \\ -3 + b = 4 \\ b = 7\end{matrix}\)
Dus \(l{:}\,y = x + 7 \text{.}\)

\(f(1) = 5 \cdot 1^{3}+\sqrt{1+3} = 7 \text{,}\) dus \(A (1 , 7)\)

\(f(x) = 5 x^{3}+\sqrt{x+3}\) geeft
\(f'(x) = 15 x^{2}+\frac{1}{2 \sqrt{x+3}}\)

\(l{:}\,y = a x + b\) met \(a = f'(1) = 15 \cdot 1^{2}+\frac{1}{2 \sqrt{1+3}} = {61 \over 4}\)

\(\begin{rcases}y = {61 \over 4} x + b \\ \text{door } A (1 , 7)\end{rcases} \begin{matrix}{61 \over 4} 1 + b = 7 \\ {61 \over 4} + b = 7 \\ b = {-33 \over 4}\end{matrix}\)
Dus \(l{:}\,y = {61 \over 4} x + {-33 \over 4} \text{.}\)