\(f(2)=8\text{,}\) dus \(A(2, 8)\text{.}\)

\(f(x)=-x^3+3x^2+x+2\) geeft \(f'(x)=-3x^2+6x+1\text{.}\)

Stel \(l{:}\,y=ax+b\) met \(a=f'(2)=1\text{.}\)

\(\begin{rcases}y=x+b \\ \text{door }A(2, 8)\end{rcases}\begin{matrix}1⋅2+b=8 \\ 2+b=8 \\ b=6\end{matrix}\)
Dus \(l{:}\,y=x+6\text{.}\)

\(f(3)=3^2+(2⋅3-5)^3=10\text{,}\) dus \(A(3, 10)\)

\(f(x)=x^2+(2x-5)^3\) geeft
\(f'(x)=2x+6(2x-5)^2\)

\(l{:}\,y=ax+b\) met \(a=f'(3)=2⋅3+6(2⋅3-5)^2=12\)

\(\begin{rcases}y=12x+b \\ \text{door }A(3, 10)\end{rcases}\begin{matrix}12⋅3+b=10 \\ 36+b=10 \\ b=-26\end{matrix}\)
Dus \(l{:}\,y=12x-26\text{.}\)