Die ersten beiden Ableitungen von ${\displaystyle {}f}$ sind

${\displaystyle {}{\begin{aligned}f'(x)&={\frac {(2x-7)(x-4)-(x^{2}-7x)}{(x-4)^{2}}}\\&={\frac {x^{2}-8x+28}{(x-4)^{2}}}\end{aligned}}}$

und

${\displaystyle {}{\begin{aligned}f^{\prime \prime }(x)&=\left({\frac {x^{2}-8x+28}{(x-4)^{2}}}\right)'\\&={\frac {(2x-8)(x-4)^{2}-2(x^{2}-8x+28)(x-4)}{(x-4)^{4}}}\\&=2{\frac {(x-4)^{2}-(x^{2}-8x+28)}{(x-4)^{3}}}\\&=2{\frac {-12}{(x-4)^{3}}}\\&={\frac {-24}{(x-4)^{3}}}.\end{aligned}}}$

Somit ist

${\displaystyle {}f(-3)=-{\frac {30}{7}}\,,}$

${\displaystyle {}f'(-3)={\frac {9+24+28}{49}}={\frac {61}{49}}\,,}$

${\displaystyle {}f^{\prime \prime }(-3)={\frac {-24}{(-7)^{3}}}={\frac {24}{343}}\,.}$

Somit ist die Taylorentwicklung in ${\displaystyle {}a=-3}$ bis zum Grad ${\displaystyle {}2}$ gleich

${\displaystyle -{\frac {30}{7}}+{\frac {61}{49}}(x+3)+{\frac {12}{343}}(x+3)^{2}.}$