Wyznaczmy granice jednostronne funkcji w punkcie x=-2 i x=3.   

$\underset{x\to -2^{-}}{\lim }f\left(x\right)=\underset{x\to -2^{-}}{\lim }\left(a{x}^2+b\right)=a\cdot {\left(-2\right)}^2+b=4a+b$  

$\underset{x\to -2^{+}}{\lim }f\left(x\right)=\underset{x\to -2^{+}}{\lim }\left(x-a\right)=-2-a$  

$\underset{x\to 3^{-}}{\lim }f\left(x\right)=\underset{x\to 3^{-}}{\lim }\left(x-a\right)=3-a$  

$\underset{x\to 3^{+}}{\lim }f\left(x\right)=\underset{x\to 3^{+}}{\lim }\left(\frac{a}{x}+b\right)=\frac{a}{3}+b$  

 

Zatem muszą być spełnione warunki:

$\{\begin{array}{l}4a+b=-2-a\mid -4a\\ 3-a=\frac{a}{3}+b\mid \cdot 3\end{array}$  

$\{\begin{array}{l}b=-2-5a\\ 9-3a=a+3b\end{array}$  

$\{\begin{array}{l}b=-2-5a\\ 9-3a=a+3\left(-2-5a\right)\end{array}$  

$\{\begin{array}{l}b=-2-5a\\ 9-3a=a-6-15a\mid +14a-9\end{array}$  

$\{\begin{array}{l}b=-2-5a\\ 11a=-15\mid :11\end{array}$  

$\{\begin{array}{l}b=-2-5a\\ a=-\frac{15}{11}\end{array}$  

$\{\begin{array}{l}b=-\frac{22}{11}-5\cdot \left(-\frac{15}{11}\right)\\ a=-\frac{15}{11}\end{array}$  

$\{\begin{array}{l}b=-\frac{22}{11}+\frac{75}{11}\\ a=-\frac{15}{11}\end{array}$  

$\{\begin{array}{l}b=\frac{53}{11}\\ a=-\frac{15}{11}\end{array}$