$\begin{array}{l}
 y(5) = 1 = 25 + 5p + q \Rightarrow 5p + q =  - 24 \\ 
 T\left( { - \frac{b}{{2a}}, - \frac{{{b^2} - 4ac}}{{4a}}} \right) \Rightarrow x = 5 =  - \frac{b}{{2a}} =  - \frac{p}{2} \Rightarrow p =  - 10 \\ 
  \\ 
 q =  - 24 - 5p = 26 \Rightarrow p + q = 16 \\ 
 \end{array}$