6. It is known that the graph of the function \(y=x^{2}+p x+q\) passes through the points \(A(1 ; 1)\) and \(B(3 ; 1)\). Does this graph pass through the point \(C(4 ; 5)\) ?
6. Solution. Since the graph passes through the point \(A(1 ; 1)\), then \(1=1^{2}+p \cdot 1+q\). Since the graph passes through the point 86
\(B(3 ; 1)\), then \(1=3^{2}+p \cdot 3+q\). We will find \(p\) and \(q\) by solving the system of equations:

\[
\left\{\begin{array}{l}
p+q=0, \\
3 p+q=-8,
\end{array} \quad p=-4 \text { and } q=4 .\right.
\]

The equation of the parabola: \(y=x^{2}-4 x+4\). For \(x=4, y=4=4^{2}-4 \cdot 4+4\), and not 5. Therefore, the graph does not pass through the point \(C\).