\[ y=a x^{2}+b x+c \] \begin{tabular}{|l|l|} \hline$x$ & $j(x)$ \\ \hline-1 & 13 \\ \hline 0 & 6 \\ \hline 1 & 2 \\ \hline 2 & 7 \\ \hline \end{tabular} Find a quadratic model for the function $j$. Use the model to predict the value of $j(1.5)$.

Step 1 ：Since the points (-1,13), (0,6), and (1,2) are on the graph, we can form three equations:

Step 2 ：\(-a+b+c=13\)

Step 3 ：\(c=6\)

Step 4 ：\(a+b+c=2\)

Step 5 ：Substitute \(c=6\) into the first and third equations, we get:

Step 6 ：\(-a+b+6=13\)

Step 7 ：\(a+b+6=2\)

Step 8 ：Solving these two equations, we get \(a=-3\) and \(b=-1\).

Step 9 ：So the quadratic model for the function j is \(y=-3x^2-x+6\).

Step 10 ：To predict the value of \(j(1.5)\), substitute \(x=1.5\) into the equation, we get \(j(1.5)=-3(1.5)^2-(1.5)+6\).

Step 11 ：After calculation, we get \(j(1.5)=\boxed{-1.75}\).