As you walk through Gate 4, you realize that Gate 5 is nowhere to be found. In front of you lies an expansive desert as far as you can see. You look at your map and see that there is an alert for this area. The warning states that this desert is almost entirely quicksand, with only one path safely through the desert. That path is defined by a piecewise function but requires that some parameters be determined before your computer can generate the plot. Thinking back on what you learned in calculus, you realize that your path will need to be continuous. You just need to tell your computer what the parameters below should be in order to create this continuous path. What parameters do you give your computer?
\[
f(x)=\left\{\begin{array}{cc}
3, & x \leq-2 \\
2 x+A, & -2< x \leq-1 \\
B x+3, & -1< x \leq 0 \\
6 x+C, & 0< x \leq 1 \\
D x+13, & 1< x \leq 2 \\
5, & x> 2
\end{array}\right.
\]
\begin{tabular}{|r|r|r|r|}
\hline A & B & C & ( \\
\hline Number & Number & Number & Number \\
\hline & & \\
\hline
\end{tabular}

Step 1 ：To ensure the function is continuous, the end of one piece must meet the start of the next piece. We can find the values of A, B, C, and D by setting the end of one piece equal to the start of the next piece and solving for the unknown.

Step 2 ：For the first piece, the function is 3 when \(x \leq -2\). The next piece starts at \(x = -2\), so we set 3 equal to \(2(-2) + A\) and solve for A: \[3 = -4 + A\] \[A = 3 + 4\] \[\boxed{A = 7}\]

Step 3 ：For the second piece, the function is \(2x + A\) when \(-2 < x \leq -1\). The next piece starts at \(x = -1\), so we set \(2(-1) + A\) equal to \(B(-1) + 3\) and solve for B: \[-2 + 7 = -B + 3\] \[\boxed{B = 2}\]

Step 4 ：For the third piece, the function is \(Bx + 3\) when \(-1 < x \leq 0\). The next piece starts at \(x = 0\), so we set \(B(0) + 3\) equal to \(6(0) + C\) and solve for C: \[3 = C\] \[\boxed{C = 3}\]

Step 5 ：For the fourth piece, the function is \(6x + C\) when \(0 < x \leq 1\). The next piece starts at \(x = 1\), so we set \(6(1) + C\) equal to \(D(1) + 13\) and solve for D: \[6 + 3 = D + 13\] \[\boxed{D = -4}\]

Step 6 ：So, the parameters A, B, C, and D are 7, 2, 3, and -4, respectively.