Not understanding the shifting in the index of balancing Tor's proof in Weibel

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This is related to Weibel's Homological Algebra Chpt 2, Sec 7, Balancing Tor and Ext. I do not think the question I am asking is related to $P,Q$ being $A,B$'s projective resolution as it is related to only part of construction of mapping cone.

Consider $P.\to A$, $Q.\to B$ resolution by modules where $P., Q.$ are complexes. Say $P_a\otimes Q_b$ is at position $(a,b)$ in complex $P.\otimes Q$.

Consider $C$ complex formed by $P.\otimes Q.$ adding $A\otimes Q[-1]$ at $a=-1$.

$\textbf{Q:}$ What is $A\otimes Q[-1]$?(In particular, where does $A\otimes Q[-1]$ start non-trivial?) The book treats $A$ as a complex concentrated at $0$. So $A\otimes Q[-1]$ really starts at $(0,1)$ position(corresponding to $A\otimes Q_{1-1}=A\otimes Q_0$) and goes like $(0,2),(0,3),\dots$. However $P.\otimes Q.$ starts at $(0,0)$ and filling the first quadrant of $(a,b)$. So I do not see how $A\otimes Q[-1]$ starts at $(-1,0)$ after adding complex $A\otimes Q[-1]$ at $a=-1$ position. The whole point of $A\otimes Q[-1]$ 's -1 is to make $C$ a double complex.

$\textbf{Q':}$ The book says $f:Tot(P.\otimes Q.)\to A\otimes Q$'s mapping cone is $Tot(C)[1]$ which I can check by hand. Why should one expect so obvious?