
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
We define: If point $P$ lies on the graph of a linear function $y=ax+b (a\neq 0)$, and point $Q$ lies on the graph of an inverse proportion function $y=\frac{c}{x} (c\neq 0)$, and point $P$ and point $Q$ are symmetric about the $y$-axis, then the quadratic function $y=ax^{2}+bx+c$ is called the "derived function" of the linear function $y=ax+b$ and the inverse proportion function $y=\frac{c}{x}$. Point $P$ is called the "base point" and point $Q$ is called the "target point".  
$(1)$ If the quadratic function $y=2x^{2}+4x+5$ is the "derived function" of the linear function $y=ax+b$ and the inverse proportion function $y=\frac{c}{x}$, then $a=$______, $b=$______, $c=$______.  
$(2)$ If the vertex of the "derived function" of the linear function $y=x+b$ and the inverse proportion function $y=\frac{c}{x}$ lies on the $x$-axis, and the abscissa of the "base point" $P$ is $1$, find the coordinates of the "target point".  
$(3)$ If the "derived function" of the linear function $y=ax+2b (a \gt b \gt 0)$ and the inverse proportion function $y=-\frac{2}{x}$ passes through the point $(2,6)$.  
① Explain that there exist two different "base points" on the graph of the linear function $y=ax+2b$.  
② Suppose the abscissas of the two different "base points" on the graph of the linear function $y=ax+2b$ are $x_{1}$ and $x_{2}$, find the range of $|x_{1}-x_{2}|$.

Present the answer in LaTex format: \boxed{Your answer}
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