After you identify $b$ and $c$, use the point $(1,0)$ to solve for $a$.

Compute $f(2x-2)$ first, then multiply the result by $2$, then subtract $3$.

Enter

and create sliders for $a$, $b$, and $c$.

Adjust $b$ so the graph gets very close to the same vertical dashed line as the given graph, and adjust $c$ so the graph gets very close to the same horizontal dashed line.

After matching $b$ and $c$, adjust $a$ until the curve passes through the labeled point $(1,0)$.

In a new line, enter

Then rewrite that expression by hand into a single fraction plus a constant to match one of the choices.

From the graph, the curve gets closer and closer to the dashed vertical line at $x=-1$, so the denominator must be $x+1$ and therefore $b=1$.

The curve also gets closer and closer to the dashed horizontal line at $y=2$, so the vertical shift is $c=2$.

So

The point $(1,0)$ lies on the graph, so substitute $x=1$ and $f(1)=0$:

Thus $\frac{a}{2}=-2$, so $a=-4$ and

Replace $x$ with $2x-2$:

Now compute $g(x)=2f(2x-2)-3$:

So the correct choice is $\displaystyle g(x)=-\frac{8}{2x-1}+1$.