<s>[INST] $\frac{-3x+1}{x^2-6x-16}>0$ and $y=-2/x +1$ find the interval of y The known info $\frac{-3x+1}{x^2-6x-16}>0$
 so, i find that : x not 8 nor -2. And $x \geq 1/3$
$y=-2/x +1$
For y i find that
$y > -5$ and y not 3/4 or 2. 
Based on that. So interval for $ y = -5 < y < 3/4$
is it right?
What is the interval of y?
 [/INST] It is obviously that for $x\to \infty $ we have a negative sign for the function $f(x) =\frac{-3x+1}{x^2-6x-16}$ which change the sign at every critical point, that is at $8,1/3$ and $-2$, since each if has an odd degree. So $x\in(-\infty,-2)\cup ({1\over 3},8)$.
Now since $x={2\over 1-y}$ we have to solve:
$${2\over 1-y}<-2\;\;\;{\rm and}\;\;\;{1\over 3}<{2\over 1-y}<8$$
From the first one we get $1>y-1$ so $\boxed{y<2}$ and from the second $1-y<6$ so $\boxed{y>-5}$ and $1<4-4y$ so $\boxed{ y<{3\over 4}}$. 
Thus the final result is $y\in (-5,{3\over 4})$.
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