Range of $\frac{x^{2}+2x+1}{x^{2}-8x+12}$, fast & easy method? How do I find range of 
$$\frac{x^{2}+2x+1}{x^{2}-8x+12}$$
To find range I put $y=f(x)$, solve for $x$ and figure out the possible values from that. But it's much time consuming. 
Finding vertex of parabola doesn't help either. 
Is there any other method to find range?
Thanks in advance. 
 A: You can factor the two quadratics to get $$\frac {(x+1)^2}{(x-6)(x-2)}$$  This starts at $+1$ when $x$ is very large in either direction and heads off to $+\infty$ as $x$ approaches $2^-$ or $6^+$.  It restarts at $-\infty$ at $2^+$ and $6^-$.  You can set the derivative equal to zero to find the first branch has a minimum of $0$ at $x=-1$, the second branch has a maximum of $-5.25$ at $x=3.2$, and the third branch is always above $1$, so the range is $(-\infty,-5.25]\cup [0,\infty)$
A: I think the range is $(-\infty, -21/4] \cup [0, \infty)$.
By setting the derivative to 0, you will find that the function has a minimum at $x=-1$ and at $x=16/5$. We have $f(-1)=0$ and $f(16/5)=-21/4$. Furthermore, on both ends of the functions (i.e. $x \rightarrow -\infty$ and $x \rightarrow \infty$) the function approaches 1.
The function $f(x)$ consists of three 'parts'. The first part $x \in (\infty, 2)$ there is the minimum at $x=-1$ with $f(-1)=0$, so here the range is $[0, \infty)$.
The second part $x \in (2, 6)$ there is a maximum at $x=16/5$ with $f(16/5)=-21/4$, so here the range is $(\infty, -21/4]$.
The third part, there is now minimum or maximum. We know that for $x \rightarrow \infty$ the function approaches 1, so here the range is $(1, \infty)$. 
Combining the three results gives your answer.
A: $y=f(x)=\dfrac{x^2+2x+1}{x^2-8x+12}\iff (1-y)x^2+(2+8y)x+(1-12y)=0$
The value $y=1$ can be reached for $10x=11\iff x=\dfrac{11}{10}$
For $y\neq 1$, this quadratic has obviously a solution $y$ when $x$ is in the domain of definition, this means that $\Delta\ge 0$.
$\Delta=(2+8y)^2-4(1-y)(1-12y)=84y+16y^2=16y(y+\dfrac{21}4)\ge 0$ 
So we have $y\ge 0$ or $y\le-\dfrac{21}4$.
Reciprocally for any $y\neq 1$ value in this range, we can calculate at least a value for $x=\dfrac{-2-8y+\sqrt{\Delta}}{2(1-y)}$, establishing the surjectivity from the whole intervals $]-\infty,-\frac{21}4]$ and $[0,+\infty[\setminus\{1\}$.
And since $y=1$ also has an antecedent, then $]-\infty,-\frac{21}4]\cup[0,+\infty[$ constitutes the range of $f(x)$.
A: My calculations show that it is $(-\infty, -10.15] \bigcup [0,+\infty)$.
