Let $f:R-\left\{2,6\right\}\to R$ be real valued function defined as $f\left(x\right)=\frac{x+2x+1}{x^2-8x+12}$. Then range of $f$ is

Given,

$f:R-\left\{2,6\right\}\to R$ be real valued function defined as $f\left(x\right)=\frac{x+2x+1}{x^2-8x+12}$

Now let, $y=\frac{x^2+2x+1}{x^2-8x+12}=\frac{{\left(x+1\right)}^2}{\left(x-2\right)\left(x-6\right)} \dots \left(1\right)$

Now differentiating the above function we get,

$\Rightarrow \frac{dy}{dx}=\frac{-2\left(x+1\right)\left(5x-16\right)}{{\left(x-2\right)}^2{\left(x-6\right)}^2}$

Now by wavy curve method we get,

So Graph of $y=\frac{{\left(x+1\right)}^2}{\left(x-2\right)\left(x-6\right)}$ for given domain will be,

So, from graph range is $y\in \left(-\infty ,\frac{21}{4}\right]\cup [0,\infty )$