Finding range of a function Can I get the range of the following function?
$$y=\dfrac{x^2-3x+2}{x^2-5x+6}$$
I cannot isolate $x$ to get $x=f(y)$.
Thanking you in advance.
 A: Factoring your function is helpful, and reduces the work involved:
$$y = \frac{x^2 - 3x + 2}{x^2 - 5x +6} =\frac{(x -1)(x-2)}{(x-2)(x-3)} = \frac {x-1}{x-3}\quad\text{$x\neq 2$}$$
Now, your range does not contain the value of $y = f(2)$, since $y$ is undefined there, and the value of $y = f(3)$, as $y$ is undefined there as well. That is, your function is not defined at $x = 2$ or at $x = 3$, as in those cases, the denominator of $y$ is $0$, but defined everywhere else.
There is no need to write the function as a function of $y$ to find the range. When is $y = 0$? When is $y > 0$, and when is $y \lt 0$. To see its range, it helps in cases like this, to graph the function. 
Let's discuss domain first, then range:
Let's denote by $y$ the original equation (before canceling out a common factor). 
Here, we have $\text{Domain}\,(y) = (-\infty, 2) \cup (2, 3)\cup (3, \infty)$.
Now put $f(x) = \dfrac{x - 1}{x - 3} $, the reduced function, defined on $ \text{Domain}(f) = (- \infty,3) \cup (3,\infty) $ (see graph below). 
Now, let's discuss ranges:
Note, as $x$ grows increasingly large, (very very large: $\to infty$), or increasingly small ($\to \infty$), $f(x) \to 1$ in each case. That is, we have a horizontal asymptote at $y = 1$
So, $ \text{Range}(f) = (- \infty,1) \cup (1,\infty) $, and you can view that below:
 graph of $f(x) = \dfrac{x-1}{x-3}$, a hyperbola.
Now, if you want to determine $ \text{Range}(y) $, we need to omit $ f(2) = 0 $ from $ \text{Range}(f) $ because, as we noted above, $ 2 \notin \text{Domain}(y) $ and because
$ f $ is a one-to-one function (which means that $ f $ does not attain the value $ 0 $ anywhere else other than at $ x = 2 $, and it is undefined at x = 2).
Therefore, $ \text{Range}(y) = (- \infty,0) \cup (0,1) \cup (1,\infty) $.
A: Here the problem collapses, since the two quadratics have a common linear factor.
You asked in a comment, what about if there is no such simplifying technique?  Consider the equation
$$y=\frac{P(x)}{Q(x)},$$
where $P$ and $Q$ are of degree $\le 2$, and at least one has degree $2$.  Rewrite this equation as  $yQ(x)-P(x)=0$.
By gathering like powers of $x$ together, we get a quadratic equation, with coefficients that involve $y$.
For any $y$, the quadratic equation has a real solution $x$ if the discriminant
("$b^2-4ac$") is $\ge 0$. (We have to be a bit careful about the case $a=0$.)  
The discriminant is a quadratic in $y$, so it is not hard to determine the values of $y$ for which it is $\ge 0$. 
A: The function $ f(x) = \dfrac{x^{2} - 3x + 2}{x^{2} - 5x + 6} $ has domain $ (- \infty,2) \cup (2,3) \cup (3,\infty) $. By eliminating the common factor $ x - 2 $ in the numerator and denominator, we see that it coincides with the function $ g(x) = \dfrac{x - 1}{x - 3} $ on $ \text{Dom}(g) = (- \infty,3) \cup (3,\infty) $. Next, observe that
\begin{align}
\forall x \in (- \infty,3) \cup (3,\infty): \quad g(x)
= \frac{x - 1}{x - 3} &= \frac{(x - 3) + 2}{x - 3} \\
                      &= 1 + \frac{2}{x - 3},
\end{align}
so we get $ \text{Range}(g) = (- \infty,1) \cup (1,\infty) $, which is easily obtainable if you sketch the graph of $ g $. Then if you want to determine $ \text{Range}(f) $, simply delete $ g(2) = 0 $ from $ \text{Range}(g) $ because


*

*$ 2 \notin \text{Dom}(f) $ and

*$ g $ is a one-to-one function (this is important, because it says that $ g $ does not attain the value $ 0 $ anywhere else other than at $ x = 2 $).
Therefore, $ \text{Range}(f) = (- \infty,0) \cup (0,1) \cup (1,\infty) $.
