Finding the range of $\frac {2x^2+x-3}{x^2+4x-5}$ I solved for the range as follows:
Setting $f(x)=\frac {2x^2+x-3}{x^2+4x-5} =y$, I rearranged it to get a quadratic in x.
$$(y-2)x^{2}+ (4y-1)x +(3-5y)=0$$
Next, using $\Delta \ge 0$, I got
$$(4y-1)^2-4(y-2)(3-5y) \ge 0$$
Which boiled down to
$$(6y-5)^2 \ge 0$$
And this gave me $$ y \in R$$
Next, the value of $x$ for which $y=2$  is $x=1$, for which the function isn't defined, so that gives me $$y \in R - \{2\}$$
However, the solution is $$y \in R- \{\frac{5}{6}, 2\}$$
I understand that the original function is identical to $$g(x) =\frac{2x+3}{x+5} 
  \forall x \in R - \{1\}$$
and that $\frac{5}{6} =g(1)$, but in the original function $f(x)$ I found $y=2$ corresponded to $x=1$ and therefore excluded it, but if $x= 1$ also corresponds to $y=\frac{5}{6}$ then wouldn't this $not$ be a function, as $x=1$ would then be associated with two different $y$ values?
I've read the answers in Why D≥0 while finding the range of rational functions and Finding the range of $y =\frac{x^2+2x+4}{2x^2+4x+9}$ (and $y=\frac{\text{quadratic}}{\text{quadratic}}$ in general) but I'm still unsure of how to apply the information from those to figure out what values of $y$ need to be excluded when dealing with such questions.
While I am familiar with derivatives and limits to a certain degree, we were assumed to $\underline {not}$ know calculus when we were taught this and solved such problems.
I apologise if there are any issues with formatting, this is my first time using Latex.
 A: Your quadratic equation has $x=1$ as a solution, for any $y$. Indeed, for $y\neq2$ the quadratic formula gives
$$x=\frac{-(4y-1)\pm(6y-5)}{2(y-2)}$$
$$x=\frac{-10y+6}{2y-4}=\frac{-(5y-3)}{y-2}\qquad\text{or}\qquad x=\frac{2y-4}{2y-4}=1$$
and for $y=2$ your equation reduces to $7x-7=0$.
The original function was
$$y=\frac{(x-1)(2x+3)}{(x-1)(x+5)}$$
and you rearranged this to get
$$(x-1)(x+5)y-(x-1)(2x+3)=0$$
which always has $x=1$ as a solution.
Essentially, you added a vertical line to the graph of $g$, so the resulting graph doesn't represent a function.
You should have just worked with $g$ from the beginning, and then excluded any values of $y$ which would have given $x=1$ or $x=-5$.
A: Alternative approach, along the lines of R. J. Mathar in the comments.
For all $x\in\mathbb{R}\setminus\{1,-5\},$
$$ \frac{2x^2 + x -3}{x^2 + 4x - 5} \equiv \frac{2(x^2 + 4x - 5) - 7x + 7}{x^2 + 4x - 5} \equiv 2 -  \frac{7x-7}{x^2 + 4x - 5} $$
$$ \equiv 2 - \frac{7(x-1)}{(x+5)(x-1)}  \equiv  2 - \frac{7}{(x+5)}. $$
Define $g:x\to 2 - \frac{7}{(x+5)}\ $ and $\ f:x\to\frac{2x^2 + x -3}{x^2 + 4x - 5}.$
We showed above that $f$ and $g$ are equivalent for all $x\in\mathbb{R}\setminus\{1,-5\}.$
The function $g:x\to 2 - \frac{7}{(x+5)}$ has domain $\mathbb{R}\setminus\{-5\}$ and range $\mathbb{R}\setminus\{2\}.$
But the function $ f:x\to\frac{2x^2 + x -3}{x^2 + 4x - 5}$ has domain $\mathbb{R}\setminus\{1,-5\}.$ Therefore the range of $f$ is the range of $g$ with the point $ g(1)$ (if it exists) removed.
A: As you have already understood x -1 is a factor of both numerator and denominator of f(x) . Therefore 1 can not be in the domain of the function f . Actually what is happening here is f(x) is indeterminate when x = 1 and it can take any value. Therefore if you define the domain of f as ℝ - {1} that issue doesn't arise and you can consider f as a function.
