If we have to find the range of $\frac{ax^2+bx+c}{px^2 + qx + r} = y$
We can follow this procedure
$ (py-a)x^2 + (qy-b)x + (ry -c) = 0$ Now we have to solve the discriminant of the above quadratic equation, the values of $y$ for which discriminant is greater than or equal to $0$ belongs to our range because if the discriminant is greater than or equal to $0$ then only the quadratic expression in left hand side can be equal to $0$ (thus satisfying the equation)
There is one extra condition which I can't understand. Solve the coefficient of $x^2$ such that it equals to $0$ . The value of $y$ for which we the coefficient $0$ should be verified. We need to see whether it is possible for the initial quadratic/quadratic expression to be equal to that value of $y$ . My teacher(on YouTube) said that it is because for that value of $y$ , the quadratic equation won't actually be a quadratic, so checking discriminant doesn't make any sense as anything like discriminant doesn't exist for linear equations.
But my question is that now when the expression becomes linear , there must be any value for which it becomes equal to $0$ , so how can there be a case when that value of $y$ should be excluded?
