I am new to quadratic equation and expression, Hote to find the range of this expression.
The expression is:
$ y = \frac{x^2 -2x +9}{x^2 - 2x -9} $
Where $ x $ is real.
What I've done:
Cross multiplied and got,
$ x^2(y-1) + x(2-2y) - (9y - 9 ) = 0 $
Since x was real,
$ D ≥ 0 $
$ y^2 - 2y + 1 ≥ 0 $
On
Given $y = \dfrac{x^2 -2x +9}{x^2 - 2x -9}$
Solve the equation w.r.t. $x$
$x=\dfrac{y-1\pm\sqrt{10 y^2-2y-8}}{y-1}$
Provided that $y\ne 1$, the discriminant $\Delta=10 y^2-2y-8$ of the previous equation must be $\ge 0$
Solve $10 y^2-2y-8\ge 0 \land y\ne 1$ to get the range of the function
$y\leq -\dfrac{4}{5}\lor y> 1$
$y\leq -\dfrac{4}{5}$ or $y> 1$
In interval notation it is
$(-\infty,\; -\dfrac{4}{5}]\cup (1,+\infty)$
Which is the line $\mathbb{R}$ less $(-\dfrac{4}{5};\;1]$
Hope this helps
$$y=\frac{(x^2-2x-9)+9+9}{x^2-2x-9}=1+\frac{18}{x^2-2x-9}$$ Now we know that $$x^2-2x-9=(x^2-2x+1) -1 -9 = (x-1)^2-10 \ge -10$$
Thus
$$\frac{1}{x^2-2x-9} \in (-\infty,-0.1] \cup(0, +\infty)$$
Thus
$$y \in (-\infty, -0.8] \cup (1, +\infty)$$