Function:
$f(x)= \frac{x-3}{x^2+2x-8} $
In terms of y:
$y= \frac{x-3}{x^2+2x-8} $
Then x isolated: $x= \frac{\sqrt{36y^2-16y+1}-2y+1}{2y} $
To find the Range we need to find the Domain of this 'new' Function.
1.- We must look if $\ \sqrt{36y^2-16y+1} $ is a Real number
2.- The Denominator $\ 2y $ should not be 0
1.- $\ \sqrt{36y^2-16y+1} $ will always be positive, so will be Real.
2.- We find the root: $\ 2y=0 $ so $\ y=0 $
Which left us that this Domain is all Real Numbers except 0
, so this is the Range of the original Function.
But look at the graphic:
That wasn't the range, but i've found, i've found the roots of $\ {36y^2-16y+1} $ and those roots are that numbers accordingly to the graphic: .0752
and .03692
, but i dont know how to argue this result.