Critical #s of division function?

Question

I am stuck on finding the critical #s for the following problem:

((x^2)-3)/(x-2)

Any help would be fantastic. I asked this question earlier but messed up on the details.

When I work it all out, I must use the quadratic formula. For the critical #s I get: (7+/- sqrt(13)) / 2.

Can anyone else confirm this, or am I way off?
On my graphing calculator, they look to be x=2 and x=6 where the slope is zero.

Correction above: 2 and 6 are the y values where slope = zero. The x values look to be 1 and 3.

Answer 1

f(x)= ((x^2)-3)/(x-2)
the critical numbers are the x's for which f'(x) =0

so f'(x) = ( (x-2)2x - (x^2-3) ) / (x-2)^2
= ( 2x^2 -4x -x^2 + 3 ) / (x-2)^2
= (x^2 -4x +3) / (x-2)^2
=0
when x^2 -4x +3 = 0
(x-1)(x-3)=0
so the critical points are x=1 and x=3

On the other hand, notice that f(x) is not define when x-2 =0
and also de derivative.
.
.

Answer 2

Lobis3 is a bit messed. y’ = ((x-2)*2x – x^2+3) / (x-2)^2 = 0; thus 2x^2 – 4x – x^2 +3 =0;
x^2-4x+3=0; x1=2-sqrt(4-3)=1, x2=3, while x is not = 2;