how to you find the range of f(x)=x-4/x^2-4 PLEASE HELP

Question

anonymous · Answer

$$\frac{x-4}{x^2-4}$$?

anonymous · Answer

calculus problem or something else?

anonymous · Answer

guess i'll never know...

anonymous · Answer

sorry! its pre cal and yest thats correct!

anonymous · Answer

ok without calc it is not going to be that easy

anonymous · Answer

do you have any insight on how to solve i t?

anonymous · Answer

yes, and it is a pain

anonymous · Answer

we are going to have to solve for x

anonymous · Answer

$$y=\frac{x-4}{x^2-4}$$
$$yx^2-4y=x-4$$
$$yx^2-x-4y-4=0$$ now we have a quadratic equation in x, so we can solve using the quadratic formula

mertsj · Answer

$$x=\frac{1\pm \sqrt{1+16y^2+16y}}{2y}$$

anonymous · Answer

@Mertsj if you have a snap way to do it let me know

anonymous · Answer

ok we are doing the same thing

mertsj · Answer

Graph it.

anonymous · Answer

denominator is y^2 but it makes no difference
now you have to solve 
$$16x^2+16x+1\geq 0$$ lord!

anonymous · Answer

oh no you are right, denominator is 2y

mertsj · Answer

How so?  a = y doesn't it?

anonymous · Answer

so we have to solve 
[16x^2+16y+1=0\] using the quadratic formula, another pain

oh yes you are right, i am wrong about the denominator

anonymous · Answer

$$16x^2+16y+1=0$$

anonymous · Answer

$$x=\frac{-2\pm\sqrt{3}}{4}$$

mertsj · Answer

I got 8 on the bottom

anonymous · Answer

also wrong, sorry
range is 
$$(-\infty,\frac{-2-\sqrt{3}}{4}]\cup (\frac{-2+\sqrt{3}}{4},\infty)$$

anonymous · Answer

mertsj · Answer

I know.  I found it.

mertsj · Answer

Do you really think a pre-calc student was supposed to do that?