Where is $f(x)={(x^2+2x-48)}/{x^2}$ increasing? decreasing? The question is to find where the graph is increasing or decreasing. 
The original function is $f(x)={(x^2+2x-48)}/{x^2}$
I know I need to find the prime of this function and I think it is this after using the quotient rule:
$2(-x^2-x+49)/x^3$
Finally, in order to draw a graph I need to find the points on the x axis that are either undefined and/or the slope equals 0.
Immediately I know that 0 is an undefined point because it makes the prime graph undefined. 
It's when I take the numerator and set it equal to 0 that trips me up.
$2(-x^2-x+49)=0$
$-x^2-x+49=0$
$(x)(x)=0$
I'm guessing I did something wrong finding this prime because the answers are all including the number 48. I'm stumped because I've been at this question for about 1 hour.
Thanks! 
 A: You computation of derivative is incorrect. It is easier to separate out each term and compute the derivative.
$$f(x) = \dfrac{x^2 + 2x - 48}{x^2} = 1 + \dfrac2{x} - \dfrac{48}{x^2}$$
Hence, $$f'(x) = 0 - \dfrac2{x^2} + \dfrac{2 \times 48}{x^3} = \dfrac{96-2x}{x^3}$$
Now setting, $f'(x) = 0$ gives us $96-2x = 0 \implies x = 48$.
EDIT
As FrankScience rightly points out in his comment, you do not really need to differentiate to figure out the behavior of this function. We have that 
\begin{align}
f(x) & = \dfrac{x^2 + 2x - 48}{x^2}\\
& = -48 \left( \left(\dfrac1x \right)^2 - \dfrac1{24} \dfrac1x\right) + 1\\
& = -48 \left( \left(\dfrac1x \right)^2 - 2 \dfrac1x\dfrac1{48} + \left(\dfrac1{48} \right)^2 - \left(\dfrac1{48} \right)^2\right) + 1\\
& = -48 \left( \dfrac1x - \dfrac1{48}\right)^2 + \dfrac1{48} + 1\\
& = \dfrac{49}{48} -48 \left( \dfrac1x - \dfrac1{48}\right)^2
\end{align}
Hence, note that the functions hits the maximum at $x=48$, since for all $x \neq 48$, you always subtract a positive quantity from $\dfrac{49}{48}$.
Analyze what happens when $x \to \infty$, $x \to - \infty$, $x \to 0^{\pm}$, $x \to 48^{\pm}$ to draw conclusions.
A: You may use this form and solve it elementarily:
$$f(x) = \dfrac{x^2 + 2x - 48}{x^2}=\dfrac{((x+1)-7)((x+1)+7)}{x^2}=\dfrac{((x+1)^2-49)}{x^2}=\left(\dfrac{x+1}{x}\right)^2 - \frac{49}{x^2}$$
After denoting $\dfrac{x+1}{x}=y$, we get $-48 y^2+98 y-49$ that has as a maximum the value $\frac{49}{48}$ at $y=\frac{49}{48}$, and hence $f$ reaches its maximum at $x=48$. According to our function interpretation we see that$f$ is strictly decreasing on $(-\infty,0)$ and $(48,+\infty)$ and strictly increasing on $(0,48)$. 
The proof is complete.
A: When you say "find the prime of this function" you presumably mean "find the derivative of this function"
The quotient rule  $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$ should have given you 
$$ f'(x) = \frac{(2x+2)x^2 - (x^2+2x-48)2x}{x^4} = \frac{96-2x}{x^3}$$ as Marvis found by a different route.
$f'(x)$ is clearly negative when $x$ is negative (the numerator is positive and the denominator negative) and when $x \gt 48$ (the numerator is negative and the denominator positive).
$f'(x)$ is clearly positive when $0 \lt x \lt 48$  (the numerator is positive and the denominator positive).
