Sketch $f(x) =\frac{ 2x}{x^2-5x+4}$ I have already found the domain, intercepts and critical number along with max and min points. Now when finding the intervals of concavity I understand you must take the second derivative but I can't seem to factor it. If one could justify if i derived it correctly that would be greatly appreciated. 
First Derivative: = $\frac{-2x^2+8}{(x^2-5x+4)^2}$
Second Derivative: = $\frac{4(x^3-12x+20)}{(x^2-5x+4)^3}$
 A: Just using common sense and the knowledge that the denominator, which is zero at $x=1$ and $x=4$, is negative between those points and positive outside that range, we can describe the sketch quite well:


*

*Between $1$ and $4$ $f(x)$ is always negative.  As $x$ approaches $1$ from above or $4$ from below, the curve asymptotically appoaches the lines $x=1$ and $x=4$, respectively.  Somewhere between those two lines, $f(x)$ turns over so that it can go back to negative infinity by the time it reaches $x=4$, so the curve in that region looks like an upside-down infinitely tall cup, with a maximum at roughly $x=2.5$, at which point $y$ is about $-2$.

*Slightly to the left of $x=1$, as we move further to the left,  the curve (which starts at positive infinity at $x=1$) falls rapidly, and by $x=0$ it has just fallen to the origin, which it crosses.  

*Since the curve and its derivatives are not discontinuous at $x=0$, it continues to go negative for negative $x$, and in fact $f(x) < 0$ whenever $x<0$.

*But when $x$ his a large negative number, the denominator grows faster than the numerator, so the curve approaches zero from below as $x \to -\infty$. 

*Combined with the behavior near the origin, we can deduce that there is some minimum at some negative value of $x$, at which point $f(x)<0$.

*To the right of $x=4$, $f(x)$ starts out at positive infinity, and falls, but it never becomes negative.  $f(x)$ behaves like $2/x$ for very large $x$, approaching the $x$ axis from above.
