On what interval is the curve concave downward? Getting this problem out of my text book. 
On what interval is the curve concave downward?
$$ y\quad =\quad \int _{ 0 }^{ x }{ \frac { { t }^{ 2 } }{ { t }^{ 2 }\quad +\quad t\quad +\quad 2 } dt } $$
The solution provided  is 
$$ \begin{align*}
y\quad &=\quad \int _{ 0 }^{ x }{ \frac { { t }^{ 2 } }{ { t }^{ 2 }\quad +\quad t\quad +\quad 2 } dt } \\ 
\Rightarrow \quad y'\quad &=\frac { { x }^{ 2 } }{ x^{ 2 }\quad +\quad x\quad +\quad 2 } \\ 
\Rightarrow \quad y''\quad &=\frac { \left( { x }^{ 2 }+x+2 \right) \left( 2x \right) -{ x }^{ 2 }\left( 2x+1 \right)  }{ { \left( { x }^{ 2 }+x+2 \right)  }^{ 2 } } \\
&=\frac { { 2x }^{ 3 }+2{ x }^{ 2 }+4x-{ 2x }^{ 3 }-{ x }^{ 2 } }{ { \left( { x }^{ 2 }+x+2 \right)  }^{ 2 } } \\ 
&=\quad \frac { { x }^{ 2 }+4x }{ { \left( { x }^{ 2 }+x+2 \right)  }^{ 2 } } \\ 
&=\quad \frac { x\left( x+4 \right)  }{ { \left( { x }^{ 2 }+x+2 \right)  }^{ 2 } } \\ 
\end{align*}$$
I have two questions. 


*

*I am not clear on how to justify / work the from the first noted  implication to the second noted implication. 

*Also I do cannot quite visualize how the provided answer actually answers the question. I wanted to know if someone can elaborate on this question a bit more for me. 
 A: The second implication is just the quotient rule, as noted in the comments.
The reason the solution is useful in providing an answer to the question is that we may now apply the second derivative test for concavity: Namely, a (second differentiable) function $f(x)$ is concave up (respectively, concave down) at a point $c$ if and only if $f''(c)>0$ (respectively, $f''(c)<0$). Now, let's look at the equation for the second derivative you have above: $$y''=\frac{x(x+4)}{(x^2+x+2)^2}$$ Note that the denominator of this function is always positive, as a squared real number is either zero or positive, and we can easily use the quadratic formula to see that it is nonzero for $x$ real. So, to determine whether the second derivative is positive or negative, we need only look at whether the numerator is positive or negative. $x(x+4)=0$ implies $x=0$ or $x=-4$. Plugging values in between these two into $x(x+4)$ yields that $x(x+4)<0$ if and only if $-4<x<0$, and $x(x+4)>0$ when $x<-4$ or when $x>0$. Hence, applying the second derivative test:
$y$ is concave down when $-4<x<0$ and $y$ is concave up when $x<-4$ and when $x>0$.
