[tex]\bf f(x)=x^2+4x\qquad \cfrac{df}{dx}=2x+4\qquad \boxed{f''(x)=2}\impliedby
\begin{array}{llll}
\textit{just a positive}\\
constant
\end{array}[/tex]
there are no inflection points, because it never changes concavity, is just a constant and thus for any region over the x-axis, will always be a positive value, and thus is always "concave up".
