Question

What's the concavity of f(x)=x^2+4x

Answer 1

`\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}`

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".