Answer
The curve is concave up for $t>0$.
Work Step by Step
We have
$x\left( t \right) = {t^2}$ and $y\left( t \right) = {t^3} - 4t$.
So,
$x'\left( t \right) = 2t$, $x{\rm{''}}\left( t \right) = 2$,
$y'\left( t \right) = 3{t^2} - 4$, $y{\rm{''}}\left( t \right) = 6t$.
Using Eq. (13) we get
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{x'\left( t \right)y{\rm{''}}\left( t \right) - y'\left( t \right)x{\rm{''}}\left( t \right)}}{{x'{{\left( t \right)}^3}}} = \frac{{4 + 3{t^2}}}{{4{t^3}}}$.
By Theorem 1 of Section 4.4, the curve is concave up if $\frac{{{d^2}y}}{{d{x^2}}} > 0$.
So we solve $\frac{{4 + 3{t^2}}}{{4{t^3}}} > 0$.
Suppose $t<0$, then $4+3 t^2<0$ such that $\frac{{4 + 3{t^2}}}{{4{t^3}}} > 0$. The solution are complex numbers. Since $4+3 t^2$ is always positive, for real solutions we must have $t>0$. Hence, the curve is concave up for $t>0$.
