Answer
The curve is concave up when $t\lt0$.
Work Step by Step
$x=t^{2}+1$, $y=t^{2}+t$
Step 1
Find the derivatives
$\frac{dx}{dt}=2t$ $\frac{dy}{dt}=2t+1$
Then we can find
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t+1}{2t}= 1+\frac{1}{2t}$
Step 2
Then we can find
$\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}=\frac{-1/(2t^{2})}{2t}=-\frac{1}{4t^{3}}$
The curve is concave up when $\frac{d^{2}y}{dx^{2}} \gt 0$, when $t \lt 0$.
