Answer
$\frac{dy}{dx} = \frac{2t+1}{2t}$
$\frac{d^2y}{dx^2} = -\frac{1}{4t^3}$
The curve is concave upward when $t \lt 0$
Work Step by Step
$x = t^2+1$
$y = t^2+t$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t+1}{2t}$
We can find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{\frac{4t-(2)(2t+1)}{4t^2}}{2t} = \frac{-2}{8t^3} = -\frac{1}{4t^3}$
The curve is concave upward when $\frac{d^2y}{dx^2} \gt 0$
$-\frac{1}{4t^3} \gt 0$ when $t \lt 0$
The curve is concave upward when $t \lt 0$
