[tex]\begin{cases}x(t)=t^2+7\\y(t)=t^2+5t\end{cases}[/tex]
By the chain rule,
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]
We have
[tex]\dfrac{\mathrm dx}{\mathrm dt}=2t[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dt}=2t+5[/tex]
[tex]\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2t+5}{2t}=\boxed{1+\dfrac5{2t}}[/tex]
Notice that this derivative is a function of [tex]t[/tex]. Let [tex]f(t)=\frac{\mathrm dy}{\mathrm dx}[/tex]. Then by the chain rule,
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]
We have
[tex]\dfrac{\mathrm df}{\mathrm dt}=-\dfrac5{2t^2}[/tex]
[tex]\implies\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{-\frac5{2t^2}}{2t}=\boxed{-\dfrac5{4t^3}}[/tex]
The curve is concave upward wherever the second derivative is positive. This happens whenever [tex]t^3<0[/tex], or [tex]\boxed{t<0}[/tex].
