Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.1 Parametric Equations - Exercises - Page 603: 37

The answer is $y=x^2+8 x+8$.

Since $y=x^2$ is always positive, the minimum of $y$ is zero. So, the minimum of the curve occurs at $(0,0)$. The parametric equation of $y=x^2$ is $c(t)=(t,t^2)$. To translate the minimum of $c(t)$ at $(0,0)$ to $(-4,-8)$, we replace $c(t)=(t,t^2)$ by $c(t)=(-4+t,-8+t^2)$. So, the coordinates of the translated curve are $x=t-4$ and $y=t^2-8$. Solving for $t$ from $x$ we get $t=x+4$. Substituting $t$ into $y$ gives $y=(x+4)^2-8$. So, $y=x^2+8 x+8$.