Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 710: 27

$(\frac{11}{8},\frac{3}{4})$

$f(t)=t^2+t+1$ Find derivative: $f'(t)=2t+1$ $f'(t)=0$ when $t=\frac{1}{2}$ Also $f''(t)=2$ Hence, by the second derivative test, $f(t)$ is a minimum when $t=\frac{1}{2}$. The point correspondsing to $t=-\frac{1}{2}$ is $(\frac{11}{8},\frac{3}{4})$