Answer
Point corresponding to $t=-\frac{1}{2}$ is $$(\frac{11}{8},\frac{3}{4})$$ or, $$(1.35, 0.75)$$ as shown in the attached graph.
Work Step by Step
As we are given that $$f(t)=t^2+t+1$$
It will be minimum for $$f'(t)=2t+1$$
$f'(t)=0$ when $t=-\frac{1}{2}$
and $$f''(t)=2$$
As per second derivative test , $f(t)$ is minimum when $t=-\frac{1}{2}$ .
Hence, point corresponding to $t=-\frac{1}{2}$ is: $$(\frac{11}{8},\frac{3}{4})$$ or, $$(1.35, 0.75)$$.
