Answer
$p(t)=2(t-0.03)^2+0.0982$
Vertex: $(0.03,0.0982)$
Axis of symmetry: $x=0.03$.
Work Step by Step
Factor out the coefficient of $x^2$ and square half of the coefficient of the $x$-term: $(0.06 / 2)^2=0.009$. Adding and subtracting this number after the $x$-term gives
$$
\begin{aligned}
& p(t)=2 t^2-0.12 t+0.1\\
& p(t)=2\left(t^2-0.06 t+0.05\right) \\
& p(t)=2\left(t^2-0.06 t+0.03^2+0.05-0.03^2\right) \\
& p(t) = 2\left(t^2-0.06 t+0.03^2\right)+2\cdot(0.0491)\\
& p(t)=2(t-0.03)^2+0.0982
\end{aligned}
$$ The vertex is $(0.03,0.0982)$ and the axis of symmetry is $x=0.03$.
