Answer
Vertex: $\left(-\frac{11}{2},-\frac{137}{4}\right)$
Axis of symmetry: $t=-\frac{11}{2}$.
Work Step by Step
To complete the square, we take $\frac{1}{2}$ of the coefficient of $t$ and square the result. This gives $\left(\frac{1}{2} \cdot 11\right)^2=\left(\frac{11}{2}\right)^2=\frac{121}{4}$. Using this number, we can rewrite the formula for $v(t)$ :
$$
v(t)=t^2+11 t+\left(\frac{11}{2}\right)^2-\quad \frac{121}{4} \quad-4
$$
or
$$v(t) =\left(t+\frac{11}{2}\right)^2-\frac{137}{4} $$
The vertex of $v$ is $\left(-\frac{11}{2},-\frac{137}{4}\right)$ and the axis of symmetry is $t=-\frac{11}{2}$.
