Answer
Vertex: $(-7 / 12,23 / 72)$.
No zeros
Work Step by Step
Complete the square to get the vertex form.
$$
\begin{aligned}
y &= 2 x^2+\frac{7}{3} x+1 \\
& =2\left(x^2+\frac{7}{6} x+\frac{1}{2}\right) \\
& =2\left(x^2+\frac{7}{6} x\left(\frac{7}{12}\right)^2-\left(\frac{7}{12}\right)^2+\frac{1}{2}\right) \\
& =2\left(x+\frac{7}{12}\right)^2+2\left(-\frac{49}{144}+\frac{1}{2}\right) \\
& =2\left(x+\frac{7}{12}\right)^2+\frac{23}{72}
\end{aligned}
$$
The vertex is $(-7 / 12,23 / 72)$. We can check for zeros by setting $y=0$:
$$
\begin{aligned}
2\left(x+\frac{7}{12}\right)^2+\frac{23}{72} & =0 \\
2\left(x+\frac{7}{12}\right)^2 & =-\frac{23}{72}
\end{aligned}
$$
This equation has no solution. No zeros.
