Answer
$g(x)=-6\left(x-\frac{5}{3}\right)^2-\frac{4}{3}$
Work Step by Step
Given the function $$g(x) =-6 x^2+20 x-18.$$ To complete the square of the function, we first factor out $-6$ so that the coefficient of the $x^2$ term is positive $1$. We then add and subtract half the square of the middle term and simplify: \begin{equation}
\begin{aligned}
g(x) & =-6 x^2+20 x-18 \\
& =-6\left(x^2-\frac{20}{6} x\right)-18 \\
& =-6\left[x^2-\frac{10 x}{3}+\left(\frac{10}{6}\right)^2\right]-18+6\left(\frac{10}{6}\right)^2 \\
& =-6\left(x-\frac{10}{6}\right)^2-18+6 \cdot \frac{100}{36} \\
& =-6\left(x-\frac{5}{3}\right)^2-18\cdot\frac{6}{6}+\frac{100}{6}\\
&=-6\left(x-\frac{5}{3}\right)^2-\frac{8}{6}\\
&=-6\left(x-\frac{5}{3}\right)^2-\frac{4}{3}.
\end{aligned}
\end{equation} Hence the function can be written: $$g(x)=-6\left(x-\frac{5}{3}\right)^2-\frac{4}{3}.$$
