a.For \(\displaystyle{x}^{{2}}−{6}{x}\) to be a perfect square trinomial, you must add \(\displaystyle{\left(\frac{{b}}{{2}}\right)}{2}={\left(−\frac{{6}}{{2}}\right)}^{{2}}={9}\). This leads us to:

\(\displaystyle{y}−{1}={3}{\left({x}^{{2}}−{6}{x}+{9}\right)}\)

To balance what we added to the right side, consider the expanded form:

\(\displaystyle{y}−{1}={3}{x}^{{2}}−{18}{x}+{27}\)

Hence, you must add 27 to the left side:

\(\displaystyle{y}−{1}+{27}={3}{x}^{{2}}−{18}{x}+{27}\)

\(\displaystyle{y}+{26}={3}{\left({x}^{{2}}−{6}{x}+{9}\right)}\)

b.Write the parabola in vertex form y−k=a(x−h)2 where (h,k) is the vertex. Here, we have:

y+26=3(x−3)2

or

y−(−26)=3(x−3)2

So, the vertex is at (3, -26).

\(\displaystyle{y}−{1}={3}{\left({x}^{{2}}−{6}{x}+{9}\right)}\)

To balance what we added to the right side, consider the expanded form:

\(\displaystyle{y}−{1}={3}{x}^{{2}}−{18}{x}+{27}\)

Hence, you must add 27 to the left side:

\(\displaystyle{y}−{1}+{27}={3}{x}^{{2}}−{18}{x}+{27}\)

\(\displaystyle{y}+{26}={3}{\left({x}^{{2}}−{6}{x}+{9}\right)}\)

b.Write the parabola in vertex form y−k=a(x−h)2 where (h,k) is the vertex. Here, we have:

y+26=3(x−3)2

or

y−(−26)=3(x−3)2

So, the vertex is at (3, -26).