Describe the transformed function for the original equation is \(\displaystyle{y}={x}^{{{2}}}\) for the following terms:

\(\displaystyle{\left({a}\right)}{y}=-{2}{\left({x}-{1}\right)}^{{{2}}}+{23}\)

\(\displaystyle{a}=-{2}\ {k}={1}\ {d}={1}\ {c}={23}\)

\(\displaystyle{\left({b}\right)}{y}={\left[{\frac{{{12}}}{{{13}}}}{\left({x}+{9}\right)}\right]}^{{{2}}}-{14}\)

\(\displaystyle{k}={\frac{{{12}}}{{{13}}}}\ {d}=-{9}\ {c}=-{14}\)

\(\displaystyle{\left({c}\right)}{y}={x}^{{{2}}}-{8}{x}+{16}\)

\(\displaystyle={\left({x}-{4}\right)}^{{{2}}}\)

\(\displaystyle{d}={4}\)

\(\displaystyle{\left({d}\right)}{y}={\left({x}+{\frac{{{3}}}{{{7}}}}\right)}{\left({x}+{\frac{{{3}}}{{{7}}}}\right)}\)

\(\displaystyle={\left({x}+{\frac{{{3}}}{{{7}}}}\right)}^{{{2}}}\)

\(\displaystyle{d}=-{\frac{{{3}}}{{{7}}}}\)

\(\displaystyle{\left({e}\right)}{y}={40}{\left[-{7}{\left({x}-{10}\right)}\right]}^{{{2}}}+{9}\)

\(\displaystyle{a}={40}\ {k}=-{7}\ {d}={10}\ {c}={9}\)