\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{2}}}−{2}{x}+{3}\)The standard form of quadratic equation is \(\displaystyle{f{{\left({x}\right)}}}={a}{\left({x}−{h}\right)}{2}+{k}\)

(1)(a)\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{2}}}−{2}{x}+{1}+{2}.{f{{\left({x}\right)}}}={\left({x}−{1}\right)}{2}+{2}\)

On comparing it with equation (1) we get:\(h=1\),\(k=2(b)\) The vertex V is given by \(V=(h,k)=(1,2)\) The y−intercept is given by \(x=0\)

\(f(0)=3\)

So, we have (0,3) as the y−intercept.

Now the x−intercept is given by equating \(\displaystyle{f{{\left({x}\right)}}}={0}\)

\(\displaystyle{x}^{{{2}}}−{2}{x}+{3}={0}\)

\(\displaystyle{x}^{{{2}}}−{3}{x}+{x}−{3}={0}{x}\)

\(\displaystyle{\left({x}−{3}\right)}+{1}{\left({x}−{3}\right)}={0}\)

\(\displaystyle{x}=−{1},{3}{S}{o},{x}−int{e}{r}{c}{e}{p}{t}{s}\ {o}{f}\ {f}{a}{r}{e}{\left(−{1},{0}\right)}{\quad\text{and}\quad}{\left({3},{0}\right)}\)