Standard vertex form of quadratic function is

\(\displaystyle{y}={a}{\left({x}−{h}\right)}^{{2}}+{k}\)

Where point (h,k) is vertex point.

\(\displaystyle\because\) vertex point is (1,6)

\(\displaystyle\because\) Substitute this point in standard vertex form

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

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

Also function pass through point (−2,−3).

\(\displaystyle\because−{3}={a}{\left(−{2}−{1}\right)}^{{2}}+{6}\)

\(\displaystyle\Rightarrow−{3}={a}\cdot{\left(−{3}\right)}{2}+{6}\)

\(\displaystyle\Rightarrow−{3}={a}\cdot{9}+{6}\)

\(\displaystyle\Rightarrow{9}{a}=−{3}−{6}\)

\(\displaystyle\Rightarrow{9}{a}=−{9}\)

\(\displaystyle\Rightarrow{a}=−\frac{{9}}{{9}}\)

=-1

Substitute 'a' value in equation (1)

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

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

\(\displaystyle=−{x}^{{2}}+{2}{x}−{1}+{6}\)

\(\displaystyle={x}^{{2}}+{2}{x}+{5}\)

Hence quadratic function is \(\displaystyle{y}=−{x}^{{2}}\)

Where

Quadratic coefficient =−1

Linear coefficient=2

Constant term\(=x^2+2x+5\)