# Question # A quadratic function has its vertex at the point (1,6). The function passes through the point (-2, -3). Find the quadratic and linear coefficients and the constant team of the function. -The quadratic coefficients is -The linear coefficients is -The constant term is

ANSWERED A quadratic function has its vertex at the point (1,6). The function passes through the point (-2, -3). Find the quadratic and linear coefficients and the constant team of the function.
-The linear coefficients is
-The constant term is 2020-12-23

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
Constant term$$=x^2+2x+5$$