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

Answer 1

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