Respuesta :
The equation is
[tex]y=3x^2-30x+63[/tex]Explanation:
Given:
The roots of the function is 3 and 7.
The function passes through the point (6,-9).
The objective is to find the quadratic function.
Consider the roots as,
[tex]\begin{gathered} a=3 \\ b=7 \end{gathered}[/tex]The quadratic function with the roots can be written as,
[tex]y=k(x-a)(x-b)[/tex]Substitute the values of roots in the above equation.
[tex]y=k\lbrack(x-3)(x-7)\rbrack\text{..}...(1)[/tex]The value of k can be calculated by substituting the point (x,y) = (6,-9) in the above equation.
[tex]\begin{gathered} -9=k(6-3)(6-7) \\ -9=3k(-1) \\ -9=-3k \\ k=\frac{-9}{-3} \\ k=3 \end{gathered}[/tex]Now substitute the value of k in equation (1).
[tex]\begin{gathered} y=3\lbrack(x-3)(x-7)\rbrack \\ =3\lbrack x^2-7x-3x+21\rbrack \\ =3\lbrack x^2-10x+21\rbrack \\ =3x^2-30x+63 \end{gathered}[/tex]Hence, the required quadratic function is obtained.
