Using the vertex of a quadratic function, it is found that the constants are given by: a = -12, b = 29.
A quadratic equation is modeled by:
[tex]y = ax^2 + bx + c[/tex]
The vertex is given by:
[tex](x_v, y_v)[/tex]
In which:
[tex]x_v = -\frac{b}{2a}[/tex]
[tex]y_v = -\frac{b^2 - 4ac}{4a}[/tex]
In this problem, the coefficients are given by: a = 1, b = a, c = b, in which a and b are constants.
The vertex is at point (6,7), hence:
[tex]x_v = -\frac{b}{2a}[/tex]
[tex]-\frac{a}{2} = 6[/tex]
[tex]a = -12[/tex]
Then:
[tex]y_v = -\frac{b^2 - 4ac}{4a}[/tex]
[tex]-\frac{(-12)^2 - 4b}{4} = 7[/tex]
[tex]144 - 4b = 28[/tex]
[tex]4b = 116[/tex]
[tex]b = 29[/tex]
The constants are given by: a = -12, b = 29.
More can be learned about the vertex of a quadratic function at https://brainly.com/question/24737967
