Using the vertex of a quadratic function, it is found that the constants are given by: a = -12, b = 29.
What is the vertex of a quadratic equation?
A quadratic equation is modeled by:
\(y = ax^2 + bx + c\)
The vertex is given by:
\((x_v, y_v)\)
In which:
\(x_v = -\frac{b}{2a}\)
\(y_v = -\frac{b^2 - 4ac}{4a}\)
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:
\(x_v = -\frac{b}{2a}\)
\(-\frac{a}{2} = 6\)
\(a = -12\)
Then:
\(y_v = -\frac{b^2 - 4ac}{4a}\)
\(-\frac{(-12)^2 - 4b}{4} = 7\)
\(144 - 4b = 28\)
\(4b = 116\)
\(b = 29\)
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