Answer:
Quadratic coefficient = -14, linear coefficient = 168, constant term = -494
Step-by-step explanation:
Its easiest to first start out with a vertex form equation because it can then be converted to a standard quadratic.
given a vertex at (6,10) , and point of intersection at (7,-4), the equation can be set up like this:
y = a(x-6)^2+10, because we know h and k from the vertex.
We also know y and x from the given point of intersection so a can be solved.
-4 = a(7-6)^2+10
-4 = a + 10
-4-10 = a
a = -14 which is also known as the quadratic coefficient because it is part of a second degree quantity and a Trionomial as a whole(quadratic).
since all the variables are known, you can
expand the equation and set it to standard form
y = -14(x-6)^2+10
y = -14(x^2-12x+36)+10
y = -14x^2+168x-504+10
y = -14x^2+168x-494
For reference, the standard form equation is ax^2+bx+c where a is the quadratic coefficient, b is the linear coefficient, and c is the constant coefficient.
In this instance, -14x^2+168x-494 = ax^2+bx+c
