A quadratic function has its vertex at the point (6,10). The function passes through the point (7, – 4). Find the quadratic and linear coefficients and the constant term of the function.

The quadratic coefficient:
The linear coefficient is:
The constant term is:

I’m confused

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

a. is correct