We are asked to state the transformations in the following quadratic equation: f(x) = -3(x-3)^2 + 4.

This is an equation for a parabola.

The parent graph of this equation is f(x) = x^ 2, where the vertex of a parabola is at (0,0) and the parabola opens upward.

The equation in this problem is in the form of f(x) = a(x-h)^2 + k.

The transformed vertex is given by (h,k) in the equation.

**The vertex of the parabola in the given problem is at (3,4).**

**The parabola opens downward because the "a" value in the transformed equation is negative. **

**Since the a value is a whole number the parabola will "shrink," meaning that its graph will be narrower than the parent graph.**

First, we'll raise to square the binomial, using the formula:

(a-b)^2 = a^2 - 2ab + b^2

(x-3)^2 = x^2 - 6x + 9

Now, we'll multiply the expansion by -3:

-3(x-3)^2 = -3(x^2 - 6x + 9)

-3(x-3)^2 = -3x^2 + 18x - 27

Now, we'll add 4:

f(x) = -3x^2 + 18x - 27 + 4

We'll combine like terms:

f(x) = -3x^2 + 18x - 23

**The expresison of the function is a quadratic, whose leading coefficients is -3: f(x) = -3x^2 + 18x - 23.**