state the transformations: f(x)= -3(x-3)^2+4
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.