The problem is asking for the domain and range of the quadratic function:

$f(x) = -4x^2 + 3x - 3$

1. Domain:

The domain of a quadratic function is all real numbers, because it is defined for any input $x$. There's no restriction on $x$ like division by zero or square roots of negative numbers that would limit the domain.

Thus, the domain is: $\mathbb{R} \quad \text{(All real numbers)}$

2. Range:

For the range, we need to determine the possible output values of the function. Since the coefficient of $x^2$ is negative ($-4$), the parabola opens downward. The function has a maximum value, and we can find that maximum using the vertex formula.

Step 1: Find the vertex

The $x$-coordinate of the vertex for a quadratic function $ax^2 + bx + c$ is given by: $x = \frac{-b}{2a}$ For the function $f(x) = -4x^2 + 3x - 3$, we have:

• $a = -4$
• $b = 3$

Substitute these values: $x = \frac{-3}{2(-4)} = \frac{-3}{-8} = \frac{3}{8}$

Step 2: Find the maximum value by substituting the $x$-coordinate of the vertex back into the function

$f\left(\frac{3}{8}\right) = -4\left(\frac{3}{8}\right)^2 + 3\left(\frac{3}{8}\right) - 3$

First, calculate $\left(\frac{3}{8}\right)^2$: $\left(\frac{3}{8}\right)^2 = \frac{9}{64}$

Now substitute into the function: