The problem asks to find the range of the quadratic function $f(x) = -2x^2 + 16x - 33$ and express it in interval notation.

Steps to find the range:

1. Determine the vertex:
   Since the quadratic is in standard form $f(x) = ax^2 + bx + c$, we use the vertex formula to find the x-coordinate of the vertex: $x = -\frac{b}{2a}$ For $f(x) = -2x^2 + 16x - 33$, we have $a = -2$, $b = 16$, and $c = -33$.

   So, $x = -\frac{16}{2(-2)} = \frac{16}{4} = 4$

2. Find the y-coordinate of the vertex:
   Substitute $x = 4$ into the quadratic function to get the corresponding y-value: $f(4) = -2(4)^2 + 16(4) - 33$ $f(4) = -2(16) + 64 - 33 = -32 + 64 - 33 = -1$ So, the vertex is $(4, -1)$.

3. Determine the range:
   Since the parabola opens downward (because $a = -2$ is negative), the vertex represents the maximum point of the function. Therefore, the maximum value of $f(x)$ is $-1$, and the function continues decreasing to $-\infty$ as $x$ moves away from the vertex.

   The range is all y-values less than or equal to $-1$.

Final answer in interval notation:

The range is $(-\infty, -1]$.

Would you like more details on this explanation or have any questions?

Here are 5 related questions:

1. How do you find the axis of symmetry for a quadratic function?
2. What is the significance of the sign of the leading coefficient in a quadratic equation?
3. How can the discriminant help determine the number of real roots of a quadratic equation?
4. How would the range change if the quadratic function opened upwards?
5. What are the properties of the graph of a quadratic function?

Tip: For any quadratic function, the vertex is a critical point to identify the function's range and direction.