Let's break down the problem step by step to find the correct answer.

1. Given Function:

The function is: $f(x) = -x^2 + 6$ This is a quadratic function where the coefficient of $x^2$ is negative, meaning the parabola opens downward.

2. Standard Form Comparison:

The standard form of a quadratic function is: $f(x) = ax^2 + bx + c$ For the given function:

• $a = -1$
• $b = 0$
• $c = 6$

3. Find the Vertex:

The vertex of a parabola for a quadratic function $f(x) = ax^2 + bx + c$ can be found using the formula for the $x$-coordinate: $x = \frac{-b}{2a}$ Substitute the values of $a$ and $b$: $x = \frac{-0}{2(-1)} = 0$ Now, substitute $x = 0$ into the original function to find the $y$-coordinate of the vertex: $y = f(0) = -(0)^2 + 6 = 6$ Thus, the vertex is at $(0, 6)$.

4. Range of the Function:

Since the parabola opens downward, the maximum value of the function is the $y$-coordinate of the vertex, which is 6. The function's values are all less than or equal to 6, because the parabola decreases as you move away from the vertex.

Thus, the range of the function is: $\{y \mid y \leq 6\}$ This means the function can take any $y$-value less than or equal to 6.

5. Correct Answer:

The range of the function is given by: $\{y \mid y \leq 6\}$

Conclusion:

The correct answer is: $\boxed{\{y \mid y \leq 6\}}$

Would you like further clarification on any part of this solution?

Related Questions:

1. What is the domain of the function $f(x) = -x^2 + 6$?
2. How would the range change if the equation were $f(x) = x^2 + 6$?
3. What is the axis of symmetry for this parabola?
4. How do you find the range of a function that opens upwards?
5. What is the significance of the vertex in the context of the range?

Tip:

When identifying the range of quadratic functions, always check if the parabola opens upwards (minimum $y$-value) or downwards (maximum $y$-value) by looking at the sign of the coefficient of $x^2$.