To determine the range of the quadratic function $y = -x^2 - 12x - 32$, we can follow these steps:

Step 1: Determine the orientation of the parabola

Since the coefficient of $x^2$ is negative ($-1$), the parabola opens downwards. This means that the function will have a maximum value.

Step 2: Find the vertex

The vertex of the parabola will give us the maximum $y$-value since the parabola opens downwards. The $x$-coordinate of the vertex can be found using the formula:

$x = -\frac{b}{2a}$

Here, $a = -1$ and $b = -12$. Substituting these values:

$x = -\frac{-12}{2(-1)} = \frac{12}{-2} = -6$

Now, substitute this $x$-value back into the original function to find the corresponding $y$-value:

$y = -(-6)^2 - 12(-6) - 32$

$y = -36 + 72 - 32$

$y = 4$

So, the vertex of the parabola is at $(-6, 4)$.

Step 3: Determine the range

Since the parabola opens downwards and the maximum value of $y$ is $4$, the range of the function is all $y$-values less than or equal to 4.

Final Answer:

The range of the quadratic function $y = -x^2 - 12x - 32$ is $\{ y \mid y \leq 4 \}$.

Would you like more details or have any questions?

Here are 5 related questions:

1. How does the sign of the leading coefficient affect the range of a quadratic function?
2. How do you find the x-intercepts of a quadratic function?
3. What role does the vertex play in determining the range of a quadratic function?
4. How can you determine the axis of symmetry from the quadratic equation?
5. What does it mean for a parabola to have a maximum or minimum point?

Tip: Always check the sign of the leading coefficient first to understand whether the quadratic function has a maximum or minimum value.