To find the range of the quadratic function $y = -3x^2 - 30x - 72$, let's analyze the function.

Step 1: Identify the form of the function

The function is a quadratic equation in the standard form $y = ax^2 + bx + c$, where:

• $a = -3$
• $b = -30$
• $c = -72$

Since $a$ is negative, this parabola opens downward, meaning it has a maximum point at its vertex.

Step 2: Find the vertex of the parabola

The $x$-coordinate of the vertex for a quadratic function $y = ax^2 + bx + c$ is given by: $x = -\frac{b}{2a}$ Plugging in the values of $a$ and $b$: $x = -\frac{-30}{2 \times -3} = \frac{30}{6} = 5$

Now, substitute $x = 5$ back into the function to find the $y$-coordinate (the maximum value of $y$): $y = -3(5)^2 - 30(5) - 72$ $y = -3(25) - 150 - 72$ $y = -75 - 150 - 72 = -297$

Step 3: State the range

Since the parabola opens downward, $y \leq -297$. Thus, the range of the function is: $y \leq -297$

Would you like more details on this solution, or do you have any questions?

Here are some related questions to expand your understanding:

1. How would the range change if the leading coefficient $a$ were positive?
2. What are the coordinates of the vertex for this function?
3. How would you find the axis of symmetry for this parabola?
4. How do you determine if a quadratic function has a maximum or minimum value?
5. How would you rewrite this quadratic function in vertex form?

Tip: The range of a quadratic function can be found easily by identifying whether the parabola opens upwards or downwards and locating the maximum or minimum value at the vertex.