To find the range of the quadratic function $f(x) = -x^2 - 10x - 23$, we need to analyze its vertex and behavior.

Step 1: Rewrite the quadratic function in vertex form

The general form of a quadratic function is: $f(x) = ax^2 + bx + c$ where $a = -1$, $b = -10$, and $c = -23$.

To find the vertex, we can use the formula for the x-coordinate of the vertex: $x = \frac{-b}{2a}$ Substitute the values of $a$ and $b$: $x = \frac{-(-10)}{2(-1)} = \frac{10}{-2} = -5$

Step 2: Find the y-coordinate of the vertex

Substitute $x = -5$ into the function to find $f(-5)$: $f(-5) = -(-5)^2 - 10(-5) - 23 = -25 + 50 - 23 = 2$

So the vertex of the quadratic function is $(-5, 2)$.

Step 3: Analyze the direction of the parabola

Since the coefficient of $x^2$ is negative ($a = -1$), the parabola opens downward, meaning the function has a maximum value at the vertex. The maximum value is $f(-5) = 2$.

Step 4: Determine the range

Because the parabola opens downward, the range of the function includes all values less than or equal to 2. Therefore, the range is: $(-\infty, 2]$

Final Answer (in interval notation):

$\boxed{(-\infty, 2]}$

Would you like more details or have any questions?

Here are 5 follow-up questions you might find interesting:

1. How do you convert a quadratic function to vertex form using completing the square?
2. What is the significance of the coefficient of $x^2$ in determining the direction of the parabola?
3. How do you determine if a parabola has a maximum or minimum value?
4. Can you find the domain of any quadratic function, and how does it differ from the range?
5. How can you use the discriminant to find the number of real solutions for a quadratic equation?

Tip: Always check the sign of the quadratic term ($x^2$) to know if the parabola opens upwards or downwards.