To find the range of the quadratic function $y = -8x^2 - 32x + 40$, we need to follow these steps:

1. Identify the vertex of the parabola:

The general form of a quadratic function is $y = ax^2 + bx + c$. In this case, $a = -8$, $b = -32$, and $c = 40$.

The x-coordinate of the vertex is given by: $x = \frac{-b}{2a}$ Substituting the values of $a$ and $b$: $x = \frac{-(-32)}{2(-8)} = \frac{32}{-16} = -2$

2. Find the y-coordinate of the vertex:

Substitute $x = -2$ back into the equation to find $y$: $y = -8(-2)^2 - 32(-2) + 40$ $y = -8(4) + 64 + 40$ $y = -32 + 64 + 40 = 72$

3. Determine the range:

Since the coefficient of $x^2$ (which is $a = -8$) is negative, the parabola opens downward. This means the vertex represents the maximum point of the parabola.

Therefore, the range of the function is: $(-\infty, 72]$

Would you like more details or have any questions?

5 Related Questions:

1. How do you determine if a parabola opens upward or downward?
2. What is the significance of the vertex in the range of a quadratic function?
3. How do you find the axis of symmetry in a quadratic function?
4. Can you explain how the sign of the leading coefficient affects the shape of the parabola?
5. What are other methods to find the range of a quadratic function?

Tip:

Remember that the vertex form of a quadratic function can make it easier to determine the range and vertex of the function directly.