The function given is $f(x) = 4x^2 - 32x + 66$, which is a quadratic function of the form $f(x) = ax^2 + bx + c$, where $a = 4$, $b = -32$, and $c = 66$.

Step 1: Determine the Range

Since this is a parabola that opens upwards (because $a > 0$), the minimum value of the function occurs at the vertex.

To find the vertex of the parabola, we use the formula for the $x$-coordinate of the vertex:

$x_{\text{vertex}} = \frac{-b}{2a} = \frac{-(-32)}{2(4)} = \frac{32}{8} = 4$

Next, substitute $x = 4$ into the function to find the $y$-coordinate of the vertex, which gives the minimum value of $f(x)$:

$f(4) = 4(4)^2 - 32(4) + 66 = 4(16) - 128 + 66 = 64 - 128 + 66 = 2$

So, the vertex is at $(4, 2)$. Since the parabola opens upwards, the minimum value is $2$, and the range of the function is:

$\boxed{[2, \infty)}$

Step 2: Determine Intervals of Increase and Decrease

To determine where the function is increasing or decreasing, we look at the vertex and the direction the parabola opens:

• Decreasing: The function decreases on the interval $(-\infty, 4]$.
• Increasing: The function increases on the interval $[4, \infty)$.

Final Answer:

• Range: $[2, \infty)$
• Interval of decrease: $(-\infty, 4]$
• Interval of increase: $[4, \infty)$

Would you like further explanation on any step, or have any other questions?

Related Questions:

1. How do you find the vertex of a general quadratic function?
2. Why does a parabola open upwards or downwards depending on the coefficient of $x^2$?
3. Can you explain how to derive the formula for the range of a quadratic function?
4. What would change in the range if the parabola opened downwards?
5. How can you graph a quadratic function based on its vertex and direction?

Tip:

The vertex form of a quadratic function $f(x) = a(x - h)^2 + k$ makes it easier to identify the vertex and determine the range directly.