The equation $y = 4x^2 - 16x + 12$ is a quadratic function. Since the coefficient of $x^2$ is positive (4), the parabola opens upwards, which means it has a minimum value, not a maximum.

To find the minimum value, we first determine the vertex, as the vertex gives us the minimum (or maximum) point for a parabola. The x-coordinate of the vertex can be found using the formula:

$x_{\text{vertex}} = \frac{-b}{2a}$

For the quadratic equation $y = ax^2 + bx + c$, we have:

• $a = 4$
• $b = -16$
• $c = 12$

Now, calculate the x-coordinate of the vertex:

$x_{\text{vertex}} = \frac{-(-16)}{2(4)} = \frac{16}{8} = 2$

Next, substitute $x = 2$ back into the equation to find the y-coordinate of the vertex (which will be the minimum value):

$y = 4(2)^2 - 16(2) + 12 = 4(4) - 32 + 12 = 16 - 32 + 12 = -4$

Thus, the minimum value of the function is $y = -4$.

The correct description is:

• It has a minimum value of -4.

Would you like more details or explanations? Here are five related questions:

1. How do you determine if a quadratic function has a minimum or maximum?
2. How can you graph the function $y = 4x^2 - 16x + 12$ to verify its minimum value?
3. What happens to the graph if the coefficient of $x^2$ changes from positive to negative?
4. How do you complete the square to find the vertex of a quadratic equation?
5. What are the real-life applications of quadratic functions and their minimum or maximum points?

Tip: The vertex form of a quadratic equation, $y = a(x - h)^2 + k$, makes it easy to see the vertex at $(h, k)$.