To analyze the function $f(x) = x^2 - 8x + 12$, we'll find the following:

1. Vertex (to determine whether the function has a maximum or minimum, and the value at that point).
2. Range (the set of possible values of $f(x)$).

1. Maximum or Minimum

The function $f(x) = x^2 - 8x + 12$ is a quadratic function in the form $ax^2 + bx + c$, where:

• $a = 1$ (positive, indicating the parabola opens upwards).
• $b = -8$
• $c = 12$

Since $a = 1 > 0$, the parabola opens upwards, meaning the function has a minimum.

The vertex occurs at $x = -\frac{b}{2a}$. For this function: $x = -\frac{-8}{2(1)} = \frac{8}{2} = 4$ Now, substitute $x = 4$ back into the function to find the minimum value: $f(4) = (4)^2 - 8(4) + 12 = 16 - 32 + 12 = -4$ So, the minimum value of $f(x)$ is $-4$, and it occurs at $x = 4$.

2. Range

Since the function has a minimum value at $x = 4$, and it opens upwards, the range is all values of $f(x)$ greater than or equal to $-4$. Thus, the range is: $\text{Range} = [-4, \infty)$

Summary:

• Minimum value: $-4$ at $x = 4$
• Range: $[-4, \infty)$

Would you like further details or have any questions?

Here are 5 related questions:

1. How do you determine if a quadratic function opens upwards or downwards?
2. What is the formula for finding the vertex of a quadratic function?
3. How do you find the axis of symmetry in a quadratic function?
4. What does the discriminant tell us about the roots of a quadratic function?
5. Can you describe how to complete the square for the function $f(x) = x^2 - 8x + 12$?

Tip: Always check the sign of $a$ in a quadratic function to quickly determine if it has a maximum (if $a < 0$) or a minimum (if $a > 0$).