Let's analyze the given function:

$f(x) = \frac{1}{2}x^2 - 4x + 8$

a. Does the function have a minimum or maximum value?

The function is quadratic, and its general form is $f(x) = ax^2 + bx + c$. For a quadratic function, the nature of the vertex (whether it's a minimum or maximum) depends on the coefficient of $x^2$, which is $a$.

• If $a > 0$, the parabola opens upward, and the function has a minimum value.
• If $a < 0$, the parabola opens downward, and the function has a maximum value.

In our function, $a = \frac{1}{2}$, which is positive. Therefore, the function has a minimum value.

b. Find the minimum value.

To find the minimum value, we need to first determine the x-coordinate of the vertex. The formula for the x-coordinate of the vertex of a quadratic function $f(x) = ax^2 + bx + c$ is:

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

For the given function, $a = \frac{1}{2}$ and $b = -4$, so:

$x_{\text{vertex}} = \frac{-(-4)}{2 \times \frac{1}{2}} = \frac{4}{1} = 4$

Now, substitute $x = 4$ into the function to find the minimum value:

$f(4) = \frac{1}{2}(4)^2 - 4(4) + 8$ $f(4) = \frac{1}{2}(16) - 16 + 8$ $f(4) = 8 - 16 + 8$ $f(4) = 0$

So, the minimum value of the function is 0.

---

Summary:

a. The function has a minimum value.
b. The minimum value is 0.

---

Do you need further details or clarifications? Here are some additional questions related to the topic:

1. What is the general form of a quadratic function and how do you identify its vertex?
2. How do you calculate the axis of symmetry for a quadratic function?
3. What happens to the graph of a quadratic function if $a$ is negative?
4. How would you solve for the roots of the quadratic function $f(x) = \frac{1}{2}x^2 - 4x + 8$?
5. Can you apply the quadratic formula to solve for the roots of a quadratic function?

Tip: Always check the coefficient of $x^2$ to quickly determine if the parabola opens up or down, which directly tells you whether the function has a minimum or maximum value.