The function given in the image is:

$y(x) = (x + 3)(x - 2)$

To find the value of $x$ at which the function has a minimum, we can follow these steps:

Step 1: Expand the function

First, let's expand the expression for $y(x)$.

$y(x) = (x + 3)(x - 2) = x^2 - 2x + 3x - 6 = x^2 + x - 6$

Step 2: Find the critical points

To locate the minimum or maximum, we need to find the derivative of the function and set it equal to zero.

The derivative of $y(x) = x^2 + x - 6$ is:

$y'(x) = 2x + 1$

Now, set $y'(x) = 0$ to find the critical points:

$2x + 1 = 0$ $x = -\frac{1}{2}$

Step 3: Determine whether it's a minimum or maximum

To determine if $x = -\frac{1}{2}$ is a minimum, we need to check the second derivative:

$y''(x) = 2$

Since the second derivative is positive, the function has a minimum at $x = -\frac{1}{2}$.

Thus, the value of $x$ at which the function has a minimum is:

$x = -\frac{1}{2}$

This matches the option:

$\boxed{-\frac{1}{2}}$

Would you like further clarification or have any other questions?

Here are five related questions to expand on this topic:

1. How do you find the maximum of a quadratic function?
2. What is the vertex form of a quadratic function, and how do you convert from standard form to vertex form?
3. How can you use the discriminant of a quadratic equation to determine the number of real roots?
4. What is the role of the second derivative test in optimization problems?
5. How can we find the critical points for higher-order polynomials?

Tip: For any quadratic equation of the form $y = ax^2 + bx + c$, the vertex (maximum or minimum) occurs at $x = -\frac{b}{2a}$.