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In mathematics, especially in calculus, determining the relative extreme points of a function is crucial. For quadratic functions, these extreme points are always at the vertex of the parabola. To find the relative extreme point, you need to understand how the function behaves.

The relative extreme point can be either a minimum or a maximum. For the quadratic function given by \( f(x) = \frac{1}{2} x^2 + \frac{1}{2} \), the coefficient \( a = \frac{1}{2} \) is positive. This means the function has a relative minimum point at the vertex. The parabola opens upwards, so the lowest point on the graph represents this relative minimum. The minimum point provides significant information about the function's behavior and helps in sketching its graph accurately.

The vertex of a parabola for a quadratic function \( f(x) = a x^2 + b x + c \) is an important feature. It determines where the parabola changes direction.

For any quadratic function, the vertex can be found using the formula \( x = - \frac{b}{2a} \). In this exercise, the quadratic function \( f(x) = \frac{1}{2} x^2 + \frac{1}{2} \) has \( a = \frac{1}{2} \) and \( b = 0 \). By substituting these values into the vertex formula, we get \( x = - \frac{0}{2 \cdot \frac{1}{2}} = 0 \). This gives us the x-coordinate of the vertex.

To find the y-coordinate, substitute this x-value back into the function: \( f(0) = \frac{1}{2} (0)^2 + \frac{1}{2} = \frac{1}{2} \). Therefore, the vertex is at the point \( (0, \frac{1}{2}) \). This point is crucial as it represents the minimum value of the function, given that \( a > 0 \).

The concavity of a function indicates the direction in which a parabola opens. It is determined by the sign of the coefficient \( a \) in the quadratic function \( f(x) = a x^2 + b x + c \).

- If \( a > 0 \), the parabola opens upwards, and the function has a relative minimum at the vertex.
- If \( a < 0 \), the parabola opens downwards, and the function has a relative maximum at the vertex.

In this exercise, the coefficient \( a = \frac{1}{2} \) is positive, so the parabola opens upwards. This confirms that the connectivity at the vertex, \( (0, \frac{1}{2}) \), is concave upwards. Knowing the concavity is essential for sketching the graph correctly.

Sketching the graph of a quadratic function becomes easier once you've identified key characteristics like the vertex and the concavity. Here’s how we can sketch the graph for \( f(x) = \frac{1}{2} x^2 + \frac{1}{2} \).

- Start by plotting the vertex \( (0, \frac{1}{2}) \) on the coordinate plane.
- Because we know that the parabola opens upwards, we draw a symmetric curve passing through the vertex.
- Ensure that the graph is smooth and extends infinitely in both left and right directions.

To get more points for accuracy, you can substitute other x-values into the function and plot the corresponding points. However, for a simple sketch, just identifying the vertex and the direction of concavity provides a clear picture.