To solve the given quadratic function problem, we need to analyze the function:

$y = \frac{1}{2}x^2 + 3x - \frac{5}{2}$

Step 1: Determine the Vertex

The quadratic function in the form $ax^2 + bx + c$ can have its vertex determined using the formula $x = -\frac{b}{2a}$.

Here, $a = \frac{1}{2}$ and $b = 3$.

$x = -\frac{3}{2 \cdot \frac{1}{2}} = -\frac{3}{1} = -3$

Step 2: Calculate the Value of the Function at the Vertex

Substitute $x = -3$ back into the function to find the value of $y$:

$y = \frac{1}{2}(-3)^2 + 3(-3) - \frac{5}{2}$ $y = \frac{1}{2}(9) - 9 - \frac{5}{2}$ $y = \frac{9}{2} - 9 - \frac{5}{2}$ $y = \frac{9}{2} - \frac{18}{2} - \frac{5}{2}$ $y = \frac{9 - 18 - 5}{2}$ $y = \frac{-14}{2}$ $y = -7$

Step 3: Determine if this is a Maximum or Minimum

Since the coefficient of $x^2$ (which is $\frac{1}{2}$) is positive, the parabola opens upwards. This means that the vertex represents the minimum value of the function.

Thus, the function has a minimum value of $-7$ when $x = -3$.

Final Answer

The function $y = \frac{1}{2}x^2 + 3x - \frac{5}{2}$ has a minimum value $-7$ when $x = -3$.

Would you like any further details or have any questions?

Suggested Questions

1. Can you explain how to find the vertex of a quadratic function?
2. How do you determine if a quadratic function has a maximum or minimum value?
3. What are the steps to convert a quadratic function into vertex form?
4. How do you solve a quadratic equation using the quadratic formula?
5. Can you provide an example of a quadratic function with a maximum value?

Tip

When dealing with quadratic functions, always check the sign of the coefficient of $x^2$ to determine if the parabola opens upwards (minimum value) or downwards (maximum value).