What is the vertex of the function f(x) = 1/2 x^2 – 3x + 3/2?

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Determining the Vertex of the Function

To find the vertex of the quadratic function f(x) =
\frac{1}{2} x^{2} – 3x + \frac{3}{2}
, we can utilize the vertex formula for a quadratic equation in the standard form f(x) = ax^2 + bx + c. In our case, the coefficients are:

  • a = 1/2
  • b = -3
  • c = 3/2

The x-coordinate of the vertex can be calculated using the formula:

x = -\frac{b}{2a}

Plugging in our values:

  • b = -3
  • a = 1/2

We substitute these into the formula:

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

Now that we have the x-coordinate of the vertex, we need to find the y-coordinate by substituting x = 3 back into the original function:

f(3) = \frac{1}{2}(3)^{2} – 3(3) + \frac{3}{2}

Calculating this:

  • \(f(3) = \frac{1}{2}(9) – 9 + \frac{3}{2} = \frac{9}{2} – 9 + \frac{3}{2}\)

Combine these terms:

  • \(f(3) = \frac{9}{2} – \frac{18}{2} + \frac{3}{2} = \frac{-6}{2} = -3\)

Thus, the vertex of the function f(x) = \frac{1}{2} x^{2} – 3x + \frac{3}{2} is:

(3, -3)

In conclusion, the vertex of the quadratic function f(x) is at the point (3, -3), representing the minimum point of the parabola.

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