To determine the vertex of the quadratic function [tex]\( f(x) = \frac{1}{2} x^2 + 3 x + \frac{3}{2} \)[/tex], we follow these steps:
1. Identify the coefficients: In the quadratic function of the form [tex]\( ax^2 + bx + c \)[/tex],
- [tex]\( a \)[/tex] is the coefficient of [tex]\( x^2 \)[/tex],
- [tex]\( b \)[/tex] is the coefficient of [tex]\( x \)[/tex],
- [tex]\( c \)[/tex] is the constant term.
For our quadratic function, we have:
[tex]\[
a = \frac{1}{2}, \quad b = 3, \quad c = \frac{3}{2}
\][/tex]
2. Calculate the x-coordinate of the vertex: The x-coordinate of the vertex for the quadratic function [tex]\( ax^2 + bx + c \)[/tex] is given by the formula:
[tex]\[
x = -\frac{b}{2a}
\][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[
x = -\frac{3}{2 \cdot \frac{1}{2}} = -\frac{3}{1} = -3
\][/tex]
3. Calculate the y-coordinate of the vertex: Substitute the x-coordinate [tex]\( x = -3 \)[/tex] back into the original quadratic function to find the y-coordinate:
[tex]\[
f(-3) = \frac{1}{2}(-3)^2 + 3(-3) + \frac{3}{2}
\][/tex]
Simplify each term:
[tex]\[
\frac{1}{2}(-3)^2 = \frac{1}{2}(9) = 4.5
\][/tex]
[tex]\[
3(-3) = -9
\][/tex]
[tex]\[
\frac{3}{2} = 1.5
\][/tex]
Adding these values together:
[tex]\[
f(-3) = 4.5 - 9 + 1.5 = -3
\][/tex]
Therefore, the vertex of the function [tex]\( f(x) = \frac{1}{2}x^2 + 3x + \frac{3}{2} \)[/tex] is [tex]\((-3, -3)\)[/tex].
