Question

what is the vertex of the function $f(x)=\frac{1}{2}x^{2}+3x+\frac{3}{2}$?

Explanation:

Step1: Recall vertex formula for parabola

For a quadratic function \( f(x) = ax^2 + bx + c \), the x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \). Here, \( a = \frac{1}{2} \) and \( b = 3 \).
\[
x = -\frac{3}{2\times\frac{1}{2}} = -\frac{3}{1} = -3
\]

Step2: Find the y-coordinate of the vertex

Substitute \( x = -3 \) into the function \( f(x) = \frac{1}{2}x^2 + 3x + \frac{3}{2} \).
\[
f(-3) = \frac{1}{2}(-3)^2 + 3(-3) + \frac{3}{2} = \frac{1}{2}(9) - 9 + \frac{3}{2} = \frac{9}{2} - 9 + \frac{3}{2} = \frac{9 + 3}{2} - 9 = \frac{12}{2} - 9 = 6 - 9 = -3
\]

Answer:

\((-3, -3)\)