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

Explanation:

Step1: Recall vertex - x formula

For a quadratic function $y = ax^{2}+bx + c$, the x - coordinate of the vertex is $x=-\frac{b}{2a}$. Here, $a=\frac{1}{2}$ and $b = 3$. $x=-\frac{3}{2\times\frac{1}{2}}$

Step2: Calculate x - coordinate

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

Step3: Find y - coordinate

Substitute $x = - 3$ into the function $f(x)=\frac{1}{2}x^{2}+3x+\frac{3}{2}$. $f(-3)=\frac{1}{2}\times(-3)^{2}+3\times(-3)+\frac{3}{2}$ $=\frac{1}{2}\times9-9+\frac{3}{2}$ $=\frac{9}{2}-9+\frac{3}{2}$ $=\frac{9 + 3}{2}-9$ $=6 - 9=-3$

Answer:

$(-3,-3)$