determining the vertex of a function\nwhat is the vertex of the function $f(x)=\frac{1}{2}x^{2}+3x+\frac{3}{2…

Explanation:

Step1: Identify coefficients

For a quadratic function $f(x)=ax^{2}+bx + c$, here $a=\frac{1}{2}$, $b = 3$, $c=\frac{3}{2}$.

Step2: Calculate x - coordinate of vertex

The formula for the x - coordinate of the vertex of a quadratic function is $x=-\frac{b}{2a}$. Substitute $a=\frac{1}{2}$ and $b = 3$ into it: $x=-\frac{3}{2\times\frac{1}{2}}=- 3$.

Step3: Calculate y - coordinate of vertex

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

Answer:

$(-3,-3)$