Answer:
Step-by-step explanation:
Hello, please consider the following.
The "vertex form" is as below.
[tex]y=a(x-h)^2+k\\\\\text{Where (h, k) is the vertex of the parabola.}\\[/tex]
Let's do it!
[tex]f(x)=\dfrac{1}{2}x^2+3x-2\\\\f(x)=\dfrac{1}{2}\left(x^2+3*2*x\right) -2\\\\f(x)=\dfrac{1}{2}\left( (x+3)^2-3^2\right)-2\\\\f(x)=\dfrac{1}{2}(x+3)^2-\dfrac{9}{2}-\dfrac{4}{2}\\\\f(x)=\dfrac{1}{2}(x+3)^2-\dfrac{9+4}{2}\\\\\large \boxed{\sf \bf \ \ f(x)=\dfrac{1}{2}(x+3)^2-\dfrac{13}{2} \ \ }[/tex]
Thank you.
