Answer and Explanation :

Given : Function f(x)=frac{x^2}{2}+2x+1

To find : The vertex, axis of symmetry, maximum or minimum value, and the graph of the function.

Solution :

The quadratic function is in the form, y=ax^2+bx+c

On comparing, a=frac{1}{2} , b=2 and c=1

The vertex of the graph is denote by (h,k) and the formula to find the vertex is

For h, The x-coordinate of the vertex is given by,

h=-frac{b}{2a}

h=-frac{2}{2(frac{1}{2})}

h=-frac{2}{1}

h=-2

For k, The y-coordinate of the vertex is given by,

k=f(h)

k=frac{h^2}{2}+2h+1

k=frac{(-2)^2}{2}+2(-2)+1

k=2-4+1

k=-1

The vertex of the function is (h,k)=(-2,-1)

The x-coordinate of the vertex i.e. x=-frac{b}{2a} is the axis of symmetry,

So, x=-frac{b}{2a}=-2 (solved above)

So, The axis of symmetry is x=-2.

The maximum or minimum point is determine by,

If a > 0 (positive), then the parabola opens upward and the graph has a minimum at its vertex.

a=frac{1}{2} >0 so,  the parabola opens upward and the graph has a minimum at its vertex.

The Minimum value is given at (-2,-1)

Now, We plot the graph of the function

At different points,

x          y

-4        1

-2        -1

0         1

Refer the attached figure below.