`f(x) = 1/2(x-3)^2-1 `

when x = 0 then f(x) = 3.5

When f(x) = 0;

`1/2(x-3)^2-1= 0`

`(x-3)^2 = 2`

` x-3 = +-sqrt2`

` x = 3+-sqrt2`

so when f(x) = 0 then `x = 3+sqrt2` and `x = 3-sqrt2`

`f(x) = 1/2(x-3)^2-1 `

`f'(x) = (x-3)`

`f''(x) = 1`

When `f'(x) = 0` ;

`x-3 = 0`

`x = 3`

At x = 3 then f(x) = -1

Since `f''(x) > 0` we have a minimum for f(x).

- when x = 0 then f(x) = -3.5
- when f(x) = 0 then `x = 3+sqrt2` and `x = 3-sqrt2`
- f(x) has a minimum at x = 3

Using these data you can plot the graph.

The look of the graph is as follows.

Graph `y=1/2(x-3)^2-1` :

The base function is `y=x^2` .

(1) Multiplying by 1/2 performs a vertical compression of factor 1/2.

(2) Subtracting 3 inside the function shifts the graph 3 units to the right.

(3) Subtracting 1 from the function shifts the graph 1 unit down.

The graph shows the base function in black, the vertical compression in blue,the horizontal translation is green, and the vertical translation in red which is the graph you seek.

Alternatively, the form of the quadratic is the vertex form `y=a(x-h)^2+k` where the vertex is (h,k). So the vertex is at (3,-1); the parabola opens up a little wider than normal.