This is basic parabola of the form:
[tex]y=ax^2-b[/tex]
So, this one is shifted 1 units down.
The vertex is at (0, -1).
To take 2 points to the left of vertex, we find coordinates for x = -2 and x = -4.
To take 2 points to the right of vertex, we find coordinates for x = 2 and x = 4.
Let's find it:
[tex]\begin{gathered} \text{When x = -2,} \\ y=\frac{1}{4}x^2-1 \\ y=\frac{1}{4}(-2)^2^{}-1 \\ y=0 \\ When\text{ x = -4,} \\ y=\frac{1}{4}x^2-1 \\ y=\frac{1}{4}(-4)^2-1 \\ y=3 \\ \text{When x = 2,} \\ y=\frac{1}{4}x^2-1 \\ y=\frac{1}{4}(2)^2-1 \\ y=0 \\ \text{When x= 4,} \\ y=\frac{1}{4}x^2-1 \\ y=\frac{1}{4}(4)^2-1 \\ y=3 \end{gathered}[/tex]
So, the 4 coordinates are:
(-2,0), (-4,3), (2,0), (4,3)
The graph: