Does anyone know how to do graphing a step function? ? if so, write a message to me, because i'm really not getting it

Basically you can graph a function, for example a parabola by following the step pattern 1,35...

If you take the "standard" parabola, y = x², which has it's vertex at the origin (0, 0), then: 

➊ one way you can use a "step pattern" is as follows: 

Starting from the vertex as "the first point" ... 

OVER 1 (right or left) from the vertex point, UP 1² = 1 from the vertex point 
OVER 2 (right or left) from the vertex point, UP 2² = 4 from the vertex point 
OVER 3 (right or left) from the vertex point, UP 3² = 9 from the vertex point 
OVER 4 (right or left) from the vertex point, UP 4² = 16 from the vertex point 
and so on ... 

where the "UP" numbers are the sequence of "PERFECT SQUARE" numbers ... 

but always counting from the VERTEX EACH time. 

➋ another way you can use a "step pattern" is just as you were doing: 

Starting with the vertex as "the first point" ... 

over 1 (right or left) from the LAST point, up 1 from the LAST point 
over 1 (right or left) from the LAST point, up 3 from the LAST point 
over 1 (right or left) from the LAST point, up 5 from the LAST point 
over 1 (right or left) from the LAST point, up 7 from the LAST point 
and so on ... 

where the "UP" numbers are the sequence of "ODD" numbers ... 

but always counting from the LAST point EACH time. 

The reason why both Step Patterns Systems work is that set of PERFECT SQUARE numbers has the feature that the difference between consecutive members is the set of ODD numbers. 

For your set of points, the vertex (and all the others) are simply "down 3" from the "standard places": 

Standard {..., (-3, 9), (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (3, 9), ...} 
shift ↓ 3 : {..., (-3, 6), (-2, 1), (-1, -2), (0, -3), (1,-2), (2, 1), (3, 6), ...} 

That is so true, i can’t believe how true this is

21/2 or 5/2