y = x² + 2x + 3

Clear of fractions:

3y = x² + 6x + 9

Factor 1 out of first two terms on the right:

3y = 1(x² + 6x) + 9

Change the parentheses to brackets so it can contain parentheses:

3y = 1[x² + 6x] + 9

To complete the square inside the bracket:

1.  Take one-half of the coefficient of x.   ·(6) = 3
2.  Square the result.                       (3)² = 9
3.  Add it then subtract it in the bracket:   Add + 9 - 9

3y = 1[x² + 6x + 9 - 9] + 9

Factor the first three terms inside the bracket as a perfect square:

3y = 1[(x+3)² - 9] + 9

Remove the bracket by distributing the 1 leaving the parentheses intact:

3y = 1(x+3)² - 9 + 9

The -9 and the +9 gives 0

3y = 1(x+3)² + 0

Solve for y by dividing through by 3

 y = (x+3)² + 0

Compare that to the standard vertex form:

 y = a(x-h)² + k

We see that the vertex (h,k) is (-3,0) and from the original
equation y = x² + 2x + 3 tells us that the y-intercept
is (0,3) 



The axis of symmetry is the vertical line through the vertex, and
is therefore the vertical line whose equation is x=-3:



The point that matches the y-intercept on the other side of the
axis of symmetry is (-6,3), so we plot that point:

and we sketch in the parabola:

Edwin