f(x) = 2x² + 12x + 15

We want to get it to the standard form

f(x) = a(x - h)² + k

f(x) = 2x² + 12x + 15

Factor 2 out of the first two terms:

f(x) = 2[x² + 6] + 15

Skip a space on the right of the bracketed expression

f(x) = 2[x² + 6x        ] + 15

To the side, complete the square:

        6(1/2) = 3  <--half of the coefficient of x
        3² = 9      <--square that result

Add and subtract the value inside the bracket:

f(x) = 2[x² + 6x + 9 - 9] + 15

Factor the first three terms inside the bracket:

f(x) = 2[(x+3)(x+3) - 9] + 15

Write the factorization as the square of (x+3)

f(x) = 2[(x+3)² - 9] + 15

Distribute the 2 inside the bracket leaving the (x+3)² intact:

f(x) = 2(x+3)² - 18 + 15

Combine like terms:

f(x) = 2(x+3)² - 3

Compare to

f(x) = a(x-h)² + k

The vertex is (h,k) = (-3,-3)

Get a couple of points on each side of
the vertex (-3,-3)

  x | y = f(x) 
  --|---------
 -5 |  5
 -4 | -1
 -3 | -3   <-- vertex
 -2 | -1
 -1 |  5

          

Edwin