Find the vertex of the parabola y = 6x 2 + 3x + 1 by completing the square.

HELP!

——————————————————————————————————————

y = 6x² + 3x + 1
y = 6[x² + ½ x] + 1
y = 6[x² + ½ x + (¼)² - (¼)²] + 1
y = 6[(x + ¼)² - (¼)²] + 1
y = 6(x + ¼)² - 6(¼)² + 1
y = 6(x + ¼)² - 6(¹∕₁₆) + 1
y = 6(x + ¼)² - ⅜ + 1
y = 6(x + ¼)² + ⅝

        ANSWER
         V(-¼,⅝)

——————————————————————————————————————

Here are the steps for completing the square.
y = 6x² + 3x + 1

factor out the leading coefficient
y = 6(x² + ½x) + 1

coefficient of the x term: ½
divide coefficient in half: ¼
square it: (¼)²
use the result to complete the square, keeping the equation balanced:
y = 6(x² + ½x + (¼)²) - 6(¼)² + 1
y = 6(x + ¼)² - ⅜ + 1
y = 6(x + ¼)² + 5/8

vertex of parabola: (-¼, 5/8)