Identify the vertex and the axis of symmetry for y=-(x+3)^2+1

Vertex
y = -(x + 3)² + 1
y = -(x + 3)(x + 3) + 1
y = -(x² + 3x + 3x + 9) + 1
y = -(x² + 6x + 9) + 1
y = -x² - 6x - 9 + 1
y = -x² - 6x - 8
-x² - 6x - 8 = 0
x = -(-6) +/- √((-6)² - 4(-1)(-8))
                      2(-1)
x = 6 +/- √(36 - 32)
                -2
x = 6 +/- √(4)
           -2
x = 6 +/- 2
          -2
x = 6 + 2       x = 6 - 2
        -2                  -2
x = 8             x = 4
     -2                 -2
x = -4            x = -2
y = -x² - 6x - 8
y = -(-4)² - 6(-4) - 8
y = -(16) + 24 - 8
y = -16 + 24 - 8
y = 8 - 8
y = 0
(x, y) = (-4, 0)
or
y = -x² - 6x - 8
y = -(-2)² - 6(-2) - 8
y = -(4) + 12 - 8
y = -4 + 12 - 8
y = 8 - 8
y = 0
(x, y) = (-2, 0)

Axis of Symmetry
The axis of symmetry is equal to -3.

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