Question

Y=-2(x-1)^2+8 how do I factor this parabola? I need to find the zeros, vertex, and axis of symmetry

Answer 1

Y = -2(x - 1)² + 8
y = -2((x - 1)(x - 1)) + 8
y = -2(x² - x - x + 1) + 8
y = -2(x² - 2x + 1) + 8
y = -2(x²) + 2(2x) - 2(1) + 8
y = -2x² + 4x - 2 + 8
y = 2x² + 4x + 6
2x² + 4x + 6 = 0
x = -4 +/- √(4² - 4(2)(6))
                   2(2)
x = -4 +/- √(16 - 48)
                  4
x = -4 +/- √(-32)
               4
x = -4 +/- 5.6568i
                4
x = -4 +/- 1.4142i
x = -4 + 1.4142i                x = -4 - 1.4142
y = -2x² + 4x + 6
y = -2(-4 + 1.4142i)² + 4(-4 + 1.4142i) + 6
y = -2((-4 + 1.4142i)(-4 + 4.4142i) - 16 + 5.6568i + 6
y = -2(16 - 5.6568i - 5.6568i + 1.99996164i²) - 16 + 5.6568i + 6
y = -2(16 - 11.3136i + 1.9996164) - 16 + 5.6568i + 6
y = -32 + 22.6272i - 3.99992328 - 16 + 5.6568i + 6
y = -32 - 3.99992328 - 16 + 6 + 22.6272i + 5.6568i
y = -35.99992328 - 16 + 6 + 28.284i
y = -51.99992328 + 6 + 28.284i
y = -45.99992328 + 28.284i
(x, y) = (-4 + 1.4142i, -45.99992328 + 28.284i)
y = -2x² + 4x + 6
y = -2(-4 - 1.4142i)² + 4(-4 - 1.4142i) + 6
y = -2((-4 - 1.4142i)(-4 - 1.4142i)) - 16 - 5.6568i + 6
y = -2(16 + 5.6568i + 5.6568i + 1.99996164i²) - 16 - 5.6568i + 6
y = -2(16 + 11.3136i + 1.99996164) - 16 - 5.6568i + 6
y = -32 - 22.6272i - 3.99992328 - 16 - 5.6568i + 6
y = -32 - 3.99992328 - 16 + 6 - 22.6272i - 5.6568i
y = -35.99992328 - 16 + 6 - 28.284i
y = -51.99992328 + 6 - 28.284i
y = -45.99992328 - 28.284i
(x, y) = (-4 - 1.4142i, -45.99992328 - 28.284i)
zeros: -4 + 1.4142i or -4 - 1.4142i
vertex: (-4 + 1.4142i, -45.99992328 + 28.284i) and (-4 - 1.4142i, -45.99992328 - 28.284i)
axis of symmetry: 0 + 2.8284i