How do you write #y = 6(x - 2)^2 - 8# in standard form?
1 Answer
May 19, 2015
Just develop this vertex form:
y = 6(x^2 - 4x + 4) - 8 = 6x^2 - 24x + 24 - 8 = 6x^2 - 24x + 16
Just develop this vertex form:
y = 6(x^2 - 4x + 4) - 8 = 6x^2 - 24x + 24 - 8 = 6x^2 - 24x + 16