Hi,

I hope someone can help.

Remember that vertex form isy = a(x-h)^2+kI'm trying to understand why the function

f(x) = 3x^2-4x-4in vertex form isf(x) = (x-2/3)^2 - 16/9. Really what I'm wondering is why would k = -16/9 given that the vertex has a y-value of -5.3? I always was taught that the k value is the y-value for the vertex, so I'm wondering why this function has a k value which doesn't correspond to the vertex's y-value.

Sincerely,

Olivia