1. ## Reciprocal functions

Hi,

I hope someone can help.

Remember that vertex form is y = a(x-h)^2+k

I'm trying to understand why the function f(x) = 3x^2-4x-4 in vertex form is f(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

2. ## Re: Reciprocal functions

\begin{align*}f(x) & = 3x^2-4x-4 \\ & = 3\left(x^2-\dfrac{4}{3}x-\dfrac{4}{3}\right) \\ & = 3\left(x^2-\dfrac{4}{3}x + \dfrac{4}{9} - \dfrac{4}{9} - \dfrac{4}{3}\right) \\ & = 3\left(\left(x-\dfrac{2}{3}\right)^2 - \dfrac{16}{9}\right) \\ & = 3\left(x-\dfrac{2}{3}\right)^2 - \dfrac{48}{9}\end{align*}

3. ## Re: Reciprocal functions

$\displaystyle -\frac{48}9=-\frac{16}3=-5.333\ldots$