The quadratic ax^2+bx+c can be expressed in the form 2(x-4)^2+8. When the quadratic 3ax^2+2bx+c is expressed in the form n(x-h)^2+k, what is h?
+2666 Convert vertex form to standard form: \(2(x-4)^2 + 8 = 2(x^2 - 8x + 16) + 8 = 2x^2- 16x + 40\)
This means that \(a = 2\), \(b = -16\), and \(c = 40\). So, the quadratic we need to convert to vertex form is \(6x^2 - 32x + 40\).
Note that in vertex form (which we need to convert it to), the vertex occurs at \((h,k)\).
This means we need to find the x-coordinate of the vertex, which occurs at \({-b \over 2a}={32 \over 12} = \color{brown}\boxed{8 \over 3}\).
