# How do I find vertex of a parabola with 2 x-intercepts and 1 y-intercept: x intercepts: (-2,0), (8,0) and y intercepts: (0, -16)?

The function is $y = \frac{2}{3} {x}^{2} - \frac{10}{3} x - 16$
Any quadratic function can be written as $y = a {x}^{2} + b x + c$.
In this case youy can easily find, that $c = - 16$ because any quadratic function intercepts Y axis in point $\left(0 , c\right)$.
$0 = 4 a - 4 b - 16$ and $0 = 64 a + 8 b - 16$
From these equations you find that $a = \frac{2}{3}$ and $b = - \frac{10}{3} = - 3 \frac{1}{3}$.