Question

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)?

Answer 1

The function is `y=2/3x^2-10/3x-16`

Any quadratic function can be written as `y=ax^2+bx+c`.
In this case youy can easily find, that `c=-16` because any quadratic function intercepts Y axis in point `(0,c)`.

The remaining 2 points can be used to form aa set of equations
`0=4a-4b-16` and `0=64a+8b-16`
From these equations you find that `a=2/3` and `b=-10/3=-3 1/3`.