How to solve this, i need to find the zeros, vertex, y-intercept, max or min, and the equation in vertex form

The standard form of a quadratic equation is: ax^2 + bx + c = 0
a = 1 ; b = -2 ; c = -8
To find the zeros, you put in x=0 which gives (0,-8) and use the quadratic formula to find the other zeros; f(x) = 0

Quadratic formula:
x = (-b+/-√(b^2-4ac))/2a

The x-value for the vertex is -b/2a
x = 1
put this in for x and solve for the f(x)
1 - 2  - 8 = -9
vertex: (1, -9)

The vertex form of a quadratic equation: f(x) = a(x-h)^2 + k
where (h, k) is the vertex.
If you don't know the vertex, you would need to factor it into this form. Look at the center term bx, what you could be added to the equation to make a perfect square.

f(x) = (x^2 - 2x + 1) - 9

By splitting the -8 into 1 - 9 we've created a perfect square (x-1)^2

f(x) = (x-1)^2 - 9

Gotta be d the music of previous eras