Question

What is the vertex of `y = (1/8)(x – 5)^2 - 3`?

Answer 1

The vertex is `v(1/20, 201/3200)`

Explanation:

`f(x) = y=1/8(x–5)^2−3`
`f(x) = 1/8 (x^2 - 10x + 1)`
Now for any `f(x) = y = ax^2 + bx + c " "` form of the equation
The vertex, `v(h, k)`
`h = -b/(2a); and k = f(h)`
`h = 1/20; k = f(1/20) = 201/3200 ~~ .063`
The vertex is `v(1/20, 201/3200)`