Question

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

Answer 1

`(2, 5)`

Explanation:

The equation:

`y = 1/8(x-2)^2+5`

is in vertex form:

`y = a(x-h)^2+k`

with `a=1/8` and `(h, k) = (2, 5)`

So we simply read the coordinates of the vertex `(h, k) = (2, 5)` from the coefficients of the equation.

Notice that for any Real value of `x`, the resulting value of `(x-2)^2` is non-negative, and it is only zero when `x=2`. So this is where the vertex of the parabola is.

When `x=2`, the resulting value of `y` is `0^2+5 = 5`.

graph{(1/8(x-2)^2+5-y)((x-2)^2+(y-5)^2-0.03)=0 [-14.05, 17.55, -1.89, 13.91]}