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

##### 1 Answer

May 2, 2016

#### Explanation:

The equation:

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

is in vertex form:

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

with

So we simply read the coordinates of the vertex

Notice that for any Real value of

When

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]}