The function y = x2 - 4x + 5 approximates the height, y, of a bird, and its
horizontal distance, x, as it flies from one fence post to another. all distances
are in feet. complete the square to find and interpret the extreme value
(vertex).

a vertex of (2, 1) means at time 2 it is at its lowest height of 1

Step-by-step explanation:

I'm assuming it's y = x^2 - 4x + 5

Completeing the square follows this pattern.

ax^2 + bx + c = 0

a(x^2 + (b/a)x + c/a) = 0

a(x^2 + (b/a)x + (b/(2a))^2 - (b/(2a)^2 + c/a) = 0

a((x + b/2)^2 - (b/(2a)^2 + c/a) = 0

a(x + b/2)^2 + a(b/(2a)^2 + c/a) = 0

So doing that on y = x^2 - 4x + 5 looks like this

x^2 - 4x + 5 = 0

x^2 - 4x + (-4/2)^2 - (-4/2)^2 + 5 = 0

x^2 - 4x + 4 - 4 + 5

(x -2)^2 + 1 = 0

This is vertex form.  let me know if you don't see how I did any of this.  I will say b is -4 so b/2 is -4/2 = -2.

Now, vertex form uses graph transformations of the graph x^2 to tell where the vertex is.

x^2 of course has the vertex at (0,0)

x^2 + a moves the graph up a units so the vertex is at (0, a)

x^2 - b moves the graph down a so the vertex is at (0, -b)

(x + c) moves the graph left c units , so the vertex would be at (-c, 0)

(x - d) moves the graph right d units , so the vertex would be at (d, 0)

You can combine them too so (x + a)^2 + b will move the vertex left a units (or right if it's negative) and up b units (or down  if this is negative)

NOW!  let's look at the vertex form of what we have.

(x -2)^2 + 1

Right 2 and up 1.  Also, since there is no negative o the outside of the parenthesis we have a minimum.  if it was something like -(x+a)^2+b with the minus on the outside we would have a maximum at (-a, b)

Now, for your problem involving the height of a bird, you have a minimum, which means that is where the bird is at its lowest.  The x value of the vertex is the time also.  So a vertex of (2, 1) means at time 2 it is at its lowest height of 1