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

50x5=$250

step-by-step explanation:

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!