Working With A Square

[nycmathguy]

See attachment.

Let A = area of square.

A = (side)^2

Side = 2x + 8

A(x) = (2x + 8)^2

Domain = (-infinity, infinity).

Yes?

 

[MathLover1]

A=8*8-2*x*x
A=64-2x^2

domain: -infinity, infinity
range:-infinity,64

 

[nycmathguy]

A=8*8-2*x*x
A=64-2x^2

domain: -infinity, infinity
range:-infinity,64

1. What is wrong with my effort?

2. How did you determine the domain and range for this problem?

3. Am I bombarding you with too many questions?

I think less questions lead to a more detailed reply, which is what I need to really understand precalculus. Otherwise, I am just wasting my time and yours.

 

[MathLover1]

you say
A(x) = (2x + 8)^2 which is wrong
the area of the square is 8^2=64
from that you need to deduct the area x*x=x^2 from both sides, means minus 2x^2
pay attention to the graph above: you see 4 little isosceles triangles, when you cut them off you can make 2 little squares whose side length is x
these two little squares (2x^2) you need to deduct from total area (64)

A(x) = 64-2x^2

 

[nycmathguy]

you say
A(x) = (2x + 8)^2 which is wrong
the area of the square is 8^2=64
from that you need to deduct the area x*x=x^2 from both sides, means minus 2x^2

A(x) = 64-2x^2

Ok.