Here's one way to figure out, for example, what is 10 ÷ 2
__  ___  ____  
10    2  =  8  count = 1  
8    2  =  6  count = 2  
6    2  =  4  count = 3  
4    2  =  2  count = 4  
2    2  =  0  count = 5 
We had to subtract 2 a total of 5 times before we couldn't subtract it any more (without getting a number < 0)
Or we can say that 2 "fits into" 10 a total of 5 times.
So 10 ÷ 2 = 5
Now let's try to use the same method to figure out what is 10 ÷ 0
__  ___  ____  
10    0  =  10  count = 1  
10    0  =  10  count = 2  
10    0  =  10  count = 3  
10    0  =  10  count = 4  
10    0  =  10  count = 5  
10    0  =  10  count = 6  
10    0  =  10  count = 7  
10    0  =  10  count = 8  
10    0  =  10  count = 9  
. . .  . . . 
We keep subtracting 0 over and over again, but we aren't getting anywhere! This is an endless loop.
There is not a number of times that we can subtract 0 from 10 to reach a number < 0.
In other words, there is not a number that is equal to 10 ÷ 0
Because the exterior angle of an octagon is 45°, we can extend  
sides BA and GH to point J to create 454590 triangle AJH  
In △AJH,  
the side across from the 90° angle = AH = 2 

the sides across from the 45° angles = AJ = JH = 2 / √2 = √2  
By the Pythagorean Theorem, 

AG^{2} = AJ^{2} + JG^{2} 

AG^{2} = AJ^{2} + (JH + HG)^{2} 

AG^{2} = (√2)^{2} + (√2 + 2)^{2} 

AG^{2} = 2 + 2 + 4√2 + 4 

AG^{2} = 8 + 4√2 

AG^{2} ≈ 13.657 

That is in square cm. 

For a function to be invertible, it must pass the "horizontal line test," which means you must not be able to draw a horizontal line that intersects the graph more than once. We can see the graph of y = f(x) by itself does not pass the horizontal line test and so is not invertible.
Let's imagine what g(x) would look like if a = 1
g(x) = f(x) + ax If a = 1 then
g(x) = f(x) + x
Let's see what g(1) would be.
g(1) = f(1) + 1 And from the picture we can see f(1) = 5
g(1) = 5 + 1
g(1) = 4 So the point (1, 4) is on the graph of g(x)
Adding the xvalue to the yvalue ended up shifting that point up.
Let's see what g(3) would be.
g(3) = f(3) + 3 And from the picture we can see f(3) = 2
g(3) = 2 + 3
g(3) = 5 So the point (3, 5) is on the graph of g(x)
Adding the xvalue to the yvalue ended up shifting that point up (even more than last time).
Let's see what g(1) would be.
g(1) = f(1)  1 And from the picture we can see f(1) = 3
g(1) = 3  1
g(1) = 2 So the point (1, 2) is on the graph of g(x)
Adding the xvalue to the yvalue ended up shifting the point down.
Let's see what g(2) would be.
g(2) = f(2)  2 And from the picture we can see f(2) = 5
g(2) = 5  2
g(2) = 3 So the point (2, 3) is on the graph of g(x)
Adding the xvalue to the yvalue ended up shifting that point down (even more than last time).
And if we increase the value of a , that will only exaggerate the changes.
By now we can see what g(x) should look like compared to f(x)
Here is a graph of f(x) and g(x) with a slider for different values of a:
https://www.desmos.com/calculator/eoqgwwn620
Moving the slider to the right until the graph "straightens out" enough to pass the horizontal line test, we can see that the smallest positive integer that makes g(x) invertible is 5. Moving the slider to the left (and zooming out as necessary), we can see that the largest negative integer that makes g(x) invertible is 4.
(4)^{2} + 5^{2} = 16 + 25 = 41