+130518 f(x) = x - √[x - 2]
So f(11) just means to put 11 into the function and evaluate the result......so we have
f(11) = 11 - √[11- 2] = 11 - √9 = 11 - 3 = 8
f(6) , f(2) and f(0) are done similarly
b. Here's the graph.......https://www.desmos.com/calculator/lok5hw3jkm
c. Note that x cannot be less than 2, because √[x - 2] would result in taking the square root of a negative number and this isn't a "real" result.....so.....the domain is x ≥ 2
d. x - √[x - 2 ] = 0 add √[x - 2 ] to both sides
x = √[x - 2 ] square both sides
x^2 = x - 2 rearrange
x^2 - x + 2 = 0 Note that, in the quadratic formula.....b^2 - 4ac = (-1)^2 - 4(1)(2) = 1 - 8 = -7........and this would mean that we would have two non-real solutions......thus...... x - √[x - 2 ] = 0 doesn't exist in real terms.......also, look at the graph.......the function never touches or crosses the x axis.....thus, it is never 0
+130518 ( x + 6)^2 + ( y - 9)^2 = 52
x^2 + y^2 = 13
The Algebra for this one could get a little messy......I might use a graphical approach....
https://www.desmos.com/calculator/cnkpzxaqrm
These are two circles that are tangent to each other....
The only "solution" point occurs at (-2, 3)....... And this is the point of tangency.......
. But then that would mean I'd have to start again. 
