We have been given a quadratic function 
and we need to restrict the domain such that it becomes a one to one function.
We know that vertex of this quadratic function occurs at (5,2).
Further, we know that range of this function is 
.
If we restrict the domain of this function to either ![(-\infty,5]](https://tex.z-dn.net/?f=%28-%5Cinfty%2C5%5D)
or 
, it will become one to one function.
Let us know find its inverse.
Upon interchanging x and y, we get:
Let us now solve this function for y.
Hence, the inverse function would be 
if we restrict the domain of original function to and the inverse function would be 
if we restrict the domain to .
It is neither.
To be even, f(x) must equal f(-x).
If you substitute -x for x, you'd get
y = (-x)^2 - 2(-x) -8
y = x^2 +2x -8
This is not the same as the original, so this is not even.
To be odd, f(x) must equal -f(-x).
If you take the -x substitution from the last step and then multiply it by -1, you'd have:
y = -1 (x^2 +2x -8)
y = -x^2 -2x +8
This is not the same as the original either.
The function is neither even nor odd.
Addition will be the last operation
Answer:
86
Step-by-step explanation:
Si 1,2,4,5,8,10,16,20,40,80
