It is 6 units because you find the absolute value of the two different coordinates, in this case it is 4 and -2. Since they belong in different quadrants (one x or y value is positive and the other is negative) you add them. If they are both in the same quadrant, you subtract them.
Given the following function:
![\text{ f(x) = }\sqrt[]{-x\text{ -2}}\text{ + 2}](https://tex.z-dn.net/?f=%5Ctext%7B%20f%28x%29%20%3D%20%7D%5Csqrt%5B%5D%7B-x%5Ctext%7B%20-2%7D%7D%5Ctext%7B%20%2B%202%7D)
y = f(x)
Therefore, let's complete the data table by substituting each x-values to be able to get the respective y-values,
We get,
At x = -2,
![\text{ y = f(x) = }\sqrt[]{-x-2}\text{ + 2}](https://tex.z-dn.net/?f=%5Ctext%7B%20y%20%3D%20f%28x%29%20%3D%20%7D%5Csqrt%5B%5D%7B-x-2%7D%5Ctext%7B%20%2B%202%7D)
![\text{f(-2) = }\sqrt[]{-(-2)-2}\text{ + 2}](https://tex.z-dn.net/?f=%5Ctext%7Bf%28-2%29%20%3D%20%7D%5Csqrt%5B%5D%7B-%28-2%29-2%7D%5Ctext%7B%20%2B%202%7D)
![\text{ = }\sqrt[]{2\text{ - 2}}\text{ + 2}](https://tex.z-dn.net/?f=%5Ctext%7B%20%3D%20%7D%5Csqrt%5B%5D%7B2%5Ctext%7B%20-%202%7D%7D%5Ctext%7B%20%2B%202%7D)
![\text{ = }\sqrt[]{0}\text{ + 2}](https://tex.z-dn.net/?f=%5Ctext%7B%20%3D%20%7D%5Csqrt%5B%5D%7B0%7D%5Ctext%7B%20%2B%202%7D)
Therefore, y = 2 at x = -2.
At x = -3,
![\text{f(-3) = }\sqrt[]{-(-3)-2}\text{ + 2}](https://tex.z-dn.net/?f=%5Ctext%7Bf%28-3%29%20%3D%20%7D%5Csqrt%5B%5D%7B-%28-3%29-2%7D%5Ctext%7B%20%2B%202%7D)
![\text{= }\sqrt[]{3-2}\text{ + 2}](https://tex.z-dn.net/?f=%5Ctext%7B%3D%20%7D%5Csqrt%5B%5D%7B3-2%7D%5Ctext%7B%20%2B%202%7D)
![\text{= }\sqrt[]{1}\text{ + 2}](https://tex.z-dn.net/?f=%5Ctext%7B%3D%20%7D%5Csqrt%5B%5D%7B1%7D%5Ctext%7B%20%2B%202%7D)
Therefore, y = 3 at x = -3.
At x = -6,
Hey! So, here's a tip. When writing exponents, an easier way is to write a^b, rather than a to the b power. Besides that, here is your answer!
So-------
9^3=729
3^2=9
6^3=216
15^2=225
Now that we have that figured out, we can add them together, wish is simple. 729 + 9 + 216 + 225= 1,179.
Therefore, your final answer will be 1,174.
If you have any questions on this, I'm happy to help you. :)
