# Question #f332a

##### 1 Answer

Jul 14, 2017

See a solution process below:

#### Explanation:

For each of these values of

#color(red)(x) = color(red)(-2)#

#color(red)(x) = color(red)(-1)#

#color(red)(x) = color(red)(0)#

#color(red)(x) = color(red)(1)#

Now that you know the pattern, see if you can do:

#color(red)(x) = color(red)(2)#