Answer:
Neither linear nor exponential
Step-by-step explanation:
To check for a linear relationship. Find slope.
slope= (-1 - (-2)) / ( 5 - 2) = 1/3
check other points
slope = (1 - (-1) )/ (8 - 5) = 2/3
check more
slope = (4 - 1) / (11 - 8) = 3/ 3 = 1
Nope.
try assuming an exponential:
y = c * (a^x)
-2 = c* (a^2); -2/c = a^2
-1 = c *(a ^5); -1/c = a^5
1 = c * (a^8), 1/c = a^8
(-2/c)^4 = a^8 = 1/c
16/(c^4) = 1/c
c^3 = 16, then a = root (-2/ cube-root(16) )
The change from negative to postive would not work for y = c(a^x)
so...
assume y = a^x + k
-2 = a^2 + k
-1 = a^5 + k
... I would say neither..
