Is the relationship linear, exponential, or neither? X-Values: 2, 5, 8, 11 Y-Values: -2, -1, 1 , 4 Question Is the relationship linear, exponential, or neither? X-Values: 2, 5, 8, 11 Y-Values: -2, -1, 1 , 4 in progress 0 Math Eliza 2 weeks 2021-09-27T14:28:02+00:00 2021-09-27T14:28:02+00:00 1 Answer 0

## Answers ( )

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..