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Is the relationship linear, exponential, or neither?

X-Values: 2, 5, 8, 11
Y-Values: -2, -1, 1 , 4

1 Answer

5 votes

Answer:

Neither linear nor exponential

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..

User Benjamin Poignant
by
8.5k points

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