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

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Eliza 2 weeks 2021-09-27T14:28:02+00:00 1 Answer 0

Answers ( )

    0
    2021-09-27T14:29:05+00:00

    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..

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