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nekit [7.7K]
4 years ago
8

Which table represents a quadratic relationship?

Mathematics
2 answers:
Marianna [84]4 years ago
7 0

Answer:  D. x -1 0 1 2 3 4

f(x) 90 56 26 0 -22 -40

Step-by-step explanation:

svp [43]4 years ago
4 0

In each case, the x-values are equally-spaced. Thus looking at second differences will tell you if the relation is quadratic. If the second differences are non-zero and constant, then the values have a quadratic relationship.

A. First differences are 2-4 = -2, 1-2 = -1, 0.5-1 = -0.5. Second differences are -1-(-2) = 1, -0.5-(-1) = 0.5. Since 1 ≠ 0.5, this relation is not quadratic. (It is exponential with a base of 1/2.)

B. First differences are 128-135 = -7, 105-128 = -23, 72-105 = -33. Second differences are -23-(-7) = -16, -33-(-23)=-10. Since -16 ≠ -10, this relation is not quadratic. (It is cubic, since 3rd differences are constant at +4.)

C. First differences are -23.2-(-23.4) = 0.2, -23.0-(-23.2) = 0.2, -22.8-(-23.0) = 0.2. Second differences are zero, so this is not a quadratic relation. (It is linear, with a slope of 0.2.)

D. First differences are 56-90 = -34, 26-56 = -30, 0-26 = -26. Second differences are -30-(-34) = 4, -26-(-30) = 4. These are constant (=4), so the relation is quadratic.

The appropriate choice is ...

... D. x -1 0 1 2 3 4

... f(x) 90 56 26 0 -22 -40

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-x/8<-1/2. Use the reverse order of operations to solve this. So, as both sides are negative, multiply them by -1 to get you to x/8>1/2, as multiplying by a negative flips the symbol. Solve that by multiplying 1/2 to 8/1. Multiply the numerators and denominators, to give you 8/2. Simplified, x>4
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