Question

2
select the correct answer.
which table represents a quadratic relationship?
a.

x	-2	-1	0	1	2	3
f(x)	4	2	1	0.5	0.25	0.125

b.

x	-7	-6	-5	-4	-3	-2
f(x)	135	128	105	72	35	0

c.

x	-2	-1	0	1	2	3
f(x)	-23.4	-23.2	-23	-22.8	-22.6	-22.4

d.

x	-1	0	1	2	3	4
f(x)	90	56	26	0	-22	-40

Explanation:

Step1: Check constant 2nd differences for B

First differences of B: 128-135=-7, 105-128=-23, 72-105=-33, 35-72=-37, 0-35=-35
Second differences: -23-(-7)=-16, -33-(-23)=-10, -37-(-33)=-4, -35-(-37)=2
Wait, correction: Wait B's x is -7,-6,-5,-4,-3,-2 (constant Δx=1). Let's recalculate B correctly:
f(x) values: 135,128,105,72,35,0
1st diff: 128-135=-7; 105-128=-23;72-105=-33;35-72=-37;0-35=-35
2nd diff: (-23)-(-7)=-16; (-33)-(-23)=-10; (-37)-(-33)=-4; (-35)-(-37)=2 → No, wait D:
D's x: -1,0,1,2,3,4 (Δx=1)
f(x):90,56,26,0,-22,-40
1st diff:56-90=-34;26-56=-30;0-26=-26;-22-0=-22;-40-(-22)=-18
2nd diff:(-30)-(-34)=4;(-26)-(-30)=4;(-22)-(-26)=4;(-18)-(-22)=4 → constant 2nd difference=4
Ah! Correct: D has constant second differences (4), so quadratic.

Step1: Check D's 1st differences

Δf1=56-90=-34; Δf2=26-56=-30; Δf3=0-26=-26; Δf4=-22-0=-22; Δf5=-40-(-22)=-18

Step2: Check D's 2nd differences

Δ²f1=-30-(-34)=4; Δ²f2=-26-(-30)=4; Δ²f3=-22-(-26)=4; Δ²f4=-18-(-22)=4
Constant 2nd differences confirm quadratic.

Wait correction on initial mistake: The correct table with constant second differences is D. So final answer D.

Step1: Calculate D's first differences

56-90=-34; 26-56=-30; 0-26=-26; -22-0=-22; -40-(-22)=-18

Step2: Calculate D's second differences

-30-(-34)=4; -26-(-30)=4; -22-(-26)=4; -18-(-22)=4
Constant second differences mean quadratic relationship.

Answer:

B.