Question

question 1 of 10
which regression equation best fits these data?

x	y
-4	8
-3	12
-2	14
-1	16
1	15
2	12
3	9
4	5

a. y = 10.72·0.95^x
b. y = 0.58x² + 0.43x + 15.75
c. y = -0.43x + 11.34
d. y = -0.58x² - 0.43x + 15.75

Explanation:

Step1: Analyze the data trend

The data first increases and then decreases as \(x\) changes, which suggests a quadratic - like trend rather than a linear or exponential trend.

Step2: Test the quadratic equations

For option B \(y = 0.58x^{2}+0.43x + 15.75\), when \(x=-4\), \(y=0.58\times(-4)^{2}+0.43\times(-4)+15.75=0.58\times16 - 1.72+15.75=9.28-1.72 + 15.75=23.31\) (not a good fit).
For option D \(y=-0.58x^{2}-0.43x + 15.75\), when \(x = - 4\), \(y=-0.58\times(-4)^{2}-0.43\times(-4)+15.75=-0.58\times16 + 1.72+15.75=-9.28+1.72 + 15.75=8.19\approx8\); when \(x=-3\), \(y=-0.58\times(-3)^{2}-0.43\times(-3)+15.75=-0.58\times9+1.29 + 15.75=-5.22+1.29+15.75 = 11.82\approx12\); and it fits other data - points relatively well.
For option A \(y = 10.72\times0.95^{x}\), it is an exponential decay function, which does not match the data's up - and - down trend.
For option C \(y=-0.43x + 11.34\), it is a linear function, which does not match the data's non - linear trend.

Answer:

D. \(y=-0.58x^{2}-0.43x + 15.75\)