Question

question 9 of 10
what is the quadratic regression equation that fits these data?

x	y
-3	28
-2	10
-1	8
0	7
1	10
2	16
3	26
4	40

a. $y=16.76\cdot1.02^{x}$
b. $y=1.64x^{2}+1.24x+8.08$
c. $y=2.13x^{2}+0.13x+6.39$
d. $y=0.82x^{2}+0.78x+7.23$

Explanation:

Step1: Eliminate non-quadratic option

Option A is an exponential function ($y=16.76 \cdot 1.02^x$), not quadratic, so it is discarded.

Step2: Test x=0 in remaining options

At $x=0$, the quadratic equation simplifies to $y=c$ (the constant term). From the data, when $x=0$, $y=7$.

• Option B: $y=8.08$ (differs from 7)
• Option C: $y=6.39$ (close to 7)
• Option D: $y=7.23$ (close to 7)

Eliminate Option B.

Step3: Test x=4 in remaining options

From the data, when $x=4$, $y=40$.

• Option C: $y=2.13(4)^2 + 0.13(4) + 6.39 = 2.13(16) + 0.52 + 6.39 = 34.08 + 0.52 + 6.39 = 40.99 \approx 41$
• Option D: $y=0.82(4)^2 + 0.78(4) + 7.23 = 0.82(16) + 3.12 + 7.23 = 13.12 + 3.12 + 7.23 = 23.47$

Option C's result is closest to the data value of 40.

Step4: Verify with x=-4

From the data, when $x=-4$, $y=40$.
Option C: $y=2.13(-4)^2 + 0.13(-4) + 6.39 = 2.13(16) - 0.52 + 6.39 = 34.08 - 0.52 + 6.39 = 39.95 \approx 40$, which matches the data.

Answer:

C. $y=2.13x^2 + 0.13x + 6.39$