Question

polynomial regression: cubic
regression: line of best fit to model a certain phenomenon

• cubic regression:
• a) models the daily volume of ice in the arctic

insert the table on desmos.com. zoom in until the points are clearly in view.
use the following quadratic regression model to find the function for the points.
$y1\sim ax^3+bx^2+cx+d$
(options:
$y=-0.15x^3+12.56429x+37.02$
$y=0.15x^3+2.38571x^2+12.56429x+37.02$
$y=-0.15x^3+2.38571x^2-12.56429x+37.02$
$y=0.15x^3+12.56429x+37.02$)

Explanation:

Step1: Identify the model type

The problem uses a cubic regression model \( y_1 = ax^3 + bx^2 + cx + d \), so we need to check which option has a cubic (degree 3) polynomial form.

Step2: Analyze the options

• Option 1: \( y = -0.15x^2 + 12.56429x + 37.02 \) is quadratic (degree 2), so eliminate.
• Option 2: \( y = 0.15x^3 + 2.38571x^2 + 125.6429x + 37.02 \) – let's check the trend. The \( x \) values are 1,2,3,4,5 and \( y \) values are decreasing (26.7,20.2,16.8,15.3,15.1), so a positive cubic coefficient might not fit (positive cubic would increase for large \( x \), but here \( x \) is small, but let's check signs. Wait, maybe miscalculation. Wait, the \( x \) is 1,2,3,4,5 (small), \( y \) is decreasing. Let's check the third option: \( y = -0.15x^3 + 2.38571x^2 - 12.56429x + 37.02 \). Let's plug \( x = 1 \): \( -0.15 + 2.38571 - 12.56429 + 37.02 \approx (-0.15 -12.56429)+(2.38571 + 37.02) \approx -12.71429 + 39.40571 \approx 26.69 \), which matches \( y=26.7 \) (x=1).
• Option 3: \( y = -0.15x^3 + 2.38571x^2 - 12.56429x + 37.02 \) – plug \( x=1 \): as above, ~26.7. \( x=2 \): \( -0.15(8) + 2.38571(4) - 12.56429(2) + 37.02 = -1.2 + 9.54284 - 25.12858 + 37.02 = (-1.2 -25.12858)+(9.54284 + 37.02) = -26.32858 + 46.56284 \approx 20.23 \), which matches \( y=20.2 \) (x=2).
• Option 4: \( y = 0.15x^3 + 12.56429x + 37.02 \) – quadratic term missing, and cubic coefficient positive. Plug \( x=1 \): \( 0.15 + 12.56429 + 37.02 \approx 49.73 \), which is way higher than 26.7, so eliminate.

Wait, maybe I messed up signs. Wait the \( y \) values are decreasing, so the cubic term should be negative (since for small \( x \), a negative cubic would start decreasing? Wait, no: cubic with negative leading coefficient: as \( x \) increases, if \( x^3 \) term is negative, for \( x>0 \), \( -ax^3 \) (a>0) would decrease as \( x \) increases. Let's check option 3: leading coefficient -0.15 (negative), so cubic term is negative. Plug \( x=3 \) into option 3: \( -0.15(27) + 2.38571(9) - 12.56429(3) + 37.02 = -4.05 + 21.47139 - 37.69287 + 37.02 \approx (-4.05 -37.69287)+(21.47139 + 37.02) \approx -41.74287 + 58.49139 \approx 16.75 \), which matches \( y=16.8 \) (x=3).

• Option 2: positive cubic coefficient, plug \( x=1 \): \( 0.15 + 2.38571 + 125.6429 + 37.02 \approx 165.1986 \), which is way off. So option 3 fits the \( x=1,2,3 \) values.

Answer:

\( y = -0.15x^3 + 2.38571x^2 - 12.56429x + 37.02 \) (the third option)