Question

use quadratic regression to find the equation for the parabola going through these 3 points. (-9, -175), (3, 29), and (10, -118) y = ?x² + x +

Explanation:

Step1: Recall quadratic regression formula

A quadratic function is of the form \( y = ax^2 + bx + c \). We substitute the three points \((-9, -175)\), \((3, 29)\), and \((10, -118)\) into this equation to get a system of equations.

For point \((-9, -175)\):
\( -175 = a(-9)^2 + b(-9) + c \)
\( -175 = 81a - 9b + c \) --- (1)

For point \((3, 29)\):
\( 29 = a(3)^2 + b(3) + c \)
\( 29 = 9a + 3b + c \) --- (2)

For point \((10, -118)\):
\( -118 = a(10)^2 + b(10) + c \)
\( -118 = 100a + 10b + c \) --- (3)

Step2: Subtract equation (2) from equation (1)

\( (81a - 9b + c) - (9a + 3b + c) = -175 - 29 \)
\( 81a - 9b + c - 9a - 3b - c = -204 \)
\( 72a - 12b = -204 \)
Divide both sides by 12: \( 6a - b = -17 \) --- (4)

Step3: Subtract equation (2) from equation (3)

\( (100a + 10b + c) - (9a + 3b + c) = -118 - 29 \)
\( 100a + 10b + c - 9a - 3b - c = -147 \)
\( 91a + 7b = -147 \)
Divide both sides by 7: \( 13a + b = -21 \) --- (5)

Step4: Solve equations (4) and (5)

Add equation (4) and equation (5):
\( (6a - b) + (13a + b) = -17 + (-21) \)
\( 19a = -38 \)
\( a = -2 \)

Substitute \( a = -2 \) into equation (4):
\( 6(-2) - b = -17 \)
\( -12 - b = -17 \)
\( -b = -17 + 12 \)
\( -b = -5 \)
\( b = 5 \)

Step5: Find c using equation (2)

Substitute \( a = -2 \) and \( b = 5 \) into equation (2):
\( 29 = 9(-2) + 3(5) + c \)
\( 29 = -18 + 15 + c \)
\( 29 = -3 + c \)
\( c = 29 + 3 \)
\( c = 32 \)

Answer:

\( y = -2x^2 + 5x + 32 \)