Question

use quadratic regression to find the equation for the parabola going through these 3 points. (-7, 134), (-3, 10), and (5, 50) y = ?x² + □x + □

Explanation:

Step1: Recall Quadratic Regression Formula

A quadratic equation is of the form \( y = ax^2 + bx + c \). We substitute the three points \((-7, 134)\), \((-3, 10)\), and \((5, 50)\) into this equation to get a system of linear equations.

For point \((-7, 134)\):
\( 134 = a(-7)^2 + b(-7) + c \)
\( 134 = 49a - 7b + c \) --- (1)

For point \((-3, 10)\):
\( 10 = a(-3)^2 + b(-3) + c \)
\( 10 = 9a - 3b + c \) --- (2)

For point \((5, 50)\):
\( 50 = a(5)^2 + b(5) + c \)
\( 50 = 25a + 5b + c \) --- (3)

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

\( (49a - 7b + c) - (9a - 3b + c) = 134 - 10 \)
\( 49a - 7b + c - 9a + 3b - c = 124 \)
\( 40a - 4b = 124 \)
Divide both sides by 4: \( 10a - b = 31 \) --- (4)

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

\( (25a + 5b + c) - (9a - 3b + c) = 50 - 10 \)
\( 25a + 5b + c - 9a + 3b - c = 40 \)
\( 16a + 8b = 40 \)
Divide both sides by 8: \( 2a + b = 5 \) --- (5)

Step4: Solve Equations (4) and (5)

Add Equation (4) and Equation (5):
\( (10a - b) + (2a + b) = 31 + 5 \)
\( 12a = 36 \)
\( a = 3 \)

Substitute \( a = 3 \) into Equation (5):
\( 2(3) + b = 5 \)
\( 6 + b = 5 \)
\( b = -1 \)

Step5: Find \( c \) using Equation (2)

Substitute \( a = 3 \) and \( b = -1 \) into Equation (2):
\( 10 = 9(3) - 3(-1) + c \)
\( 10 = 27 + 3 + c \)
\( 10 = 30 + c \)
\( c = -20 \)

Answer:

\( y = 3x^2 - x - 20 \)