Question

use quadratic regression to find the equation for the parabola going through these 3 points. (3, -12), (-2, 43), and (2, -5) y = ?x² + x +

Explanation:

Step1: Recall Quadratic Form

The general form of a quadratic equation is \( y = ax^2 + bx + c \). We substitute the three points \((3, -12)\), \((-2, 43)\), and \((2, -5)\) into this equation to get a system of linear equations.

For \((3, -12)\): \( -12 = a(3)^2 + b(3) + c \) → \( 9a + 3b + c = -12 \) (Equation 1)

For \((-2, 43)\): \( 43 = a(-2)^2 + b(-2) + c \) → \( 4a - 2b + c = 43 \) (Equation 2)

For \((2, -5)\): \( -5 = a(2)^2 + b(2) + c \) → \( 4a + 2b + c = -5 \) (Equation 3)

Step2: Subtract Equations to Eliminate \(c\)

Subtract Equation 2 from Equation 3:

\( (4a + 2b + c) - (4a - 2b + c) = -5 - 43 \)

Simplify: \( 4b = -48 \) → \( b = -12 \)

Step3: Substitute \(b = -12\) into Equations 1 and 2

Substitute \(b = -12\) into Equation 1: \( 9a + 3(-12) + c = -12 \) → \( 9a - 36 + c = -12 \) → \( 9a + c = 24 \) (Equation 4)

Substitute \(b = -12\) into Equation 2: \( 4a - 2(-12) + c = 43 \) → \( 4a + 24 + c = 43 \) → \( 4a + c = 19 \) (Equation 5)

Step4: Subtract Equation 5 from Equation 4

\( (9a + c) - (4a + c) = 24 - 19 \)

Simplify: \( 5a = 5 \) → \( a = 1 \)

Step5: Find \(c\) using \(a = 1\) and Equation 5

Substitute \(a = 1\) into Equation 5: \( 4(1) + c = 19 \) → \( 4 + c = 19 \) → \( c = 15 \)

Answer:

The quadratic equation is \( y = 1x^2 - 12x + 15 \), so the coefficient of \(x^2\) (the value in the first box) is \(1\).