Question

use quadratic regression to find the equation for the parabola going through these 3 points. (5, 210), (1, 14), and (-3, 10) y = ?x² + x +

Explanation:

Step1: Recall Quadratic Form

The general quadratic equation is \( y = ax^2 + bx + c \). We substitute the three points \((5, 210)\), \((1, 14)\), and \((-3, 10)\) into this equation to form a system of linear equations.

For \((5, 210)\): \( 210 = a(5)^2 + b(5) + c \) → \( 25a + 5b + c = 210 \)
For \((1, 14)\): \( 14 = a(1)^2 + b(1) + c \) → \( a + b + c = 14 \)
For \((-3, 10)\): \( 10 = a(-3)^2 + b(-3) + c \) → \( 9a - 3b + c = 10 \)

Step2: Subtract Equations to Eliminate \( c \)

Subtract the second equation from the first:
\( (25a + 5b + c) - (a + b + c) = 210 - 14 \)
\( 24a + 4b = 196 \) → Simplify by dividing by 4: \( 6a + b = 49 \) (Equation 4)

Subtract the second equation from the third:
\( (9a - 3b + c) - (a + b + c) = 10 - 14 \)
\( 8a - 4b = -4 \) → Simplify by dividing by 4: \( 2a - b = -1 \) (Equation 5)

Step3: Solve the System of Two Equations

Now we have Equation 4: \( 6a + b = 49 \) and Equation 5: \( 2a - b = -1 \). Add them to eliminate \( b \):
\( (6a + b) + (2a - b) = 49 + (-1) \)
\( 8a = 48 \) → \( a = 6 \)

Step4: Find \( b \) and \( c \)

Substitute \( a = 6 \) into Equation 5: \( 2(6) - b = -1 \) → \( 12 - b = -1 \) → \( b = 13 \)

Substitute \( a = 6 \) and \( b = 13 \) into the second original equation (\( a + b + c = 14 \)):
\( 6 + 13 + c = 14 \) → \( 19 + c = 14 \) → \( c = -5 \)

Answer:

The coefficient of \( x^2 \) is \( \boldsymbol{6} \), the coefficient of \( x \) is \( \boldsymbol{13} \), and the constant term is \( \boldsymbol{-5} \). So the equation is \( y = 6x^2 + 13x - 5 \). For the first box (coefficient of \( x^2 \)), the answer is \( 6 \).