Question

use quadratic regression to find the equation for the parabola going through these 3 points: (2, -33), (4, 5), and (5, 39) y = ?x² + x +

Explanation:

Step1: Recall Quadratic Form

The general quadratic equation is \( y = ax^2 + bx + c \). We substitute the three points into this equation to form a system of equations.

For point \((2, -33)\):
\( -33 = a(2)^2 + b(2) + c \)
\( -33 = 4a + 2b + c \) --- (1)

For point \((4, 5)\):
\( 5 = a(4)^2 + b(4) + c \)
\( 5 = 16a + 4b + c \) --- (2)

For point \((5, 39)\):
\( 39 = a(5)^2 + b(5) + c \)
\( 39 = 25a + 5b + c \) --- (3)

Step2: Subtract Equations to Eliminate \( c \)

Subtract (1) from (2):
\( (16a + 4b + c) - (4a + 2b + c) = 5 - (-33) \)
\( 12a + 2b = 38 \)
Simplify: \( 6a + b = 19 \) --- (4)

Subtract (2) from (3):
\( (25a + 5b + c) - (16a + 4b + c) = 39 - 5 \)
\( 9a + b = 34 \) --- (5)

Step3: Solve for \( a \)

Subtract (4) from (5):
\( (9a + b) - (6a + b) = 34 - 19 \)
\( 3a = 15 \)
\( a = 5 \)

Step4: Solve for \( b \)

Substitute \( a = 5 \) into (4):
\( 6(5) + b = 19 \)
\( 30 + b = 19 \)
\( b = 19 - 30 = -11 \)

Step5: Solve for \( c \)

Substitute \( a = 5 \) and \( b = -11 \) into (1):
\( -33 = 4(5) + 2(-11) + c \)
\( -33 = 20 - 22 + c \)
\( -33 = -2 + c \)
\( c = -33 + 2 = -31 \)

Answer:

The quadratic equation is \( y = 5x^2 - 11x - 31 \), so the coefficient of \( x^2 \) is \( \boldsymbol{5} \).