Question

use quadratic regression to find the equation for the parabola going through these 3 points. (-4, -33), (1, 2), and (9, 162) y = ?x² + x +

Explanation:

Step1: Set up system of equations

A quadratic equation has the form $y = ax^2 + bx + c$. Substitute each point to get 3 equations:

1. For $(-4, -33)$: $(-4)^2a + (-4)b + c = -33 \implies 16a - 4b + c = -33$
2. For $(1, 2)$: $1^2a + 1b + c = 2 \implies a + b + c = 2$
3. For $(9, 162)$: $9^2a + 9b + c = 162 \implies 81a + 9b + c = 162$

Step2: Eliminate $c$ (Eq1 - Eq2)

Subtract equation 2 from equation 1:
$(16a - 4b + c) - (a + b + c) = -33 - 2$
$15a - 5b = -35$
Simplify by dividing by 5: $3a - b = -7$ (Equation 4)

Step3: Eliminate $c$ (Eq3 - Eq2)

Subtract equation 2 from equation 3:
$(81a + 9b + c) - (a + b + c) = 162 - 2$
$80a + 8b = 160$
Simplify by dividing by 8: $10a + b = 20$ (Equation 5)

Step4: Solve for $a$ (Eq4 + Eq5)

Add equation 4 and equation 5:
$(3a - b) + (10a + b) = -7 + 20$
$13a = 13$
$a = 1$

Step5: Solve for $b$ (Substitute $a=1$ to Eq5)

Substitute $a=1$ into $10a + b = 20$:
$10(1) + b = 20$
$b = 20 - 10 = 10$

Step6: Solve for $c$ (Substitute $a,b$ to Eq2)

Substitute $a=1, b=10$ into $a + b + c = 2$:
$1 + 10 + c = 2$
$c = 2 - 11 = -9$

Answer:

$y = 1x^2 + 10x + (-9)$ or simplified: $y = x^2 + 10x - 9$