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: Set up quadratic equations

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

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

Step2: Eliminate $c$ (Eq1 - Eq2)

Subtract Equation 2 from Equation 1:
$$(25a + 5b + c) - (a + b + c) = 210 - 14$$
$$24a + 4b = 196$$
Simplify by dividing by 4: $6a + b = 49$ (Eq4)

Step3: Eliminate $c$ (Eq3 - Eq2)

Subtract Equation 2 from Equation 3:
$$(9a - 3b + c) - (a + b + c) = 10 - 14$$
$$8a - 4b = -4$$
Simplify by dividing by 4: $2a - b = -1$ (Eq5)

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

Add Equation 4 and Equation 5:
$$(6a + b) + (2a - b) = 49 + (-1)$$
$$8a = 48$$
$$a = 6$$

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

Substitute $a=6$ into $2a - b = -1$:
$$2(6) - b = -1$$
$$12 - b = -1$$
$$b = 13$$

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

Substitute $a=6, b=13$ into $a + b + c = 14$:
$$6 + 13 + c = 14$$
$$19 + c = 14$$
$$c = -5$$

Answer:

$y = 6x^2 + 13x - 5$