Question

use quadratic regression to find the equation for the parabola going through these 3 points.
(-9, -175), (3, 29), and (10, -118)
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 $(-9, -175)$: $a(-9)^2 + b(-9) + c = -175$ → $81a - 9b + c = -175$
2. For $(3, 29)$: $a(3)^2 + b(3) + c = 29$ → $9a + 3b + c = 29$
3. For $(10, -118)$: $a(10)^2 + b(10) + c = -118$ → $100a + 10b + c = -118$

Step2: Eliminate c (Eq1 - Eq2)

Subtract equation 2 from equation 1 to eliminate $c$:
$(81a - 9b + c) - (9a + 3b + c) = -175 - 29$
$72a - 12b = -204$
Divide by 12: $6a - b = -17$ → Equation 4

Step3: Eliminate c (Eq3 - Eq2)

Subtract equation 2 from equation 3 to eliminate $c$:
$(100a + 10b + c) - (9a + 3b + c) = -118 - 29$
$91a + 7b = -147$
Divide by 7: $13a + b = -21$ → Equation 5

Step4: Solve for a (Eq4 + Eq5)

Add equation 4 and equation 5 to eliminate $b$:
$(6a - b) + (13a + b) = -17 + (-21)$
$19a = -38$
$a = \frac{-38}{19} = -2$

Step5: Solve for b (Sub a to Eq4)

Substitute $a=-2$ into equation 4:
$6(-2) - b = -17$
$-12 - b = -17$
$-b = -17 + 12 = -5$
$b = 5$

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

Substitute $a=-2, b=5$ into equation 2:
$9(-2) + 3(5) + c = 29$
$-18 + 15 + c = 29$
$-3 + c = 29$
$c = 29 + 3 = 32$

Answer:

$y = -2x^2 + 5x + 32$