Question

use quadratic regression to find the equation for the parabola going through these 3 points. (-13, -126), (-7, 24), and (3, -46) y = ?x² + □x + □

Explanation:

Step1: Recall Quadratic Regression Formula

A quadratic equation is of the form \( y = ax^2 + bx + c \). We have three points \((x_1,y_1)=(-13, -126)\), \((x_2,y_2)=(-7, 24)\), \((x_3,y_3)=(3, -46)\). We substitute each point into the equation to get a system of equations:

For \((-13, -126)\): \(-126=a(-13)^2 + b(-13)+c=169a - 13b + c\)
For \((-7, 24)\): \(24=a(-7)^2 + b(-7)+c = 49a-7b + c\)
For \((3, -46)\): \(-46=a(3)^2 + b(3)+c=9a + 3b + c\)

So the system is:

1. \(169a - 13b + c=-126\)
2. \(49a - 7b + c=24\)
3. \(9a + 3b + c=-46\)

Step2: Subtract Equations to Eliminate \(c\)

Subtract equation 2 from equation 1:
\((169a - 13b + c)-(49a - 7b + c)=-126 - 24\)
\(120a - 6b=-150\) (Equation 4)

Subtract equation 3 from equation 2:
\((49a - 7b + c)-(9a + 3b + c)=24-(-46)\)
\(40a - 10b=70\) (Equation 5)

Step3: Simplify Equations 4 and 5

Simplify Equation 4: Divide by 6: \(20a - b=-25\) (Equation 4a: \(b = 20a + 25\))

Simplify Equation 5: Divide by 10: \(4a - b=7\) (Equation 5a)

Step4: Substitute \(b\) from 4a into 5a

Substitute \(b = 20a + 25\) into \(4a - b=7\):
\(4a-(20a + 25)=7\)
\(4a - 20a - 25=7\)
\(-16a=32\)
\(a = - 2\)

Step5: Find \(b\) using \(a=-2\)

From Equation 4a: \(b=20(-2)+25=-40 + 25=-15\)

Step6: Find \(c\) using \(a=-2\) and \(b=-15\)

Substitute \(a=-2\) and \(b=-15\) into equation 3:
\(-46=9(-2)+3(-15)+c\)
\(-46=-18 - 45 + c\)
\(-46=-63 + c\)
\(c=-46 + 63 = 17\)

Answer:

The quadratic equation is \(y=-2x^2-15x + 17\), so the coefficient of \(x^2\) is \(-2\).