Question

use quadratic regression to find the equation for the parabola going through these 3 points. (-4, 7), (6, -33), and (10, -105) y = ?x² + x +

Explanation:

Step1: Set up quadratic equations

A quadratic equation has the form $y = ax^2 + bx + c$. Substitute each point into the equation:
For $(-4, 7)$: $7 = a(-4)^2 + b(-4) + c \implies 16a - 4b + c = 7$
For $(6, -33)$: $-33 = a(6)^2 + b(6) + c \implies 36a + 6b + c = -33$
For $(10, -105)$: $-105 = a(10)^2 + b(10) + c \implies 100a + 10b + c = -105$

Step2: Eliminate $c$ (Eq2-Eq1)

Subtract the first equation from the second:
$(36a + 6b + c) - (16a - 4b + c) = -33 - 7$
$20a + 10b = -40$
Divide by 10: $2a + b = -4$ (Equation 4)

Step3: Eliminate $c$ (Eq3-Eq2)

Subtract the second equation from the third:
$(100a + 10b + c) - (36a + 6b + c) = -105 - (-33)$
$64a + 4b = -72$
Divide by 4: $16a + b = -18$ (Equation 5)

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

Subtract Equation 4 from Equation 5:
$(16a + b) - (2a + b) = -18 - (-4)$
$14a = -14$
$a = \frac{-14}{14} = -1$

Step5: Solve for $b$

Substitute $a=-1$ into Equation 4:
$2(-1) + b = -4$
$-2 + b = -4$
$b = -4 + 2 = -2$

Step6: Solve for $c$

Substitute $a=-1, b=-2$ into the first original equation:
$16(-1) - 4(-2) + c = 7$
$-16 + 8 + c = 7$
$-8 + c = 7$
$c = 7 + 8 = 15$

Answer:

$y = -1x^2 + (-2)x + 15$ (or simplified: $y = -x^2 - 2x + 15$)