Question

use quadratic regression to find the equation for the parabola going through these 3 points. (-7, 134), (-3, 10), and (5, 50) y = 3x² + ?x +

Explanation:

Step1: Recall quadratic formula

The quadratic equation is \( y = ax^2 + bx + c \). We know \( a = 3 \), so the equation is \( y = 3x^2 + bx + c \). We can use one of the points to form an equation. Let's use the point \((-3, 10)\). Substitute \( x = -3 \), \( y = 10 \), and \( a = 3 \) into the equation:
\( 10 = 3(-3)^2 + b(-3) + c \)
\( 10 = 3(9) - 3b + c \)
\( 10 = 27 - 3b + c \)
Rearrange to get \( c = 3b - 17 \).

Step2: Use another point

Now use the point \((5, 50)\). Substitute \( x = 5 \), \( y = 50 \), \( a = 3 \) into \( y = 3x^2 + bx + c \):
\( 50 = 3(5)^2 + b(5) + c \)
\( 50 = 3(25) + 5b + c \)
\( 50 = 75 + 5b + c \)
Rearrange to get \( c = -5b - 25 \).

Step3: Solve for b

We have two expressions for \( c \): \( c = 3b - 17 \) and \( c = -5b - 25 \). Set them equal:
\( 3b - 17 = -5b - 25 \)
Add \( 5b \) to both sides: \( 8b - 17 = -25 \)
Add 17 to both sides: \( 8b = -8 \)
Divide by 8: \( b = -1 \)

Step4: Solve for c (optional, but let's check)

Using \( c = 3b - 17 \) with \( b = -1 \):
\( c = 3(-1) - 17 = -3 - 17 = -20 \)
We can verify with the third point \((-7, 134)\). Substitute \( x = -7 \), \( a = 3 \), \( b = -1 \), \( c = -20 \):
\( y = 3(-7)^2 + (-1)(-7) + (-20) = 3(49) + 7 - 20 = 147 + 7 - 20 = 134 \), which matches.

Answer:

The coefficient of \( x \) is \(-1\), so the middle box is \(-1\) (and the constant term is \(-20\), but the question asks for the coefficient of \( x \), so the answer for the [?] is \(-1\)).