use quadratic regression to find the equation for the parabola going through these 3 points. (-5, -60), (0…

Explanation:

Step1: Recall Quadratic Form

The general form of a quadratic equation is ( y = ax^2 + bx + c ). We substitute the given points into this equation to form a system of equations.

Step2: Substitute (0, -5)

Substitute ( x = 0 ), ( y = -5 ) into ( y = ax^2 + bx + c ):
( -5 = a(0)^2 + b(0) + c ), so ( c = -5 ).

Step3: Substitute (-5, -60)

Now with ( c = -5 ), substitute ( x = -5 ), ( y = -60 ):
( -60 = a(-5)^2 + b(-5) - 5 )
Simplify: ( -60 = 25a - 5b - 5 )
Add 5 to both sides: ( -55 = 25a - 5b ) (Equation 1)

Step4: Substitute (13, -96)

Substitute ( x = 13 ), ( y = -96 ), ( c = -5 ):
( -96 = a(13)^2 + b(13) - 5 )
Simplify: ( -96 = 169a + 13b - 5 )
Add 5 to both sides: ( -91 = 169a + 13b ) (Equation 2)

Step5: Solve the System

From Equation 1: ( 25a - 5b = -55 ), divide by 5: ( 5a - b = -11 ) → ( b = 5a + 11 )
Substitute ( b = 5a + 11 ) into Equation 2:
( -91 = 169a + 13(5a + 11) )
Expand: ( -91 = 169a + 65a + 143 )
Combine like terms: ( -91 = 234a + 143 )
Subtract 143: ( -234 = 234a )
Divide by 234: ( a = -1 )

Step6: Find b

Substitute ( a = -1 ) into ( b = 5a + 11 ):
( b = 5(-1) + 11 = -5 + 11 = 6 )

Answer:

The quadratic equation is ( y = -1x^2 + 6x - 5 ), so the coefficient of ( x^2 ) is (-1).
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