Exercise 8 Page 212

Let's start by recalling the standard form of a quadratic function. y=ax^2+bx+c To find the equation of a parabola that includes the given points, we will substitute their coordinates into the above equation and simplify. With the resulting equations, we will write a system of equations. Then, we will solve it to find the coefficients a, b, and c.

y=ax^2+bx+c
Point	Substitute	Simplify
( 1, - 2)	- 2=a( 1)^2+b( 1)+c	a+b+c=- 2
( 2, - 4)	- 4=a( 2)^2+b( 2)+c	4a+2b+c=- 4
( 3, - 4)	- 4=a( 3)^2+b( 3)+c	9a+3b+c=- 4

We can now write a system of three equations. a+b+c=- 2 & (I) 4a+2b+c=- 4 & (II) 9a+3b+c=- 4 & (III) Let's solve this system using the Elimination Method. We will start by subtracting Equation (I) from Equation (II) and Equation (III) to eliminate c.

a+b+c=- 2 4a+2b+c=- 4 9a+3b+c=- 4

(II), (III): Subtract (I)

a+b+c=- 2 4a+2b+c-( a+b+c)=- 4-( - 2) 9a+3b+c-( a+b+c)=- 4-( - 2)

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(II), (III): Simplify

(II), (III): Distribute - 1

a+b+c=- 2 4a+2b+c-a-b-c=- 4-(- 2) 9a+3b+c-a-b-c=- 4-(- 2)

(II), (III): a-(- b)=a+b

a+b+c=- 2 4a+2b+c-a-b-c=- 4+2 9a+3b+c-a-b-c=- 4+2

(II), (III): Add and subtract terms

a+b+c=- 2 3a+b=- 2 8a+2b=- 2

Now, neither Equation (II) nor Equation (III) contains c. These equations form a system in terms of only a and b. Let's solve this system by using the Elimination Method again. First, we will need to multiply or divide one equation by a number.

a+b+c=- 2 & (I) 3a+b=- 2 & (II) 8a+2b=- 2 & (III)

FactorOut

(III): Factor out 2

a+b+c=- 2 3a+b=- 2 2(4a+b)=- 2

DivEqn

(III): .LHS /2.=.RHS /2.

a+b+c=- 2 3a+b=- 2 4a+b=- 1

SysEqnSub

(III): Subtract (II)

a+b+c=- 2 3a+b=- 2 4a+b-( 3a+b)=- 1-( - 2)

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(III): Solve for a

Distr

(III): Distribute - 1

a+b+c=- 2 3a+b=- 2 4a+b-3a-b=- 1-(- 2)

SubNeg

(III): a-(- b)=a+b

a+b+c=- 2 3a+b=- 2 4a+b-3a-b=- 1+2

AddSubTerms

(III): Add and subtract terms

a+b+c=- 2 3a+b=- 2 a=1

We found our first value, allowing us to write a partial equation. y= 1x^2+bx+c ⇕ y=x^2+bx+c Let's substitute 1 for a in Equation (II) to find the value of b.

a+b+c=- 2 & (I) 3a+b=- 2 & (II) a=1 & (III)

Substitute

(II): a= 1

a+b+c=- 2 3( 1)+b=- 2 a=1

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(II): Solve for b

IdPropMult

(II): Identity Property of Multiplication

a+b+c=- 2 3+b=- 2 a=1

SubEqn

(II): LHS-3=RHS-3

a+b+c=- 2 b= - 5 a=1

With our second value, we can continue writing the equation. y=x^2+( - 5)x+c ⇕ y=x^2-5x+c Finally, to find the value of c, we will substitute a=1 and b=- 5 into Equation (I).

a+b+c=- 2 & (I) b=- 5 & (II) a=1 & (III)

SubstituteII

(I): a= 1, b= - 5

1+( - 5)+c=- 2 b=- 5 a=1

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(I): Solve for c

AddNeg

(I): a+(- b)=a-b

- 4+c=- 2 b=- 5 a=1

AddEqn

(I): LHS+4=RHS+4

c= 2 b=- 5 a=1

Now that we have all three values, we can write the equation of the parabola that passes through the given points. y=x^2-5x+ 2 To help visualize this situation, we can plot the given points and sketch the curve.