Exercise 15 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 three points given in the table, 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. Let's consider the given table.

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

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

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

SysEqnSub

(II): Subtract (I)

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

▼

(II): Solve for b

Distr

(II): Distribute - 1

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

SubNeg

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

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

AddSubTerms

(II): Add and subtract terms

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

DivEqn

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

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

We found our first value, allowing us to write a partial equation. y=ax^2+ 1x+c ⇕ y=ax^2+x+c We can now substitute 1 for b in Equation (I) and Equation (III).

a-b+c=- 4 & (I) b=1 & (II) 4a+2b+c=- 4 & (III)

(I), (III): b= 1

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

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(I), (III):Simplify

AddEqn

(I): LHS+1=RHS+1

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

IdPropMult

(III): Identity Property of Multiplication

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

SubEqn

(III): LHS-2=RHS-2

a+c=- 3 b=1 4a+c=- 6

Now, neither Equation (I) nor Equation (III) includes b. These equations form a system in terms of only a and c. Let's solve this system by using the Elimination Method again. To do so, we will subtract Equation (I) from Equation (III).

a+c=- 3 & (I) b=1 & (II) 4a+c=- 6 & (III)

SysEqnSub

(III): Subtract (I)

a+c=- 3 b=1 4a+c-( a+c)=- 6-( - 3)

▼

(III): Solve for a

Distr

(III): Distribute - 1

a+c=- 3 b=1 4a+c-a-c=- 6-(- 3)

SubNeg

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

a+c=- 3 b=1 4a+c-a-c=- 6+3

AddSubTerms

(III): Add and subtract terms

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

DivEqn

(III): .LHS /3.=.RHS /3.

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

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

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

Substitute

(I): a= - 1

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

AddEqn

(I): LHS+1=RHS+1

c= - 2 b=1 a=- 1

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