Exercise 1 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, 0)	0=a( 1)^2+b( 1)+c	a+b+c=0
( 2, - 3)	- 3=a( 2)^2+b( 2)+c	4a+2b+c=- 3
( 3, - 10)	- 10=a( 3)^2+b( 3)+c	9a+3b+c=- 10

We can now write a system of three equations. a+b+c=0 & (I) 4a+2b+c=- 3 & (II) 9a+3b+c=- 10 & (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=0 4a+2b+c=- 3 9a+3b+c=- 10

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

a+b+c=0 4a+2b+c-( a+b+c)=- 3- 0 9a+3b+c-( a+b+c)=- 10- 0

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(II), (III): Simplify

(II), (III): Distribute - 1

a+b+c=0 4a+2b+c-a-b-c=- 3-0 9a+3b+c-a-b-c=- 10-0

(II), (III): Subtract terms

a+b+c=0 3a+b=- 3 8a+2b=- 10

Now, neither Equation (II) nor Equation (III) includes c. These equations form a system in terms of only a and b. Let's solve this system by using the Elimination Method again. Since neither variable has the same or opposite coefficients, we need to multiply or divide one equation by a number first.

a+b+c=0 & (I) 3a+b=- 3 & (II) 8a+2b=- 10 & (III)

FactorOut

(III): Factor out 2

a+b+c=0 3a+b=- 3 2(4a+b)=- 10

DivEqn

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

a+b+c=0 3a+b=- 3 4a+b=- 5

SysEqnSub

(III): Subtract (II)

a+b+c=0 3a+b=- 3 4a+b-( 3a+b)=- 5-( - 3)

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(III): Solve for a

Distr

(III): Distribute - 1

a+b+c=0 3a+b=- 3 4a+b-3a-b=- 5-(- 3)

SubNeg

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

a+b+c=0 3a+b=- 3 4a+b-3a-b=- 5+3

AddSubTerms

(III): Add and subtract terms

a+b+c=0 3a+b=- 3 a= - 2

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

a+b+c=0 & (I) 3a+b=- 3 & (II) a=- 2 & (III)

Substitute

(II): a= - 2

a+b+c=0 3( - 2)+b=- 3 a=- 2

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(II): Solve for b

MultPosNeg

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

a+b+c=0 - 6+b=- 3 a=- 2

AddEqn

(II): LHS+6=RHS+6

a+b+c=0 b= 3 a=- 2

With our second value, we can continue forming the partial equation. y=- 2x^2+ 3x+c Finally, to find the value of c, we will substitute a=- 2 and b=3 into Equation (I).

a+b+c=0 b=3 a=- 2

SubstituteII

(I): a= - 2, b= 3

- 2+ 3+c=0 b=3 a=- 2

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(I): Solve for c

AddTerms

(I): Add terms

1+c=0 b=3 a=- 2

SubEqn

(I): LHS-1=RHS-1

c= - 1 b=3 a=- 2

Now that we have all three values, we can complete the equation of the parabola that passes through the given points. y=- 2x^2+3x+( - 1) ⇕ y=- 2x^2+3x-1 To help visualize this situation, we can plot the given points and sketch the curve.