Exercise 9 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, 6)	6=a( - 1)^2+b( - 1)+c	a-b+c=6
( 1, 4)	4=a( 1)^2+b( 1)+c	a+b+c=4
( 2, 9)	9=a( 2)^2+b( 2)+c	4a+2b+c=9

We can now write a system of three equations. a-b+c=6 & (I) a+b+c=4 & (II) 4a+2b+c=9 & (III) Let's solve this system using the Elimination Method. We will start by subtracting Equation (II) from Equation (I).

a-b+c=6 a+b+c=4 4a+2b+c=9

SysEqnSub

(I): Subtract (II)

a-b+c-( a+b+c)=6- 4 a+b+c=4 4a+2b+c=9

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(I): Solve for b

Distr

(I): Distribute - 1

a-b+c-a-b-c=6-4 a+b+c=4 4a+2b+c=9

SubTerms

(I): Subtract terms

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

DivEqn

(I): .LHS /(- 2).=.RHS /(- 2).

b= - 1 a+b+c=4 4a+2b+c=9

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

b=- 1 & (I) a+b+c=4 & (II) 4a+2b+c=9 & (III)

(II), (III): b= - 1

b=- 1 a+( - 1)+c=4 4a+2( - 1)+c=9

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(II), (III): Simplify

AddNeg

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

b=- 1 a-1+c=4 4a+2(- 1)+c=9

MultPosNeg

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

b=- 1 a-1+c=4 4a-2+c=9

AddEqn

(II): LHS+1=RHS+1

b=- 1 a+c=5 4a-2+c=9

AddEqn

(III): LHS+2=RHS+2

b=- 1 a+c=5 4a+c=11

Now, neither Equation (II) nor Equation (III) contains b. These equations form a system in terms of only a and c. Let's solve this system by using the Elimination Method again.

b=- 1 & (I) a+c=5 & (II) 4a+c=11 & (III)

SysEqnSub

(III): Subtract (II)

b=- 1 a+c=5 4a+c-( a+c)=11- 5

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(III): Solve for a

Distr

(III): Distribute (- 1)

b=- 1 a+c=5 4a+c-a-c=11-5

SubTerms

(III): Subtract terms

b=- 1 a+c=5 3a=6

DivEqn

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

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

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

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

Substitute

(II): a= 2

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

SubEqn

(II): LHS-2=RHS-2

b=- 1 c= 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-x+ 3 To help visualize this situation, we can plot the given points and sketch the curve.