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

We can now write a system of three equations. a+b+c=1 & (I) a-b+c=- 3 & (II) 9a-3b+c=1 & (III) Let's solve this system using the Elimination Method. We will start by subtracting Equation (II) from Equation (I).

a+b+c=1 a-b+c=- 3 9a-3b+c=1

SysEqnSub

(I): Subtract (II)

a+b+c-( a-b+c)=1-( - 3) a-b+c=- 3 9a-3b+c=1

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(I): Solve for b

Distr

(I): Distribute - 1

a+b+c-a+b-c=1-(- 3) a-b+c=- 3 9a-3b+c=1

SubNeg

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

a+b+c-a+b-c=1+3 a-b+c=- 3 9a-3b+c=1

AddSubTerms

(I): Add and subtract terms

2b=4 a-b+c=- 3 9a-3b+c=1

DivEqn

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

b= 2 a-b+c=- 3 9a-3b+c=1

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

b=2 a-b+c=- 3 9a-3b+c=1

(II), (III): b= 2

b=2 a- 2+c=- 3 9a-3( 2)+c=1

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(II), (III): Simplify

Multiply

(III): Multiply

b=2 a-2+c=- 3 9a-6+c=1

AddEqn

(II): LHS+2=RHS+2

b=2 a+c=- 1 9a-6+c=1

AddEqn

(III): LHS+6=RHS+6

b=2 a+c=- 1 9a+c=7

Now, neither Equation (II) 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 (II) from Equation (III).

b=2 a+c=- 1 9a+c=7

SysEqnSub

(III): Subtract (II)

b=2 a+c=- 1 9a+c-( a+c)=7-( - 1)

▼

(III): Solve for a

Distr

(III): Distribute (- 1)

b=2 a+c=- 1 9a+c-a-c=7-(- 1)

SubNeg

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

b=2 a+c=- 1 9a+c-a-c=7+1

AddSubTerms

(III): Add and subtract terms

b=2 a+c=- 1 8a=8

DivEqn

(III): .LHS /8.=.RHS /8.

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

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

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

Substitute

(II): a= 1

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

SubEqn

(II): LHS-1=RHS-1

b=2 c= - 2 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+2x+( - 2) ⇕ y=x^2+2x-2 To help visualize this situation, we can plot the given points and sketch the curve.