Exercise 13 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, - 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
( 2, 8)	8=a( 2)^2+b( 2)+c	4a+2b+c=8

We now have a system of three equations. a-b+c=- 1 & (I) a+b+c=3 & (II) 4a+2b+c=8 & (III) Let's solve this system using the Elimination Method. We will start by subtracting Equation (I) from Equation (II) to eliminate the a and c.

a-b+c=- 1 a+b+c=3 4a+2b+c=8

SysEqnSub

(II): Subtract (I)

a-b+c=- 1 a+b+c-( a-b+c)=3-( - 1) 4a+2b+c=8

▼

(II): Solve for b

Distr

(II): Distribute - 1

a-b+c=- 1 a+b+c-a+b-c=3-(- 1) 4a+2b+c=8

SubNeg

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

a-b+c=- 1 a+b+c-a+b-c=3+1 4a+2b+c=8

AddSubTerms

(II): Add and subtract terms

a-b+c=- 1 2b=4 4a+2b+c=8

DivEqn

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

a-b+c=- 1 b= 2 4a+2b+c=8

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

a-b+c=- 1 & (I) b=2 & (II) 4a+2b+c=8 & (III)

(I), (III): b= 2

a- 2+c=- 1 b=2 4a+2( 2)+c=8

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(I), (III):Simplify

AddEqn

(I): LHS+2=RHS+2

a+c=1 b=2 4a+2(2)+c=8

Multiply

(III): Multiply

a+c=1 b=2 4a+4+c=8

SubEqn

(III): LHS-4=RHS-4

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

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

SysEqnSub

(III): Subtract (I)

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

▼

(III): Solve for a

Distr

(III): Distribute - 1

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

SubTerms

(III): Subtract terms

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

DivEqn

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

a+c=1 b=2 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 (I) to find the value of c.

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

Substitute

(I): a= 1

1+c=1 b=2 a=1

SubEqn

(I): LHS-1=RHS-1

c= 0 b=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+ 0 ⇕ y=x^2+2x To help visualize this situation, we can plot the given points and sketch the curve.