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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

3. Modeling With Quadratic Functions

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Exercise 11 Page 212

Substitute the given points into the standard form of a quadratic function y=ax^2+bx+c to write a system of equations.

y=x^2-6x+3

Practice makes perfect

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
( 3, - 6)	- 6=a( 3)^2+b( 3)+c	9a+3b+c=- 6
( 1, - 2)	- 2=a( 1)^2+b( 1)+c	a+b+c=- 2
( 6, 3)	3=a( 6)^2+b( 6)+c	36a+6b+c=3

We can now write a system of three equations. 9a+3b+c=- 6 & (I) a+b+c=- 2 & (II) 36a+6b+c=3 & (III) Let's solve this system using the Elimination Method. We will start by subtracting Equation (II) from Equation (I) and Equation (III) to eliminate c.

9a+3b+c=- 6 a+b+c=- 2 36a+6b+c=3

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

9a+3b+c-( a+b+c)=- 6-( - 2) a+b+c=- 2 36a+6b+c-( a+b+c)=3-( - 2)

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(I), (III): Simplify

(I), (III): Distribute - 1

9a+3b+c-a-b-c=- 6-(- 2) a+b+c=- 2 36a+6b+c-a-b-c=3-(- 2)

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

9a+3b+c-a-b-c=- 6+2 a+b+c=- 2 36a+6b+c-a-b-c=3+2

(I), (III): Add and subtract terms

8a+2b=- 4 a+b+c=- 2 35a+5b=5

Now, neither Equation (I) 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. First, we need to simplify the equations a bit.

8a+2b=- 4 & (I) a+b+c=- 2 & (II) 35a+5b=5 & (III)

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(I), (III): Simplify

(I): Factor out 2

2(4a+b)=- 4 a+b+c=- 2 35a+5b=5

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

4a+b=- 2 a+b+c=- 2 35a+5b=5

(III): Factor out 5

4a+b=- 2 a+b+c=- 2 5(7a+b)=5

(III): .LHS /5.=.RHS /5.

4a+b=- 2 a+b+c=- 2 7a+b=1

(III): Subtract (I)

4a+b=- 2 a+b+c=- 2 7a+b-( 4a+b)=1-( - 2)

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(III): Solve for a

(III): Distribute - 1

4a+b=- 2 a+b+c=- 2 7a+b-4a-b=1-(- 2)

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

4a+b=- 2 a+b+c=- 2 7a+b-4a-b=1+2

(III): Add and subtract terms

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

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

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

We found our first value, allowing us to write a partial equation. y= 1x^2+bx+c ⇕ y=x^2+bx+c Let's substitute 1 for a in Equation (I) to find the value of b.

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

(I): a= 1

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

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(I): Solve for b

(I): Identity Property of Multiplication

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

(I): LHS-4=RHS-4

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

With our second value, we can continue writing our equation. y=x^2+( - 6)x+c ⇕ y=x^2-6x+c Finally, to find the value of c, we will substitute a=1 and b=- 6 into Equation (II).

b=- 6 & (I) a+b+c=- 2 & (II) a=1 & (III)

(II): a= 1, b= - 6

b=- 6 1+( - 6)+c=- 2 a=1

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(II): Solve for c

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

b=- 6 - 5+c=- 2 a=1

(II): LHS+5=RHS+5

b=- 6 c= 3 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^2-6x+ 3 To help visualize this situation, we can plot the given points and sketch the curve.

Subchapter links

Lesson Check p.212

Practice and Problem-Solving Exercises p.212-214

Standardized Test Prep p.214

Mixed Review p.214