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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 20 Page 213

Can you write the standard form of a quadratic function using the information provided? If not, there is no quadratic model for the given set of values.

A quadratic model does not exist for the given set of values.

Practice makes perfect

We want to determine whether there is a parabola that fits the given set of values.

Given information	Point
f( - 1)= - 4	( - 1, - 4)
f( 1)= - 2	( 1, - 2)
f( 2)= - 1	( 2, - 1)

To use the given points, we need to substitute their ( x, y) coordinate pairs into the standard form of a quadratic function. y=a x^2+b x+c, a≠ 0 Doing so will create a system of equations that we can solve for the values of a, b, and c. Let's start with (- 1,- 4).

y=ax^2+bx+c

x= - 1, y= - 4

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

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Simplify

Calculate power

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

Identity Property of Multiplication

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

a(- b)=- a * b

- 4=a-b+c

Rearrange equation

a-b+c=- 4

We just wrote our first equation! Now let's do the same thing for (1,- 2).

y=ax^2+bx+c

x= 1, y= - 2

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

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Simplify

1^a=1

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

Identity Property of Multiplication

- 2=a+b+c

Rearrange equation

a+b+c=- 2

To find our third and last equation, we will use (2,- 1).

y=ax^2+bx+c

x= 2, y= - 1

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

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Simplify

Calculate power

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

Multiply

- 1=4a+2b+c

Rearrange equation

4a+2b+c=- 1

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

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

(II): Subtract (I)

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

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(II): Solve for b

(II): Distribute - 1

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

(II): - (- a)=a

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

(II): Add and subtract terms

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

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

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

We found that b=1. y=ax^2+1x+c ⇔ y=ax^2+x+c Let's now substitute 1 for b in Equation (I) and Equation (III).

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

(I), (III): b= 1

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

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(I), (III): Simplify

(I): LHS+1=RHS+1

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

(III): Identity Property of Multiplication

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

(III): LHS-2=RHS-2

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

Note that now Equation (I) and Equation (III) form a system of two equations with two variables. We will solve it using the Elimination Method again. Let's subtract Equation (I) from Equation (III).

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

(III): Subtract (I)

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

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(III): Solve for a

(III): Distribute - 1

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

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

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

(III): Add and subtract terms

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

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

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

We found that a=0. y=0x^2+1x+c ⇔ y=x+c Finally, to find the value of c, we will substitute 0 for a in Equation (I).

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

(I): a= 0

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

(I): Identity Property of Addition

c=- 3 b=1 a=0

Now that we have a=0, b=1, and c=- 3, we can write the function. The standard form of a quadratic function is y=ax^2+bx+c, with a≠ 0. In our equation, we have that a=0 and thus it is not a parabola. y=0x^2+1x+(- 3) ⇔ y=x-3 If we pay close attention, we can see that the above equation represents a line, not a parabola.

Subchapter links

Lesson Check p.212

Practice and Problem-Solving Exercises p.212-214

Standardized Test Prep p.214

Mixed Review p.214