NCTM Standards: 2 & 6

Appreciation

The increase of value of an item over a period of time.

The formula for compound interest can be used to find the value after appreciation. trPA )1(

Where P is the original amount invested

r is the interest rate or rate of return,

t is the time invested (in years)

A is the value after appreciation

Example 1

tPxA

10500x 5250x181000x)(xT$1000 invested for 18 years at an unknown interest rate

$500 invested for 10 years at an unknown interest rate

$1000 invested for 5 years at an unknown interest rate

The total value of the college fund

To simplify things, let x = 1+r

trPA )1(

= + +

)(xT = + +10500x 5250x181000x

Remember: x = 1+r

So the actual value to evaluate is 1.1225 )(xT = + + 101225.1500 51225.1250 181225.11000

33.038,10)( xT

The total value of the college fund is $10,038.33 after 18 years.

How much was the total initial investment?This expression is called a

polynomial in one variable.

The degree of a polynomial in one variable is the greatest exponent of its variableS.

)(xT = + +10500x 5250x181000x

Zeros of a function: the values for x for which f(x) = 0. (These are also the x-intercepts when the function is graphed.)

Leading coefficient: the coefficient of the variable with the greatest exponent.

Example 2

a.The value of the highest exponent is:

The leading coefficient is:

3 (degree)

1

b. Evaluate at x = 4 to see if the value of the function is zero:

0

8410464

8106423

23

xxxf

Since the resulting value of the function is zero, 4 is a zero of f(x). Meaning the graph will cross or touch the x-axis at 4.

Polynomial Equation

The term for the result of replacing f(x) with zero.

81060

810623

23

xxx

xxxxf A polynomial function that can be graphed

A polynomial function that can be solved.Root: a solution for a polynomial

equation. (this term is interchangeable with the term zero)

A root can be a real number or an imaginary number such as 3i.

The x-intercepts represents the real solutions; imaginary solutions cannot be determined with a graph; you can determine how many imaginary solutions there are, but not what they are.

Example 3

0442 ixixx

16

)1(16

1644

44

2

2

22

x

x

ixixix

ixix

Work backwards:If the roots are 2, 4i, & -4i there must have been a product of three linear factors that were equal to zero.

Distribute the products & simplify. What is the special name for the last two factors?

What is true about them?

conjugates

The imaginary part will always cancel out.

032162

16223

2

xxx

xx

This is the equation with the least degree with the given roots.

b. The equation is an odd degree; it crosses the x-axis one time.

Example 4

The degree of the equation indicates how many complex roots (or solutions) an equation has.

All 4 roots are real

2 real roots and 2 imaginary roots

All 4 roots are imaginarySolve by factoring:

There are 4 complex roots. The possibilities of these root are:

Example 4

3649 04359 24 xx041369 224 xxx

Solve by factoring: Multiply the leading coefficient

and the constant

List all the factors of 36:

1 36

2 18

3 13

4 9

6 6

Look for a set of factors that will add or subtract to obtain

the middle term (- 35 )

Rewrite the original problem as four parts

0194

0414922

222

xx

xxx

( ) ( )

019 04 22 xx

42 x2,2 xx

19 2 x

9

12 x

3

ix

2,2 xx

3

ix

04359 24 xx

These are the imaginary solutions; we can only guess where they are

located.

112232160

2232161142

2

tt

tt

a

acbbx

2

42

a = - 16

b = 232

c = -112

162

112164232232 2

x

32

216232

32

46656232

x

1432

216232

.5 32

216232

x

x

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