6 Polynomials and Quadratic Applications

8/10/2019 6 Polynomials and Quadratic Applications

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Polynomials and Quadratic Applications

Operations with Polynomials - Adding, Subtracting, and MultiplyingPolynomial Expressions

Polynomials are expressions that include numbers and variables. Thevariables are separated by addition, subtraction, or multiplication.

Polynomials contain more than one term.

Monomial ("mono" means one) has one term:

8x or 2y 2 or 12.

Binomial ("bi" means two) has two terms:

5x 2 + 9 or 5x 2 4y 2

Trinomial ("tri" means three) has three terms:

6x 2 + 7x 2 or 4y 2 + x 2 2

Polynomials are usually written in descending order with the size of theexponents decreasing from left to right.

To write 8x 2 3x 3 8 + 2x indescending order, it must bearranged as shown at the right.

3x 3 + 8x 2 +2x 8


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The addition of polynomials is accomplished by grouping like termsand then simplifying. This process is equivalent to combining thecoefficients of like terms.

In the example at theright, pairs of like termsare grouped.

(5x 3 2x 2 + 7) + (9x 2x 3)

5x 3 (-2x 2 + 9x 2 ) 2x + (7 3)

5x 3 + 7x 2 2x + 4

To add the two polynomials shown below,

(3x 5 4x 4y + 3x 2 y3 y 5) + (2x 4 y + 5x 2y 3 xy 4 y 5 )

1. Group like terms.

3x 5 + ( 4x 4 y + 2x 4 y) + (3x 2 y3 + 5x 2y 3) xy 4 + ( y5 y 5)

2. Combine coefficients of like terms.

3x 5 2x 4 y + 8x 2 y 3 xy 4 2y 5


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The subtraction of polynomials always involves:

1. Changing the signs of all terms of the subtracted polynomials(subtrahend).

2. Completing the problem as the addition of two polynomials.

To subtract the polynomialsshown at the right,

1. Remove the parentheses


3. Add like terms.

(6x 2 5x 2) (2x 2 + 7x 5)

6x 2 5x 2 2x 2 7x + 5

6x 2 2x 2 5x 7x 2 + 5

4x 2 12x + 3


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In multiplying terms with exponents use the rule that(x a)(x b) = x a + b . Thus (x 3 )(x 4) = x 7 and (x)(x 2) = x 3 . In thesecond example note that x = x 1 .

When multiplying a monomial with a polynomial, multiply eachterm of the polynomial by the monomial.

For example,

4x 2(5x 3 2x 2 + 3x + 7)

(4x 2)(5x 3) (4x 2)(2x 2) + (4x 2)(3x) + (4x 2)(7)

20x 5 8x 4 + 12x 3 + 28x 2

3a(5b + 6c 2d)

(3a)(5b) + (3a)(6c) (3a)(2d)

15ab + 18ac 6ad


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To multiply two polynomials, if one or more of them has threeor more terms, use a vertical form of multiplication, beingcareful to put like terms under like terms.

For example, to multiply x2 3x + 5 by 2x

2 + x 4, place one

term under the other and multiply each term of the second byeach term of the first.

x2 3x + 5 2x 2 + x 4

4x2 + 12x 20 -4( x2 3x + 5) x 3 3 x2 + 5x x( x2 3x + 5) 2x 4 6x 3 + 10 x2 2 x2 (x2 3x + 5) 2x 4 5x 3 + 3 x2 + 17x 20


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Operations with Polynomials - Simplifying Fractional Exponents

To simplify the fraction 6 x 5 y 2

8 x 2 y 4, separate fractions for the numerical

coefficients and each of the letter bases are written. Each fraction isthen simplified.

6 x 5 y 2

8 x 2 y 4=

68

x 5

x 2

y 2

y 4=

3 x 3

4 y 2

To write a fraction with only positive exponents, move any base to itsopposite position in the fraction and change the sign of its exponent.

When the base of a negative exponent is a fraction, the exponent ischanged to positive by inverting the fraction.

x 2 y 5

z 3

=y 5 z 3

x 2

In multiplying terms with likebases, the exponents areadded.

In raising a power to a power,multiply the exponents.

In dividing terms with likebases, the exponents aresubtracted.

x 6 x 4 = x

6 + 4 = x 2 =

1

x 2

( x 3 )

5 = x 6 5 = x 15

x 6

x 4= x

6 4 = x 10 =

1

x 10

The following three examples are problems with negative exponentsand their simplification.

x 2 + y 2 =1

x 2+

1

y 2=

y 2

x 2 y 2+

x 2

x 2 y 2=

x 2 + y 2

x 2 y 2

x 3

y 2

1

= x 3

y 2

= x 3 y 2

c + 4d ( )11

c + 4d ( )10= (c + 4 d ) 11 10 = (c + 4d )21 =

1

c + 4d ( )21


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Operations with Polynomials

Work these problems to test your understanding. Click the Nextbutton when you are finished to check your answers.

1. (4x 2y 6) (3y + 10) 2y + 1 =

2. (3x 8) 2 =

3. Factor completely: 4x 3y 6 + 21x 2 y 35xz

4. (3x 7)(3x + 7) =

5. 7(x 2 x + 2) 3(6x 2 7x) 10 =

6.18 x 9

3 x 10 =

7. 6 x 2 y 4

2 x 3 y 2

2

=

8. (ab 3c 4 )( a 4 b

3c 1 ) =


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Solutions to practice problems:

1. (4x 2y 6) (3y + 10) 2y + 1 =

(4x 2y 6) (3y + 10) 2y + 1

4x 2y 6 3y 10 2y + 1

4x 7y 15

2. (3x 8) 2 =

(3x 8) 2

(3x 8)(3x 8)

3x(3x 8) 8(3x 8)

9x 2 24x 24x + 64

9x 2 48x + 64

3. Factor completely: 4x 3y 6 + 21x 2 y 35xz

4x 3 y6 + 21x 2y 35xz

x(4x 2 y6 + 21xy 35z)

4. (3x 7)(3x + 7) =

(3x 7)(3x + 7)

3x(3x + 7) 7(3x + 7)

9x 2 + 21x 21x 49

9x 2 49

5. 7(x 2 x + 2) 3(6x 2 7x) 10 =

7(x 2 x + 2) 3(6x 2 7x) 10

7x 2 7x + 14 18x 2 + 21x 10

11x 2 + 14x + 4

6.18 x 9

3 x 10 =

61

x 9 10 = 61

x 1 = 61

1

x =

6 x

7. 6 x 2 y

4

2 x 3 y 2

2

=3 x 5

y 6

2

=y 12

9 x 10

8. ( ab3c 4 )( a 4 b 3c 1 ) =

= a (14) b (3 3) c (4 1)

= c 3

a 3

= c a

3


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Operations with Polynomials - The FOIL Method

The multiplication of two binomials is usually done using the FOILmethod. F stands for the product of the First terms, O for theproduct of the O uter terms, I for the product of the I nner terms,and L for the product of the Last terms.

The method is often accomplished by drawing curved lines toindicate the terms to be multiplied. The following exampleillustrates this.

= 5x 2x + 5x 3y + ( - 4y) 2x + ( - 4y) 3y

= 10x 2 + 15xy 8xy 12y 2

= 10x 2 + 7xy 12y 2 (Combining like terms)


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Application of the FOIL Method

Note that the FOIL method is simply an organized way ofaccomplishing an underlying principle. To multiply any twobinomials, or any two polynomials for that matter, multiply each

term of one polynomial by each term of the other polynomial.

For the remainder of the binomial examples use the FOIL method.

(2x + 3)(x + 8)

(3x + 5)(3x 5)

(2x + 3) 2

Click Next when you are finished to check your answers.


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Application of the FOIL Method

Note that the FOIL method is simply an organized way ofaccomplishing an underlying principle. To multiply any twobinomials, or any two polynomials for that matter, multiply each

term of one polynomial by each term of the other polynomial.

For the remainder of the binomial examples use the FOIL method.

(2x + 3)(x + 8) = (2x)(x) + (2x)(8) + (3)(x) + (3)(8)

= 2x 2 + 16x + 3x + 24

= 2x 2 + 19x + 24

(3x + 5)(3x 5) = (3x)(3x) + (3x)(-5) + (5)(3x) + (5)(-5)

= 9x 2 15x + 15x 25

= 9x 2 25

(2x + 3) 2 = (2x + 3)(2x + 3) (multiply it by itself)

= (2x)(2x) + (2x)(3) + (3)(2x) + (3)(3)

= 4x 2 + 6x + 6x + 9

= 4x 2 + 12x + 9


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Consider the following example.

A water bottle isthrown straightdownward from a

height of 200 feet. Atthe same time, a brickfalls from a height of100 feet. Thepolynomials representthe heights (in feet) ofthe objects after x seconds.

1. Write a polynomial that represents the distance between thewater bottle and the brick after x seconds.

Water Bottle Brick( 16x 2 40x + 200) ( 16x 2 + 100)

= ( 16x 2 40x + 200) + (16x 2 100)

= ( 16x 2 + 16x 2 ) 40x + (200 100)

= 40x + 100

The polynomial 40x + 100 represents the distance between theobjects after x seconds.

2. What is the distance between the objects after 2 seconds?

Find the value of 40 x + 100 when x = 2.

40x + 100 = 40(2) + 100

= 20

Substitute 2 for x .

After 2 seconds, the distance between the objects is 20 feet.


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1. (3x 8)(x 4) =

2. (x 1)(x + 1) =

3. (w 12

v )2 =

4. The North Carolina Department of Fish and Wildlife iscalculating population growth changes in Northern Cardinalsover the past decade. They have determined the populationgrowth in 2003 to be expressed as (25 x 2 200) while the

population growth in 2013 is expressed as (20 x 2 + 300) . Writean expression for the difference in population growth from 2003to 2013.


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1. (3x 8)(x 4) = (3x)(x) + (3x)( 4) + ( 8)(x) + ( 8)( 4)

= 3x2 12x 8x + 32

= 3x 2 20x + 32

2. (x 1)(x + 1) = (x)(x) + (x)(1) + ( 1)(x) + ( 1)(1)

= x 2 + x x 1

= x 2 1

3. (w 12

v )2 = (w 12

v )( w 12

v )

= (w )( w ) + (w )( 12

v ) + ( 12

v )( w ) + (12

v )( 12

v )

= (w )2 2(w )(12

v ) + (12

v )2

= w 2 wv +14

v 2

4. The North Carolina Department of Fish and Wildlife iscalculating population growth changes in Northern Cardinalsover the past decade. They have determined the populationgrowth in 2003 to be expressed as (25 x 2 200) while the

population growth in 2013 is expressed as (20 x 2 + 300) . Writean expression for the difference in population growth from 2003to 2013.

Solution:

Find the difference. To

subtract the polynomialsshown at the right,

1. Remove the parentheses


3. Add like terms.

(25 x 2 200) (20 x 2 + 300)

25 x 2 200 20 x 2 300

25 x 2 20 x 2 300 200

5 x 2 500


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Factoring Quadratics and Polynomials

Factoring is an operation in which a polynomial is written as theproduct of two or more polynomials. In that sense, it is the inverseoperation for multiplication of polynomials. The most basic type offactoring involves a common monomial factor. If all of the terms of apolynomial have some factor in common, besides one, that factor is a

common monomial factor of the polynomial. One part of completelyfactoring a polynomial is to remove the largest possible commonmonomial factor.

To find the largest common monomial factor of 6x 4 + 8x 3 4x 2 ,

Examine the coefficientsand determine that theyhave a common factor of 2.

6x 4

2x 2= 3x 2

Examine the powers of xand determine that thehighest power common toall the terms is x 2 .

8x 3

2x 2= 4x

4x 2

2x 2= 2

Factor 2x 2 out of each term. 6x 4 + 8x 3 4x 2 = 2x 2(3x 2 + 4x 2)


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Factoring Trinomials by Considering Possible FOIL Combinations

To factor a trinomial into the product oftwo binomials, use a method that makesuse of FOIL. The FOIL multiplication of

(2x + 1) and (x 3) is shown at theright.

(2x + 1)(x 3)

2x 2 5x 3

The FOIL method of multiplying two binomials discloses the followingfactors that should be considered when factoring 2x 2 5x 3.

1. The product of the first two terms (F) in the binomials is 2x 2 , sothe factors of 2x 2 are the only possible choices for the first twoterms in the binomials.The choices are 2x and x.

2. The product of the last two terms (L) in the binomials is -3, so thefactors of -3 are the only possible choices for the last two terms inthe binomials. The choices are 3 and -1 or -3 and 1.

3. The middle term in the trinomial, -5x, is the sum of the productsof the outer and inner terms (O + I), so list all of the variouscombinations until a pair of binomials whose outer and inner termsadd up to -5x are found.

Possible Binomials(2x + 3)(x 1)(2x 1)(x + 3)

(2x 3)(x + 1)(2x + 1)(x 3)

Outer + Inner-2x + 3x = x6x x = 5x

2x 3x = -x-6x + x = -5x

The last pair in the list is the correct one, but the four pairs give somehelp in factoring this type of trinomial. Since the last term of thetrinomial was a negative number, then the last terms in the binomialsmust have opposite signs. The second pair of binomials in the list hasthe sum, 5x, but this is the wrong sign. If that happens, switch thesigns in the two binomials to get the right sum of outer and innerterms.

If none of the possible choices have the correct outer and inner sum,then the trinomial will not factor and it is prime.


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If the last term of the trinomial is positive, then the last terms in thepair of binomials will both need to have the same sign, and those signswill be the same as the sign of the middle term of the trinomial.

This means that for the trinomial 8x 2 10x + 3, the last terms in the

pair of binomials will have to be factors of 3. Furthermore, since themiddle term is -10x, the factors will both have to be negative.Therefore, the only choices for the factors of 3 are -3 and -1. For 8x 2 ,our choices are 4x and 2x, and 8x and x.

The trinomial factors as follows:

8x 2 10x + 3 = (4x 3)(2x 1)


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To factor x 2 + 8x + 16,

1. The choices for the factors of x 2 are only x and x. (x )(x )

2. The choices for the factors of 16both have to be positive since thesign of the middle term ispositive. Possible factors of 16 are16 and 1, 8 and 2, and 4 and 4.

(x + 16)(x + 1)(x + 8)(x + 2)(x + 4)(x + 4)

The correct factorization is (x + 4)(x + 4) or (x + 4) 2 . Sinceboth binomial factors are the same, this type of trinomial is aperfect square trinomial.


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The binomial a 2 b 2 is a difference of two squares and factors as:

a 2 b 2 = (a + b)(a b)

The a and b are the square roots of a 2 and b 2 , respectively.

The factors of a 2 b 2 are the sum (a + b) and difference (a b) of thesquare roots of a 2 and b 2 . Here are a couple of examples.

4x 2 25 = (2x + 5)(2x 5)

9x 4 16y 2 = (3x 2 + 4y)(3x 2 4y)


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To remember the factoring of a sum or difference of two cubes,

1. Note that in each case, the terms in the binomial factor are thecube roots of a 3 and b 3 , and the sign in the binomial factor is thesame as the sign between a 3 and b 3 .

2. Note that the trinomial factor can be obtained using the binomialfactor. In each case the first term in the trinomial factor is thesquare of the first term in the binomial factor, the second term inthe trinomial factor is the product of the terms in the binomialfactor with the sign changed, and the third term in the trinomialfactor is the square of the second term in the binomial factor.

Here are two examples of factoring sums and differences of cubes:

8x 3 + 27 = (2x + 3)[(2x) 2 (2x)(3) + 3 2 ]

= (2x + 3)(4x 2 6x + 9)

y3 125 = (y 5)[y 2 + (y)(5) + 5 2 ]

= (y 5)(y 2 + 5y + 25)


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The Zero Product Property states that if ab = 0, then eithera = 0 or b = 0 (or both). This property is particularly usefulwhen solving quadratic equations.

To solve x 2 2x 15 = 0,

1. Factor the trinomial.

2. Set each factor equal to zero.

3. Solve each equation.

x 2 2x 15 = 0

(x 5)(x + 3) = 0

x 5 = 0 or x + 3 = 0

x = 5 or x = -3

By the zero product property, either x 5 = 0 or x + 3 = 0, whichmeans either x = 5 or x = 3. These are the two solutions of theequation.

To solve 2x 2 7x + 3 = 0,

1. Factor the trinomial.

2. Set each factor equal to zero.

3. Solve each equation.

2x 2 7x + 3 = 0

(2x 1)(x 3) = 0

2x 1 = 0 or x 3 = 0

x =12

or x = 3


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If the left side of a quadratic equation cannot be factored, thenthe quadratic equation is most easily solved using the quadraticformula.

If ax 2 + bx + c = 0, where a, b, and c are real numbers, then:

x =b b 2 4 ac

2 a

To solve 2x 2 x 2 = 0,

1. The polynomial is notfactorable, and thequadratic formula isthe easiest way tosolve the problem.

2x 2 x 2 = 0

2. Note that a = 2, b = -1,and c = -2.

ax 2 + bx + c = 0

3. Replace the letters a, b,and c, in the formula. x =

( 1) ( 1) 2 4(2)( 2)2 2

4. Simplify the expression. x =

1 1 + 164

5. The two solutions are: x =

1 + 17

4 or x =

1 17

4

Notice that the notation in the quadratic formula is ashorthand notation for the two solutions.


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To solve 2x 2 5x + 1 = 0,

1. The polynomial is notfactorable, and thequadratic formula is

the easiest way tosolve the problem.

2x 2 5x + 1 = 0

2. Note that a = 2, b = -5,and c = 1.

ax 2 + bx + c = 0

3. Replace the letters a, b,and c in the formula. x =

( 5) ( 5) 2 4(2)(1)2 2

4. Simplify the expression. x =

5 174

5. The two solutions are: x =

5 + 174

or x =5 17

4


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The most common use of quadratic equations in the real world appliesto ballistics calculations. For example, throwing a ball to a friend,tossing something into a trash can, or jumping into a pool all involvethe quadratic equation.

Consider the following example.

Erin is practicing to make the varsity diveteam. She is on the high dive, bouncesonce and then launches into the air beforelanding in the pool. The high dive is 10meters above the pool surface.

The function h = 5 t 2 + 14 t + 10 givesErin's height h (in meters) after t seconds.How many seconds is Erin in the airbefore she reaches the pool?

Here the first term ( 5 t 2) represents the deceleration of Erin as shefalls back down toward the pool (m/s 2 ), the second term (14 t ) to herrate of speed when travelling upward (m/s), and the third term (10meters) to the height of the diving board. The variable t representstime and h represents height.

If zero is the height of the pool, the problem can be rewritten instandard quadratic equation form.

5 t 2 + 14 t + 10 = 0

The values a = 5, b = 14, and c = 10 can be plugged into thequadratic formula and the equation can be solved for t .

b b 2 4 ac 2 a

= 14 14 2 4( 5)10

2( 5)

= 14 196 + 200

10

= 14 19.9

10

= 0.59 seconds and 3.9 seconds

There are two real number solutions, 0.59 and 3.9 seconds, but onlyone makes sense. Since a human being cannot jump into the air andland in the water in a little over a half a second, the more reasonableanswer is 3.9 seconds.


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1. Find the real number solutions to x 2 + 7x + 4 = 0.

2. Find the real number solutions to 4x 2 8x 5 = 0.

3. Find the real number solutions to 5x 2 + 1 = -6x.

4. Find the real number solutions to x 2 + 8 = 5x.

5. You throw a stone from a height of16 feet with an upward velocity of32 feet per second.

The function h = 16 t 2 + 32 t + 16gives the height h of the stoneafter t seconds. When does thestone hit in the ground?

6. A farmer plants a rectangular

carrot patch in the northeastcorner of a square plot of land.The area of the carrot patch is600 square meters. What is thearea of the square plot of land?


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Solutions to practice problems

1. Find the real number solutions to x 2 + 7x + 4 = 0.

This is a quadratic equation. Some quadratic equation can be

solved by factoring, but since x2 + 7x + 4 is not factorable, this

equation is most easily solved by using the quadratic formula.

x =b b 2 4 ac

2 a, with a = 1, b = 7 and c = 4

x =(7) (7) 2 4(1)(4)

2(1) =

7 49 162

=7 33

2

2. Find the real number solutions to 4x 2 8x 5 = 0.

The equation can be solved using the quadratic formula or byfactoring.

4x 2 8x 5 = 0

(2x + 1)(2x 5) = 0

2x + 1 = 0 2x 5 = 0

x =-12

x =52

3. Find the real number solutions to 5x 2 + 1 = -6x.

The equation can be solved using the quadratic formula or byfactoring. Its solution requires making one side of the equationequal to zero.

5x 2 + 1 = -6x

5x 2 + 6x + 1 = 0

(5x + 1)(x + 1) = 0

5x + 1 = 0 x + 1 = 0

x =-15

x = -1

Add 6x to both sides

Factor the polynomial

Set each factor equal to 0 andsolve both linear equations

4. Find the real number solutions to x 2 + 8 = 5x.

The polynomial cannot be solved by factoring. The quadraticequation will reveal the correct result.

x2 + 8 = 5x is first written as x 2 5x + 8 = 0

x =b b 2 4 ac

2 a , with a = 1, b = -5 and c = 8

x =( 5) ( 5) 2 4(1)(8)

2(1) =

5 25 322

=5 7

2

Because there is no real number that is the square root of -7,there are no real number solutions of the equation.


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5. You throw a stone from a height of

16 feet with an upward velocity of32 feet per second.

The function h = 16 t 2 + 32 t + 16gives the height h of the stoneafter t seconds. When does thestone hit in the ground?

Find the t -values for which h = 0.So, solve 16 t 2 + 32 t + 16 = 0.

Another method of solving quadratic equations is completing thesquare. In this method, a constant c is added to the expression

x 2 + b x so that x 2 + b x + c is a perfect square trinomial. Thisproblem will be solved by completing the square.

16 t 2 + 32 t + 16 = 0

t 2 2 t 1 = 0

t 2 2 t = 1

Write the equation.

Divide each side by 16.

Add 1 to each side.

t 2 2 t + 1 = 1 + 1

( t 1) 2 = 2

Complete the square by adding

22

2

, or 1, to each side.

Factor t 2 2 t + 1.

t 1 = 2 Take the square root ofeach side.

t = 1 2 Add 1 to each side.

The solutions are t = 1 + 2 and t = 1 2 . Use the positive

solution, t = 1 + 2 2.4 .

The stone lands in the water after about 2.4 seconds.


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6. A farmer plants a rectangularcarrot patch in the northeastcorner of a square plot of land.The area of the carrot patch is600 square meters. What is thearea of the square plot of land?

The length of the carrot patch is (x 30) meters and the widthis (x 40) meters. Write and solve an equation for its area.

600 = (x 30)(x 40)

600 = x 2 40x + 1200

0 = x 2 40x + 600

0 = (x 10)(x 60)

x 10 = 0 or x 60 = 0

x = 10 or x = 60

Write an equation.

Multiply.

Subtract 600 from each side.

Factor the polynomial.

Use Zero-Product Property.

Solve for x.

The diagram shows that the side length is at least 30meters, so 10 meters does not make sense in thissituation. The width is 60 meters.

So, the area of the square plot of land is60(60) = 3,600 square meters.


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Relations and Functions

In algebra, a relation is any set, or group, of ordered pairs. Thesymbol for "set" is { }. Any group of numbers is a relation as long asthey are in the form of ordered pairs. A relation can be shown in fourdifferent ways: graph, ordered pairs, mapping diagram, or an input-output table.

Graph: Ordered Pairs:

{(-2, 4), (-1, 1), (0, 0), (2, 4)}

Mapping Diagram: Input-Output Table:

Consider the relation of ordered pairs:

{( 1 , 2 ), ( 3 , 2 ), ( 5 , 7 ), ( 9 , 8 )}

The first elements in the ordered pairs (the x-values ), form thedomain. The second elements in the ordered pairs (the y-values ), formthe range. Only the elements "used" by the relation constitute therange.

The domain is the set { 1 , 3 , 5 , 9 }.

The range is the set { 2 , 7 , 8 }.


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Relations and Functions

Functions are referred to by the notation f(x), which is read " f of x " or"f as a function of x ".

Example:

For the function, f(x) = 5 x + 7, evaluate for f ( 3 ) .

f( 3 ) = 5( 3) + 7 Replace the x -value with 3

f( 3 ) = 15 + 7 Simplify

f( 3 ) = 22

input = 3output = 22

( 3, 22)

Note that f(x) notation can be thought of as another way ofrepresenting the y -value in a function. This is especially true whengraphing. The y -axis may even be labeled as the f(x) axis.

An equation is in function form if it is solved for y .

y = x + 1 x + y = 1in function form not in function form

y = x 2 8 x + 16 x 2 8 x = 16 in function form not in function form

Example:

If the domain of a function represented by 2x + y = 8 is 2, 0, 2, 4,and 6, what is the range of the function?

Write the function infunction form.

2x + y = 8

y = 2x + 8 or f(x) = 2x + 8

Use the function form of the equation to make an input-output table.The range of the function is represented by the output.

Input x

f ( x ) = 2x + 8 Outputf(x) 2 2( 2) + 8 120 2(0) + 8 82 2(2) + 8 44 2(4) + 8 06 2(6) + 8 4

The range is 12, 8, 4, 0, and 4.


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Quadratic Functions Equations and Graphs

Your previous studies have taught you to solve quadratic equations inthe general form ax 2 + bx + c = 0 by factoring the expression on theleft- hand side of this equation to find the equations two roots thevalues of x that satisfy the equation. (Remember that the two roots

might be the same.) You should also be familiar with questionsinvolving quadratic functions in the general form f(x) = ax 2 + bx + c .(Note that a, b, and c are constants and that a is the only essentialconstant.) In quadratic functions, especially where defining a graphon the xy-plane is involved, the variable y is often used to representf(x), and x is often used to represent f(y).

The graph of a quadratic equation of the basic form y = ax 2 or x = ay 2 is a parabola, which is a U-shaped curve. The point atwhich the dependent variable is at its minimum (or maximum) value

is the vertex . In each of the following four graphs, the parabolasvertex lies at the origin (0, 0). Notice that the graphs are constructedby tabulating and plotting several (x, y) pairs, and then connectingthe points with a smooth curve:


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Any equation of the form y = ax 2 + bx + c is a quadratic equation.The graph will be a U-shaped curve, a parabola, opening up if a > 0or opening down if a < 0. To sketch the graph, find the vertex, orturning point, of the graph and a couple of points on each side of thevertex.

To sketch the graph of y = x 2 2x 3,

1. Note that the coefficient of x 2 is 1, so the graph opens up.

2. The x-coordinate of the vertex can be found using the formula:

x =b

2 a

x =( 2)2(1)

=22

= 1

3. The y-coordinate of thevertex is found bysubstituting for x in theequation and calculating y.

y = 1 2 2(1) 3y = 1 2 3y = -4

The vertex is the point (1, -4).

4. To get four more points on

the graph, choose fourvalues for x (two on eachside of the vertex).Substitute them into theequation, and solve for y.

If x = -1, y = (-1) 2 2( 1) 3 y = 1 + 2 3 y = 0 This gives the point (-1, 0).

If x = 0, y = 0 2 2(0) 3 y = 0 0 3 y = -3 This gives the point (0, -3).

If x = 2, y = 2 2 2(2) 3

y = 4 4 3 y = -3 This gives the point (2, -3).

If x = 3, y = 3 2 2(3) 3 y = 9 6 3 y = 0 This gives the point (3, 0).

In the examples at theright, the vertex was(1, -4) and thex-coordinate is 1. Twovalues less than 1 wereused and two valuesgreater than 1 were used.These substitutions for xgenerate four more pointsof the curve ofy = x 2 - 2x 3.

5. Plotting the points on a graph and sketching a smooth curvethrough the points gives this graph.


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The equation y = -2x 2 - 4x + 1 has a parabola which opens down.The process for graphing it is shown below.

1. For the x-coordinate of the

vertex use the formula x =b

2 a

x =( 4)

2( 2)=

4 4

= 1

2. For the y-coordinate of thevertex replace x by -1 in theequation.

y = - 2(-1) 2 - 4(-1) + 1y = - 2 + 4 + 1y = 3

3. Choose values of xon each side of thevertex, (-1,3).

Find four more pointson the curve ofy = -2x 2 - 4x + 1

If x = 3, y = 2( 3) 2 4( 3) + 1 y = -5 Point is ( 3, -5).

If x = 2, y = 2(-2) 2 4( 2) + 1 y = 1 Point is (-2, 1).

If x = 0, y = 2(0) 2 4(0) + 1 y = 1 Point is (0, 1).

If x = 1, y = 2(1) 2 4(1) + 1 y = 5 Point is (1, 5).

4. The graph is sketched as a smooth curve through the vertexand the other four points that were found.


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To graph a quadratic inequality, treat it as an equality and draw theparabola, making the line solid if the inequality has a > or < symbol,or using a dashed line if the inequality has a > or < symbol. Theparabola serves to divide the coordinate system into two regions, oneinside the parabola and the other outside. The solution to the

inequality will include one of these regions.

To determine which region is included, pick an ordered pair thatmakes the inequality (> or


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1. Which of the points could not be a point on the graph of afunction with points ( 4, 2), (2, 3), (6, 1), (8, 4)?

A. ( 6, 3) B. ( 3, 2)

C. ( 4, 1) D. (0, 7)

2. What is the range of this circle?

3. What is the domain of this relation?

{(-4, 2), (2, 4), (6, 9), (7, 8)}

4. What is the domain of the function shown in the graph?(Each grid line represents one unit.)


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4. Use the vertical line test to determine if the following graph is afunction.

5. The function f ( x ) = x 2 + 24 x + 25 is graphed in the xy -plane

below. For what value of x is the value of f ( x ) greatest?


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Solutions to Practice Problems:

1. Which of the points could not be a point on the graph of afunction with points ( 4, 2), (2, 3), (6, 1), (8, 4)?

A. ( 6, 3) B. ( 3, 2)

C. ( 4, 1) D. (0, 7)

C is the correct answer. A function cannot have two points withthe same x -values and different y -values.

2. What is the range of this circle?

The range includes all y -values from 3 to 3,including 3 and 3.

3. What is the domain of this relation?

{(-4, 2), (2, 4), (6, 9), (7, 8)}

The range includes the y -values: {2, 4, 9, 8}

4. What is the domain of the function shown in the graph?(Each grid line represents one unit.)

The domain represents the x -values: {x | x 5}.


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4. Use the vertical line test to determine if the following graph is afunction.

This is a function. A vertical line drawn through the graph willintersect this function in only one location.

5. The function f ( x ) = x 2 + 24 x + 25 is graphed in the xy -plane

below. For what value of x is the value of f ( x ) greatest?

The quadratic expression f ( x ) = x 2 + 24 x + 25 factors

as ( x 1)( x 25). It follows that the graph of

f ( x ) = x 2 + 24 x + 25 intersects the x -axis at x = 1 and

x = 25. The greatest value of f (x) occurs at the vertex,and the x -coordinate of the vertex of the parabola isthe point halfway between 1 and 25 on the x-axis.

This is1 + 25

2= 12 . So the value of x for which f (x) is

greatest is x = 12.

You can also use the formula x =b2 a

to find the

x -coordinate of the vertex: x = b

2 a=

(24)2( 1)

= 12 .

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6 Polynomials and Quadratic Applications