1. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Factoring CHAPTER...

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

FactoringCHAPTER

6.1 Greatest Common Factor and Factoring by Grouping

6.2 Factoring Trinomials of the Form x2 + bx + c6.3 Factoring Trinomials of the Form ax2 + bx + c,

where a 16.4 Factoring Special Products6.5 Strategies for Factoring6.6 Solving Quadratic Equations by Factoring6.7 Graphs of Quadratic Equations and Functions

66


Greatest Common Factor and Factoring by Grouping6.16.1

1. List all possible factors for a given number.2. Find the greatest common factor of a set of numbers or

monomials.3. Write a polynomial as a product of a monomial GCF and

a polynomial.4. Factor by grouping.

4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Factored form: A number or expression written as a product of factors.

Following are some examples of factored form:An integer written in factored form with integer factors:

28 = 2 • 14A monomial written in factored form with monomial factors: 8x5 = 4x2 • 2x3

A polynomial written in factored form with a monomial factor and a polynomial factor: 2x + 8 = 2(x + 4)

A polynomial written in factored form with two polynomial factors: x2 + 5x + 6 = (x + 2)(x + 3)


Objective 1

List all possible factors for a given number.


Example

List all natural number factors of 36.Solution:

To list all the natural number factors, we can divide 36 by 1, 2, 3, and so on, writing each divisor and quotient pair as a product until we have all possible combinations.

1 • 36 2 • 18 3 • 124 • 96 • 6

The natural number factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.


Objective 2

Find the greatest common factor of a set of numbers or monomials.


Greatest common factor (GCF): The largest natural number that divides all given numbers with no remainder.

Listing Method for Finding GCF To find the GCF of a set of numbers by listing: 1. List all possible factors for each given number. 2. Search the lists for the largest factor common to all lists.


Example

Find the GCF of 48 and 54.Solution:

Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54

The GCF of 48 and 54 is 6.


Prime Factorization Method for Finding GCFTo find the greatest common factor of a given set of numbers:

1. Write the prime factorization of each given number in exponential form.

2. Create a factorization for the GCF that includes only those prime factors common to all the factorizations, each raised to its smallest exponent in the factorization.

3. Multiply the factors in the factorization created in Step 2.Note: If there are no common prime factors, then the GCF is 1.


Example

Find the GCF of 45a3b and 30a2.Solution:

Write the prime factorization of each monomial, treating the variables like prime factors. 45a3b = 32 • 5 • a3 • b 30a2 = 2 • 3 • 5 • a2

The common prime factors are 3, 5, and a.

GCF = 3 • 5 • a2 = 15a2


Objective 3

Write a polynomial as a product of a monomial GCF and a polynomial.


Factoring a Monomial GCF Out of a PolynomialTo factor a monomial GCF out of a given polynomial:

1. Find the GCF of the terms that make up the polynomial.

2. Rewrite the given polynomial as a product of the GCF and parentheses that contain the result of dividing the given polynomial by the GCF.

Given polynomial = GCF Given polynomial

GCF


Example

Factor. 2 3 4 29 15 18 x yz x y x y

Solution 1. Find the GCF of

Because the first term in the polynomial is negative, we will factor out the negative of the GCF to avoid a negative first term inside the parentheses. We will factor out 3x2 y.

2 3 4 29 , 15 , and 18 .x yz x y x y


continued

2. Write the given polynomial as the product of the GCF and the parentheses containing the quotient of the given polynomial and the GCF.

2 3 4 29 15 18 x yz x y x y2 3 4 2

22

9 15 183

3

x yz x y x yx y

x y

2 3 4 22

2 2 2

9 15 183

3 3 3

x yz x y x yx y

x y x y x y

2 23 3 5 6x y z x x y


Example

Factor. 5 8 5 a b b

Solution: Notice that this expression is a sum of two products, a and (b + 5), and 8 and (b + 5). Further, note that (b + 5) is the GCF of the two products.

5 8 55

5

a b bb

b 5 8 5 a b b

5 8 55

5 5

a b bb

b b

5 8 b a


Objective 4

Factor by grouping.


Factoring by GroupingTo factor a four-term polynomial by grouping:

1. Factor out any monomial GCF (other than 1) that is common to all four terms.

2. Group together pairs of terms and factor the GCF out of each pair.

3. If there is a common binomial factor, then factor it out.

4. If there is no common binomial factor, then interchange the middle two terms and repeat the process. If there is still no common binomial factor, then the polynomial cannot be factored by grouping.


Example

Factor. 3 28 24 3 9 p p pq q

Solution: First we look for a monomial GCF (other than 1). This polynomial does not have one. Because the polynomial has four terms, we now try to factor by grouping.

3 28 24 3 9 p p pq q 3 28 24 3 9 p p pq q

28 3 3 3 p p q p

23 8 3 p p q


Factor by factoring out the GCF.

a)

b)

c)

d)

4 2 2 356 32 72 x y xy x y

3 24 14 8 18 xy x y y x

24 14 8 18 xy x y y xy

2 28 7 4 9 xy x y xy x

3 28 7 4 9 xy x y y x


Factor by grouping.

a)

b)

c)

d)

2 3 7 21 b bc b c

7 3 b c

3 7 b c b

3 7 c b

3 7 b c


Factoring Trinomials of the Form x2 + bx + c6.26.2

1. Factor trinomials of the form x2 + bx + c.2. Factor out a monomial GCF, then factor the trinomial of

the form x2 + bx + c.


Objective 1

Factor trinomials of the form x2 + bx + c.


Following are some examples of trinomials of the form x2 + bx + c.

x2 + 5x + 6 or x2 –7x + 12 or x2 – 5x – 24

Products in the form x2 + bx + c are the result of the product of two binomials.

When we factor a trinomial of the form x2 + bx + c, we reverse the FOIL process, using the fact that b is the sum of the last terms in the binomials and c is the product of the last terms in the binomials.


Factoring x2 + bx + cTo factor a trinomial of the form x2 + bx + c :

1. Find two numbers with a product equal to c and a sum equal to b.

2. The factored trinomial will have the form:(x + first number) (x + second number).

Note: The signs in the binomial factors can be minus signs, depending on the signs of b and c.


Example

Factor. x2 – 6x + 8

Solution: We must find a pair of numbers whose product is 8 and whose sum is –6. If two numbers have a positive product and negative sum, they must both be negative. Following is a table listing the products and sums:

Product Sum

(–1)(–8) = 8 –1 + (–8) = –9

(–2)(–4) = 8 –2 + (–4) = –6

This is the correct combination.


continued

Answer

Check We can check by multiplying the binomial factors to see if their product is the original polynomial.

x2 – 6x + 8 = (x – 2)(x – 4)

(x – 2)(x – 4) = x2 – 4x – 2x + 8

= x2 – 6x + 8

Multiply the factors using FOIL.

The product is the original polynomial.


Example

Factor. a2 – ab – 20b2

Solution: We must find a pair of terms whose product is 20b2 and whose sum is –1b. These terms would have to be –5b and 4b.

Answer a2 – ab – 20b2 = (a – 5b)(a + 4b)


Objective 2

Factor out a monomial GCF, then factor the trinomial of the form x2 + bx + c.


Example

Factor. 4xy3 + 12xy2 – 72xy

Solution First, we look for a monomial GCF (other than 1). Notice that the GCF of the terms is 4xy.

Factoring out the monomial, we have 4xy3 + 12xy2 – 72xy = 4xy(y2 + 3y – 18)

Now try to factor the trinomial to two binomials. We must find a pair of numbers whose product is –18 and whose sum is 3.


continued

Answer

Product Sum

(–1)(18) = –18 –1 + 18 = 17

(–2)(9) = – 18 –2 + 9 = 7

(–3)(6) = – 18 –3 + 6 = 3 This is the correct combination.

4xy3 + 12xy2 – 72xy = 4xy(y – 3)(y + 6)


Factor. x2 + 5x – 36

a) (x + 3)(x – 12)

b) (x – 3)(x + 12)

c) (x + 9)(x – 4)

d) (x – 9)(x + 4)


Factor completely. 5rs3 – 10rs2 – 40rs

a) 5rs(s2 – 2s – 8)

b) 5rs(s2 + 2s – 8)

c) 5rs(s + 2)(s – 4)

d) 5rs(s – 2)(s + 4)


Factoring Trinomials of the Form ax2 + bx + c, where a 16.36.3

1. Factor trinomials of the form ax2 + bx + c, where a 1, by trial.

2. Factor trinomials of the form ax2 + bx + c, where a 1, by grouping.


Objective 1

Factor trinomials of the form ax2 + bx + c, where a 1, by trial.


We will focus on factoring trinomials in which the coefficient of the squared term is other than 1, such as the following:

3x2 + 17x + 10 8x2 + 29x – 12

In general, like trinomials of the form x2 + bx + c, trinomials of the form ax2 + bx + c, where a 1, also have two binomial factors.


Factoring by Trial and ErrorTo factor a trinomial of the form ax2 + bx + c, where a 1, by trial and error:

1. Look for a monomial GCF in all the terms. If there is one, factor it out.

2. Write a pair of first terms whose product is ax2.

ax2


3. Write a pair of last terms whose product is c.

4. Verify that the sum of the inner and outer products is bx (the middle term of the trinomial).

c

+ Outerbx

Inner


If the sum of the inner and outer products is not bx, then try the following:

a. Exchange the first terms of the binomials from step 3, then repeat step 4.

b. Exchange the last terms of the binomials from step 3, then repeat step 4.

c. Repeat steps 2 – 4 with a different combination of first and last terms.


Example

Factor. 26 13 5x x

Solution

26 13 5x x +

The first terms must multiply to equal 6x2. These could be x and 6x, or 2x and 3x.

The last terms must multiply to equal –5. Because –5 is negative, the last terms in the binomials must have different signs. This factor pair must be 1 and 5.


continued

2 25 6 1 6 30 5 6 29 5x x x x x x x

Now we multiply binomials with various combinations of these first and last terms until we find a combination whose inner and outer products combine to equal 13x.

2 21 6 5 6 5 6 5 6 5x x x x x x x 2 23 1 2 5 6 15 2 5 6 13 5x x x x x x x 2 23 5 2 1 6 3 10 5 6 7 5x x x x x x x 2 22 1 3 5 6 10 3 5 6 7 5x x x x x x x 2 22 5 3 1 6 2 15 5 6 13 5x x x x x x x Correct

combination.

Incorrect combinations.

Answer 26 13 5 2 5 3 1x x x x


Example

Factor. 3 221 60 9x x x Solution First, we factor out the monomial GCF, 3x.

3 + x

The last terms must multiply to equal 3. Because 3 is a prime number, its factors are 1 and 3.

3 221 60 9x x x 23 7 20 3x x x

Now we factor the trinomial within the parentheses.

23 7 20 3x x x

The first terms must multiply to equal 7x2. These could be x and 7x.


continued

2 23 1 7 3 3 7 3 7 3 3 7 4 3x x x x x x x x x x

Now we multiply binomials with various combinations of these first and last terms until we find a combination whose inner and outer products combine to equal –20x.

Correct combination.

Answer

2 23 1 7 3 3 7 3 7 3 3 7 4 3x x x x x x x x x x

2 23 3 7 1 3 7 21 3 3 7 20 3x x x x x x x x x x

2 23 3 7 1 3 7 21 3 3 7 20 3x x x x x x x x x x

3 221 60 9 3 3 7 1x x x x x x


Objective 2

Factor trinomials of the form ax2 + bx + c, where a 1, by grouping.


Factoring ax2 + bx + c, where a 1, by GroupingTo factor a trinomial of the form ax2 + bx + c, where a 1, by grouping:

1. Look for a monomial GCF in all the terms. If there is one, factor it out.

2. Multiply a and c.3. Find two factors of this product whose sum is b.4. Write a four-term polynomial in which bx is written

as the sum of two like terms whose coefficients are the two factors you found in step 3.

5. Factor by grouping.


Example

Factor. 22 15 7x x Solution Notice that for this trinomial, a = 2, b = –15, and c = 7. We begin my multiplying a and c: (2)(7) = 14.

Now we find two factors of 14 whose sum is –15. Notice that these two factors must both be negative.

Factors of ac Sum of Factors of ac

(–2)(–7) = 14 –2 + (–7) = –9

(–1)(–14) = 14 –1 + (– 14) = –15 Correct


continued

22 7x

2 1 7 2 1x x x

22 7x –15x –x – 14x

2 1 7x x


Factor completely. 6x2 –33x – 63

a) 3(2x + 7)(x – 3)

b) 3(2x + 3)(x – 7)

c) 3(2x – 3)(x + 7)

d) 3(2x – 7)(x + 3)


Factor completely. 2x2 +3x – 20

a) (2x + 2)(x – 10)

b) (2x + 4)(x – 5)

c) (2x – 5)(x + 4)

d) (2x – 10)(x + 2)


Factoring Special Products6.46.4

1. Factor perfect square trinomials.2. Factor a difference of squares.3. Factor a difference of cubes.4. Factor a sum of cubes.


Objective 1

Factor perfect square trinomials.


Factoring Perfect Square Trinomials

a2 + 2ab + b2 = (a + b)2

a2 – 2ab + b2 = (a – b)2


Example

Factor. 9a2 + 6a + 1

Solution This trinomial is a perfect square because the first and the last terms are perfect squares and twice the product of their roots is the middle term.

9a2 + 6a + 1The square root of 9a2 is 3a. The square root of 1 is 1.

Twice the product of 3a and 1 is (2)(3a)(1) = 6a, which is the middle term.

Answer 9a2 + 6a + 1 = (3a + 1)2


Example

Factor. 16x2 – 56x + 49

Solution This trinomial is a perfect square.

The square root of 16x2 is 4x. The square root of 49 is 7.

Twice the product of 4x and 7 is (2)(4x)(7) = 56x, which is the middle term.

Answer 16x2 – 56x + 49 = (4x – 7)2

16x2 – 56x + 49

Use a2 – 2ab + b2 = (a – b)2, where a = 4x and b = 7.


Objective 2

Factor a difference of squares.


Factoring a Difference of Squares

a2 – b2 = (a + b)(a – b)

Warning: A sum of squares a2 + b2 is prime and cannot be factored.


Example

Factor. 9x2 – 16y2

Solution This binomial is a difference of squares because 9x2 – 16y2 = (3x)2 – (4y)2 . To factor it, we use the rule a2 – b2 = (a + b)(a – b).

a2 – b2 = (a + b)(a – b)

9x2 – 16y2 = (3x)2 – (4y)2 = (3x + 4y)(3x – 4y)


Example

Factor. n4 – 625

Solution This binomial is a difference of squares, wherea = n2 and b = 25.

(n2 + 25)(n2 – 25) n4 – 625 = Use a2 – b2 = (a + b)(a – b).

Factor n2 – 25, using a2 – b2 = (a + b)(a – b) with a = n and b = 5.

= (n2 + 25)(n + 5)(n – 5)


Objective 3

Factor a difference of cubes.


Factoring a Difference of Cubes

a3 – b3 = (a – b)(a2 + ab + b2)


Example

Factor. 216x3 – 64

Solution This binomial is a difference of cubes.

a3 – b3 = (a – b) (a2 + a b + b2)

216x3 – 64 = (6x)3 – (4)3 = (6x – 4)((6x)2 + (6x)(4) + (4)2)

= (6x – 4)(36x2 + 24x + 16)

Note: The trinomial may seem like a perfect square. However, to be a perfect square, the middle term should be 2ab. In this trinomial, we only have ab, so it cannot be factored.


Objective 4

Factor a sum of cubes.


Factoring a Sum of Cubes

a3 + b3 = (a + b)(a2 – ab + b2)


Example

Factor. 6x +162xy3

Solution The terms in this binomial have a monomial GCF, 6x.

= 6x(1 + 27y3)6x +162xy3

= 6x(1 + 3y)((1)2 – (1)(3y) + (3y)2)

= 6x(1 + 3y)(1 – 3y + 9y2)


Factor completely. 4a2 – 20a + 25

a) (2a + 5)2

b) (2a – 5)2

c) (4a + 5)2

d) (4a – 5)2


Factor completely. 9x2 – 49

a) (3x + 5)2

b) (3x + 7)(3x – 7)

c) (3x – 7)2

d) (7x + 3)(7x – 3)


Factor completely. 2n2 + 24n + 72

a) 2(n + 6)2

b) 2(n + 6)(n – 6)

c) 2(n – 6)2

d) (2n + 6)(2n – 6)


Strategies for Factoring6.56.5

1. Factor polynomials.


Objective 1

Factor polynomials.


Factoring a PolynomialTo factor a polynomial, first factor out any monomial GCF, then consider the number of terms in the polynomial. If the polynomial has:

I. Four terms, then try to factor by groupingII. Three terms, then determine if the trinomial is a perfect square or

not.A. If the trinomial is a perfect square, then consider its form.

1. If in the form a2 + 2ab + b2, then the factored form is (a + b)2.

2. If in the form a2 2ab + b2, then the factored form is (a b)2.B. If the trinomial is not a perfect square, then consider its form.

1. If in the form x2 + bx + c, then find two factors of c whose sum is b, and write the factored form as (x + first number)(x + second number).


Factoring a Polynomial continued

2. If in the form ax2 + bx + c, where a 1, then use trial and error. Or, find two factors of ac whose sum is b; write these factors as coefficients of two like terms that, when combined, equal bx; and then factor by grouping.

III. Two terms, then determine if the binomial is a difference of squares, sum of cubes, or difference of cubes.A. If given a binomial that is a difference of squares, a2 – b2,

then the factors are conjugates and the factored form is (a + b)(a – b). Note that a sum of squares cannot be

factored.B. If given a binomial that is a sum of cubes, a3 + b3, then the

factored form is (a + b)(a2 – ab + b2).C. If given a binomial that is a difference of cubes, a3 – b3, then

the factored form is (a – b)(a2 + ab + b2).

Note: Always look to see if any of the factors can be factored.


ExampleFactor. 12x2 – 8x – 15 Solution

There is no GCF.Not a perfect square, since the first and last terms are not perfect squares.Use trial and error or grouping.

(x – 3)(12x + 5) = 12x2 + 5x – 36x – 15 No(6x – 3)(2x + 3) = 12x2 + 18x – 6x – 9 No(6x + 5)(2x – 3) = 12x2 – 18x + 10x – 15

12x2 – 8x – 15 Correct


ExampleFactor. 5x3 – 10x2 – 120xSolution

5x(x2 – 2x – 24) Factored out the monomial GCF, 5x.

Look for two numbers whose product is –24 and whose sum is 2.

5x(x + 4)(x – 6)

Product Sum

(1)(24) = 24 1 + 24 = 23

4(6) = 24 4 + (6) = 2 Correct combination.


Example

Factor. 8a4 – 72n2

Solution 8a4 – 72n2 = 8(a4 – 9n2) Factor out the monomial GCF, 8.

a4 – 9n2 is a difference of squares

= 8(a2 – 3n)(a2 + 3n)


Example

Factor. 12y5 + 84y3

Solution

12y3(y2 + 7) Factor out the monomial GCF, 12y3.


Example

Factor. 150x3y – 120x2y2 + 24xy3

Solution

6xy(25x2 – 20xy + 4y2)6xy(5x – 2y)2

Factor out the monomial GCF, 6xy.

Factor the perfect square trinomial.


Example

Factor. x5 – 2x3 – 27x2 + 54SolutionNo common monomial, factor by grouping.

(x5 – 2x3)(– 27x2 + 54)x3(x2 – 2)–27(x2 – 2)(x2 – 2)(x3 – 27) Difference of cubes

(x2 – 2)(x – 3)(x2 + 3x + 9)


Factor. 6x2 + 17x + 5

a) (6x + 1)(x + 5)

b) (3x + 1)(2x + 5)

c) (6x + 1)(x – 5)

d) (3x – 1)(2x – 5)


Factor. 7y4 + 49y2

a) 7y(y3 + 7y)

b) 7y2(y2 + 49)

c) y2(7y2 + 49)

d) 7y2(y2 + 7)


Solving Quadratic Equations by Factoring6.66.6

1. Use the zero-factor theorem to solve equations containing expressions in factored form.

2. Solve quadratic equations by factoring.3. Solve problems involving quadratic equations.4. Use the Pythagorean theorem to solve problems.


Objective 1

Use the zero-factor theorem to solve equations containing expressions in factored form.


Zero-Factor Theorem

If a and b are real numbers and ab = 0, then a = 0 or b = 0.


ExampleSolve. (x + 4)(x + 5) = 0SolutionAccording to the zero-factor theorem, one of the two factors, or both factors, must equal 0.

x + 4 = 0 or x + 5 = 0 Solve each equation.

x = 4 x = 5CheckFor x = 4: For x = 5: (x + 4)(x + 5) = 0 (x + 4)(x + 5) = 0 (4 + 4)(4 + 1) = 0 (5 + 4)(5 + 5) = 0

0(3) = 0 (1)(0) = 0


Solving Equations with Two or More Factors Equal to 0To solve an equation in which two or more factors are equal to 0, use the zero-factor theorem:1. Set each factor equal to zero.2. Solve each of those equations.


Example

Solve.a. y(5y + 2) = 0 b. x(x + 2)(5x – 4) = 0Solutiona. y(5y + 2) = 0

y = 0 or 5y + 2 = 0 5y = 2

2

5y

This equation is already solved.

b. ( 2)(5 4) 0 x x x

0 2 0 5 4 0 x x x2x 5 4x

4

5x

To check, we verify that the solutions satisfy the original equations.


Objective 2

Solve quadratic equations by factoring.


Quadratic equation in one variable: An equation that can be written in the form ax2 + bx + c = 0, where a, b, and c are all real numbers and a 0.


Solving Quadratic Equations Using FactoringTo solve a quadratic equation:1. Write the equation in standard form (ax2 + bx + c = 0).2. Write the variable expression in factored form. 3. Use the zero-factor theorem to solve.


Example

Solve. 2x2 – 5x – 3 = 0 SolutionThe equation is in standard form, so we can factor.

2x2 – 5x – 3 = 0 (2x + 1)(x – 3) = 0 Use the zero-factor theorem to

solve.

2x + 1 = 0 or x – 3 = 0

2 1x 3x1

2x To check, we verify that the

solutions satisfy the original equations.


Example

Solve. 6y2 + 11y = 10 + 4ySolutionWrite the equation in standard form.

6y2 + 11y = 10 + 4y 6y2 + 7y = 10 Subtract 4y from both sides.

6y2 + 7y – 10 = 0 Subtract 10 from both sides.

(6y – 5)(y + 2) = 0 Factor.

6y – 5 = 0 or y + 2 = 0 Use the zero-factor theorem.

6 5y5

6y

2y


Objective 3

Solve problems involving quadratic equations.


Example

The product of two consecutive odd natural numbers is 323. Find the numbers.

Understand Odd numbers are 1, 3, 5,… Let x = the first odd number Let x + 2 = consecutive odd number

The word product means that two numbers are multiplied to equal 323.Plan Translate to an equation, then solve.


continuedExecute x(x + 2) = 323

x(x + 2) – 323 = 0 x2 + 2x – 323 = 0(x + 19)(x – 17) = 0

x + 19 = 0 x – 17 = 0x = –19 x = 17

Answer Because –19 is not a natural number and 17 is, the first number is 17. This means that the consecutive odd natural number is 19.

Check 17 and 19 are consecutive odd natural numbers and their product is 323.


Objective 4

Use the Pythagorean theorem to solve problems.


The Pythagorean TheoremGiven a right triangle, where a and b represent the lengths of the legs and c represents the length of the hypotenuse, then a2 + b2 = c2.

c (hypotenuse)

b (leg)

a (leg)


ExampleFind the length of the missing side.SolutionUse the Pythagorean theorem, a2 + b2 = c2

152 + 362 = c2 Substitute.

225 + 1296 = c2 Simplify exponential forms.

1521 = c2 Add.

c2 – 1521 = 0 Standard form.

(c – 39)(c + 39) = 0 Factor.

c – 39 = 0 or c + 39 = 0 c = 39 or c = –39 Only the positive solution is sensible.

?

36

15


Solve. x2 = 6x – 8

a) 2 and 4

b) 2 and 4

c) 2 and 4

d) 1 and 8


Solve. One natural number is four times another. The product of the two numbers is 900. Find the larger number.

a) 15

b) 30

c) 35

d) 60


Find the length of the hypotenuse.

a) 15

b) 46

c) 50

d) 62

?

48

14


Graphs of Quadratic Equations and Functions6.76.7

1. Graph quadratic equations in the form y = ax2 + bx + c.2. Graph quadratic functions.


Objective 1

Graph quadratic equations in the form y = ax2 + bx + c.


Quadratic equation in two variables: An equation that can be written in the form y = ax2 + bx + c, where a, b, and c are real numbers and a 0.

Axis of symmetry: A line that divides a graph into two symmetrical halves.

Vertex: The lowest point on a parabola that opens up or the highest point on a parabola that opens down.

vertex

(0, 0)

axis of symmetry

x = 0 (y-axis)


Graphing Quadratic EquationsTo graph a quadratic equation:1. Find ordered pair solutions and plot them in the coordinate plane. Continue finding and plotting solutions until the shape of the parabola can be clearly seen.2. Connect the points to form a parabola.


ExampleGraph. y = 2x2 + 1

SolutionComplete a table of solutions.

x y

2 9

1 3

0 1

1 3

2 9

Plot the points.

Connect the points.


ExampleGraph. y = 3x2 + 4


x y

2 8

1 1

0 4

1 1

2 8

Plot the points.

Connect the points.


Opening of a Parabola

Given an equation in the form y = ax2 + bx + c, if a > 0, then the parabola opens upward; if a < 0, then the parabola opens downward.


Objective 2

Graph quadratic functions.


Graphing Quadratic FunctionsTo graph a quadratic function:1. Find enough ordered pairs by evaluating the function for various values of x so that when those ordered pairs are plotted, the shape of the

parabola can be clearly seen. 2. Connect the points to form the parabola.


ExampleGraph. f(x) = 2x2 + 8x – 1


x y

1 11

0 1

1 5

2 7

3 5

4 1

Plot the points.

Connect the points.

This parabola opens downward

since a < 0.


Graph. y = x2 – 2

a) b)

c) d)


Graph. f(x) = x2 + 2x – 2

a) b)

c) d)

1. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Factoring CHAPTER...

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