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HAWKES LEARNING SYSTEMS

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Copyright © 2010 Hawkes Learning Systems. All rights reserved.

Hawkes Learning Systems:Intermediate Algebra

Section 7.1a: Quadratic Equations: The Square Root Method




Objectives

o Solve quadratic equations by factoring

o Solve quadratic equations by using the definition of a square root.




Solving Quadratic Equations by Factoring

When the solutions of a polynomial equation can be found by factoring, the method depends on the zero-factor property.

Zero-Factor PropertyIf the product of two factors is 0, then one or both of the factors must be 0. For factors a and b,

If 0, then 0 or 0 or both.a b a b

For example,

1 2 0x x 1 0x 2 0x





An equation that can be written in the form

2 0ax bx c

Where a, b, and c are real numbers and is called a quadratic equation.

0a




To Solve an Equation by Factoring

1. Add or subtract terms so that one side of the equation is 0.

2. Factor the polynomial expression.3. Set each factor equal to 0 and solve each of

the resulting equations. Note: If two of the factors are the same, then the

d solution is said to be a double root or a root d of multiplicity two.




Example 1: Solving Quadratic Equations by Factoring

Solve the following quadratic equation by factoring. 2 3 18x x

Subtract 18 from both sides. One side must be zero.

2 3 18 0x x

3 6 0x x Factor the left-hand side.

3 0x 6 0x

3x 6x

Set each factor equal to 0.

Solve each linear equation.

or




Example 2: Solving Quadratic Equations by Factoring

Solve the following quadratic equation by factoring.22 16 32x x

22 16 32 0x x

22 8 16 0x x

22 4 0x

4 0x

4x

The solution is a double root.





o Quadratic equations may have nonreal complex solutions.

o In particular, the sum of two squares can be factored as the product of complex conjugates.

o For example,

2 81 9 9 .x x i x i




Example 3: Quadratic Equations Involving the Sum of Two Squares

Solve the following equation by factoring.2 49 0x

7 7 0x i x i

7 0x i 7 0x i 7x i 7x i

Check: 2

97 4 0i 249 49 0i

49 49 0

0 0

297 4 0i

249 49 0i

49 49 0

0 0




Square Root Property

v 2x c .x c

2x a c x a c x a c If, then or , or .

If , then

Note: If c is negative, then the solutions will be nonreal.




Using the Definition of Square Root and the Square Root Property

Consider the equation2 8x

Allowing that the variable, x, might be positive or negative, we use the definition of square root, 2 .x x

Taking the square root of both sides of the equation gives:

8x

So we have two solutions,

8 and 8, or 8.x x x




Using the Definition of Square Root and the Square Root Property

Similarly, for the equation 2

4 3x

the definition of square root gives3.4x

This leads to the two equations and the two solutions, as follows:

4 3x 4 3x

4 3x 4 3x

4 3x




Example 4: Using the Square Root Property

Solve the following quadratic equations by using the Square Root Property.

25 3x

5 3x

5 3x

a.

b. 2 16y

16y

4y i





Section 7.1b: Quadratic Equations: Completing the Square




Objectives

o Solve quadratic equations by completing the square

o Find polynomials with given roots.




Completing the Square

o Recall that a perfect square trinomial is the result of squaring a binomial.

o Our objective here is to find the third term of a perfect square trinomial when the first two terms are given. This is called completing the square.





Perfect Square Trinomials Equal Factors Square of a

Binomial

2 12 36x x 6 6x x 26x

2 8 16x x 4 4x x 24x

2 22x ax a x a x a 2x a

2 22x ax a x a x a 2x a





Step 1: Write the equation in the form .

Step 2: Divide by if , so that the coefficient of

2 0ax bx c 2ax bx c

a 1a 2x 2 b c

x xa a

is : .1

Step 3: Divide the coefficient of by , square the result, and add this to both sides.

Step 4: The trinomial on the left side will now be a perfect square. That is, it can be written as the square of an algebraic expression.

x 2




Example 1: Completing the Square

Add the constant that will complete the square for the expression. Then write the new expression as the square of a binomial.

2 18x xSolution: Since the leading coefficient is 1, we can begin

to complete the square. We take half of the coefficient of the x term and square the result.

118

29

29 81





Adding 81 to the expression yields the perfect square trinomial

2 18 81x x

This can be factored as 2.9x

Thus completing the square gives us:

22 18 8 9 .1x x x

2 18x x





Solve the following quadratic equation by completing the square.

2 6 10x x

1

23

2 9

Take half of the coefficient of the x term.

6

Square the result.

3




Example 2: Completing the Square (Cont.)

Adding 9 to both sides of the original equation will result in a perfect square trinomial.

Add 9 to both sides (completing the square).

2 96 10 9x x

23 19x

2 6 10x x

Factor the left-hand side.

Use the Square Root Property to solve.

3 19x 3 19x or

3 19x 3 19x or

3 19x We write these two solutions as:





Solve the following quadratic equation by completing the square.

Divide each term by 2 so that the leading coefficient will be 1.

Isolate the constant term.

22 8 24 0x x 2 4 12 0x x

2 4 12x x

Take half of the coefficient of the x term, and square it.

1

2

4

2

2

2

4




Example 3: Completing the Square (Cont.)

2 16x

2 4x 2 4x

2x 6x

Use the Square Root Property.

or

or

2 44 12 4x x

22 16x

Add 4 to both sides of 2 1 .4 2x x

Factor the left-hand side of the equation.




Writing Equations with Known Roots

To find the quadratic equation that has the given roots and , first get 0 on one side of each equation.

4x i 4x i

4 0x i 4 0x i

Set the product of the two factors equal to 0 and simplify.

4 4 0x i x i

Continued on the next slide…




Writing Equations with Known Roots

4 4 0x i x i

Regroup the terms to present the product of complex conjugates. This makes the multiplication easier.

4 4 0x i x i

2 24 0x i 2 8 16 1 0x x

2 8 17 0x x

2 1.i




Example 4: Equations with Known Roots

Find the quadratic equation that has the given roots.

7 5x 7 5x and

7 5 0x 7 5 0x Get 0 on one side of each equation.

Set the product of the two factors equal to 0.

7 5 7 5 0x x

7 5 7 5 0x x




Example 4: Equations with Known Roots (Cont.)

7 5 7 5 0x x

Simplify and solve.

227 5 0x

2 14 49 5 0x x

2 14 44 0x x




Hawkes Learning SystemsIntermediate Algebra

Section 7.2: Quadratic Equations: The Quadratic Formula




Objectives

o Write quadratic equations in standard form.o Identify the coefficients of quadratic equations in

standard form.o Solve quadratic equations using the quadratic

formula.o Determine the nature of the solutions (one real, two

real, or two non-real) for quadratic equations using the discriminant.




Standard Form of the Quadratic Equation

We are interested in developing a formula that can solve quadratic equations of any form. This formula will always work, though other techniques may be more convenient to use.We want to solve the standard form equation of the quadratic formula for x in terms of a, b and c.

The standard form of the quadratic equation is

where , , and are real numbers and .

2 0ax bx c a b c 0a




Development of the Quadratic Formula2 0ax bx c

2ax bx c

2 b cx x

a a

2 22

2 24 4

b b c bx x

a a a a

Continued on the next page…

Standard form.

Add –c to both sides.

Divide both sides by a so that the leading coefficient is 1.

Complete the square.

Simplify.

2 2

22 4

b c bx

a a a




Development of the Quadratic Formula

2 2

2

4

2 4

b b acx

a a

2 4

2 2

b b acx

a a

2 4

2

b b acx

a

Combine the fractions on the right side of the equation.

Square Root method.

The quadratic formula.

2 2

2 2

4

2 4 4

b ac bx

a a a

Find LCM of the denominators on the right side.




Development of the Quadratic Formula

The quadratic formula,

,

can ALWAYS solve quadratic equations of any form. Because it is so useful, you should memorize the quadratic formula.

2 4

2

b b acx

a




Discriminant

The expression is called the discriminant. The discriminant determines the number of solutions to the given quadratic equation.If the discriminant is:

Positive Zero NegativeDiscriminantNumber of Solutions 2 1 2

Solution(s) Real or Non-real Real Real Non-real

2 4b ac

2 4 0b ac 2 4 0b ac 2 4 0b ac




Solve Quadratic Equations Using the Quadratic Formula

Ex: Solve the quadratic equation using the quadratic formula.

2 11 02x x b = , , c 121 = a

Compare it to the standard quadratic equation to find a, b and c.2 0ax bx c

22 14 1

Solve the discriminant by plugging a, b and c into . Since the discriminant is 0, there is one real solution.

2 4b ac

Use the quadratic formula

to find the solution, x.

We know that the discriminant is 0, so we can just plug that in.

2 4

2

b b acx

a

2 1

02x

21

2x

Solution: G 1

4 4 0




Example 1: Solve Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation using the quadratic formula.

Discriminant:

Number of solutions:Real or non-real:

2 1 0x x Compare it to the standard quadratic equation to find a, b and

c.

2 0ax bx c

21 14 1 Solve the discriminant by plugging a, b

and c into . 2 4b ac1 4

3


What does it mean for a discriminant to be negative?

2non real

b = , , c =11 1a




Example 1: Solve Quadratic Equation Using the Quadratic Formula (Cont.)

Solutions:

2 1

31x


to find the solution,

x. We know that the discriminant is G 3, so we can plug that in.

2 4

2

b b acx

a

1 3

2

ix

and 1 3 1 3

2 2

i ix x

1 3 1 3

2 2x




Example 2: Solve Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation using the quadratic formula.

Discriminant:

Number of solutions:Real or non-real:

2 3 2 0x x 1a

Compare it to the standard quadratic equation to find a, b and

c.

2 0ax bx c

23 24 1 Solve the discriminant by plugging a, b

and c into . 2 4b ac

, 3b , 2c

9 8 1


What does it mean for a discriminant to be positive?

2real




Example 2: Solve Quadratic Equation Using the Quadratic Formula (Cont.)

Solutions:

2 1

13x


to find the solution,

x. We know that the discriminant is 1, so we can plug that in.

2 4

2

b b acx

a

3 1

2x

and 1 2x x





Section 7.3: Applications- Quadratic Equations




Objective

Solve applied problems by using quadratic equations.




Strategy for Solving Word Problems

1. Read the problem carefully.2. Decide what is asked for and assign a variable to

the unknown quantity.3. Draw a diagram or set up a chart whenever

possible.4. Form an equation (or inequality) that relates the

information provided. 5. Solve the equation or inequality.6. Check your solution with the wording of the

problem to be sure that it makes sense.




The Pythagorean Theorem

o Problems involving right triangles often require the use of quadratic equations.

o In a right triangle, one of the angles is a right angle (measures ), and the side opposite this (the longest side) is called the hypotenuse.

o The other two sides are called legs.

90

hypotenuseleg

leg




The Pythagorean Theorem

In a right triangle, the square of the hypotenuse ( ) is equal to the sum of the squares of the legs ( and ).

22 2c ba

a

bc

ca b




Example 1: The Pythagorean Theorem

The width of a rectangle is 5 yards less than its length. If one diagonal measures 25 yards, what are the dimensions of the rectangle?

Solution: Draw a diagram for problems involving geometric figures whenever possible.

Let x length of the rectangle

5x width of the rectangle

Then, by the Pythagorean Theorem,5x

x

25

2 22 5 25x x




Example 1: The Pythagorean Theorem (Cont.)

2 22 5 25x x Now, solve the quadratic equation.

2 2 10 25 625x x x 22 10 25 625 625 625x x

22 10 600 0x x

22 5 300 0x x

2 20 15 0x x 20x 15x

5 15x

A negative number does not fit the conditions of the problem.

Length:Width:




Projectiles

The formula is used in physics and relates to the height of a projectile such as a thrown ball, a bullet or a rocket.

ht

0v

0h

0 0h

height of object, in feet.

time object is in the air, in seconds.

beginning velocity, in feet per second.

beginning height.

if object is initially at ground level.

20 016h t v t h




Example 2: Projectiles

A bullet is fired straight up from 6 feet above ground level with a muzzle velocity of 420 ft per sec. When will the bullet hit the ground?

Solution: 0 420v ft per sec.

0 6h The bullet hits the ground when 0.h

20 016h t v t h

20 16 420 6t t 20 8 210 3t t Divide both sides

by 2.




Example 2: Projectiles

20 8 210 3t t Use the quadratic formula to solve the equation.

2 4

2

b b act

a

2210 210 4 8 3

2 8t

210 44196

16t

26.26t 0.01t

Therefore, the bullet hits the ground in 26.26 seconds.




Example 3: Cost per Person

The Ski Club is planning to charter a bus to a ski resort. The cost will be $200 and each member will share the cost equally. At the last minute, 30 more member decide to go on the trip. The cost to each of the members will be $6 less. How many members are going to the ski resort now?

Let x the original number of club members going on the trip.

the actual number of club members that took the trip. 30x

200

x

200

30x 6

Initial cost per member

Final cost per member

Difference in cost per member




Example 3: Cost per Person (Cont.)

Multiplying each term by the LCM of the denominators results in the following:

200 20030 30 6 30

30x x x x x x

x x

200 30 200 6 30x x x x

200 6000 200 6 30x x x x

26000 6 180x x

20 6 180 6000x x 20 30 1000x x

Factor out 6.




Example 3: Cost per Person (Cont.)

Use the quadratic formula to solve.

2 30 1000 0x x

2 4

2

b b acx

a

230 30 4 1 1000

2 1x

30 70

2x

20x 50x The number of members that actually went to the ski resort is

3020 0.5




Example 4: Geometry

The Smiths have a rectangular swimming pool that is 5 ft longer than it is wide. The pool is completely surrounded by a concrete deck that is 3 ft wide. The total area of the pool and the deck is 1 . Find the dimensions of the pool.

3 ft

w

5w

Letwidth of the pool and the deck

length of the pool and the deck

6w

11w

6 11 1750w w

21750 ft

3 ft3 ft

3 ft




Example 4: Geometry (Cont.)

Use the quadratic equation to solve. 26 11 1750 17 1684 0w w w w

2 4

2

b b acw

a

217 17 4 1 1684

2 1w

17 7025

2w

33.41w 50.41w

5 38.41w

Therefore, the length of the pool is 33.41 feet, and the width of the pool is 38.41 feet.

or




Hawkes Learning SystemsIntermediate AlgebraSection 7.4: Equations in Quadratic Form




Objectives

o Make substitutions that allow equations to be written in quadratic form.

o Solve equations that can be written in quadratic form.

o Solve equations that contain rational expressions.




Solving Equations in Quadratic Form

The general quadratic equation is where .The equations

and are not quadratic equations, but they are in quadratic form because the degree of the middle term is one-half the degree of the first term. Specifically,

and

2 0ax bx c 0a

4 27 12 0x x 2 1

3 34 21 0x x

1

24 2

1

2

2

3

1

3

Degree of the first term

Degree of the first term

Degree of the second term

Degree of the second term




Solving Equations in Quadratic Form

Solving Equations in Quadratic Form by Substitution1.Look at the middle term.2.Substitute a first-degree variable, such as u, for the

variable expression in the middle term.3.Substitute the square of this variable, u², for the

variable expression in the first term.4.Solve the resulting quadratic equation for u.5.Substitute the results back for u in the beginning

substitution and solve for the original variable.




Example 1: Solving Equations in Quadratic Form

Solve the equation in quadratic form by substitution.6 37 12 0x x Step 1: Look at the middle term.

2 7 12 0u u

3 4 0u u 3 or 4u u

3 33 or 4x x

33 3 or 4x x

Step 2: Substitute a first-degree variable, such as u, for the variable expression in the middle term. Step 3: Substitute the square of this variable, u², for the variable expression in the first term.

Step 4: Solve the resulting quadratic equation for u.

Step 5: Substitute the results back for u in the beginning substitution and solve for the original variable.

3u x





Solve the equation in quadratic form by substitution.5

5 29 18 0x x Step 1: Look at the middle term.

2 9 18 0u u

6 3 0u u 6 or 3u u

5 5

2 26 or 3x x

5 536 or 9x x



Step 5: Substitute the results back for u in the beginning substitution and solve for the original variable.5 536 or 9x x

5

2u x





Solve the equation in quadratic form by substitution.2 1

3 33 2 0x x Step 1: Look at the middle term.

2 3 2 0u u

1 2 0u u 1 or 2u u

1 1

3 31 or 2x x

1 or 8x x



Step 5: Substitute the results back for u in the beginning substitution and solve for the original variable. 3 3

1 or 2x x

1 3u x





Solve the equation in quadratic form by substitution.2 110 16 0x x Step 1: Look at the middle term.

2 10 16 0u u

2 8 0u u 2 or 8u u

1 12 or 8x x

1 1 or

2 8x x




1u x





Solve the equation in quadratic form by substitution. 2

6 4 6 4 0x x Step 1: Look at the middle term.

2 4 4 0u u

2 2 0u u 2 u

6 2 x

4 x




6u x




Solving Equations with Rational Expressions

Recall that to solve equations with rational expressions, first multiply every term on both sides of the equation by the least common multiple (LCM) of the denominators. This will “clear” the equation of fractions. Remember to check the restrictions on the variables. That is, no denominator can have a value of 0.




Example 6: Solving Equations with Rational Expressions

Solve the following equation containing rational expressions. 3 2

4 2 3 3

x

x x x

3 24 2 3 4 2 3

4 2 3 3

xx x x x

x x x

3 3 2 4 2 4 2x x x x 23 9 8 4 4 2x x x x 20 4 13 5x x

This equation is not in the quadratic form. Multiply both sides of the equation bythe LCM, (4x – 2)(x + 3), to “clear” the fractions. The restrictions on x are .

1, 3

2x

Simplify.

Continued on next slide...




Example 6: Solving Equations with Rational Expressions (Cont.)

20 4 13 5x x Use the quadratic formula to solve.

213 13 4 4 5

2 4x

13 169 80

8x

13 249

8x

There are two solutions to this

equation.




Example 7: Solving Equations with Rational Expressions

Solve the following equation containing rational expressions.1 2

2 3 2 2

x

x x x

1 22 3 2 2 3 2

2 3 2 2

xx x x x

x x x

1 2 2 2 3 2 3x x x x 22 4 6 2 3x x x x 20 2 2 4x x

This equation is not in the quadratic form. Multiply both sides of the equation bythe LCM, (2x + 3)(x – 2), to “clear” the fractions. The restrictions on x are .

3,2

2x

Simplify.

Continued on next slide...




Example 7: Solving Equations with Rational Expressions (Cont.)

20 2 2 4x x Use the quadratic formula to solve.

22 2 4 2 4

2 2x

2 4 32 2 36

4 4x

2 61

4x

2

, so the only solution is 3

,22

x

or

1x




Solve the higher degree equation.

Example 8: Solving Higher-Degree Equations

5 9 0x x

4 9 0x x

2 23 3 0x x x

2 20 or 3 or 3x x x

0, 3, 3, 3, 3 x i i

0 or 3 or 3 x x i x

Factor out x.

Factor the difference of two squares.

Solve for x.

There are five solutions for x.




Example 9: Solving Higher-Degree Equations

Solve the higher degree equation.3 8 0x

22 2 4 0x x x

2 1 3 1 3 0x x i x i

2 or 1 3 or 1 3x x i x i

2, 1 3, 1 3x i i

Factor out (x – 2).

Factor the quadratic equation.

Solve for x.

There are three solutions for x.




Hawkes Learning SystemsIntermediate Algebra

Section 7.5: Graphing Parabolas




Objectives

o Graph a parabola (a quadratic function) and determine its vertex, domain , range, line of symmetry and zeros.

o Solve applied problems by using quadratic functions and the concepts of maximum and minimum.




Introduction to Quadratic Functions

Concepts related to various types of functions and their graphs include domain, range and zeros.

In this section we will include a detailed analysis of quadratic functions, functions that are represented by quadratic expressions.





What is the graph of the function ? The nature of the graph can be investigated by plotting several points.

2 4 3y x x

x y2 4 3x x

1

0

1

2

2

3

7

2

5

21 4 1 3

20 4 0 3

21 1

4 32 2

22 4 2 3

23 4 3 3

1 21 4 1 3

27 7

4 32 2

25 4 5 3

4 24 4 4 3

8

3

5

4

0

1

0

5

4

3

8





The complete graph of is shown below.2 4 3y x x

This curve is called a parabola. The point is the “turning point” of the parabola and is called the vertex. The line is the line of symmetry or axis of symmetry for the parabola. That is, the curve is a “mirror image” of itself with respect to the line .

2, 1

2x

2x





Quadratic FunctionsA quadratic function is any function that can be written in the form

where a, b, and c are real numbers and The graph of every quadratic function is a parabola. Parabolas that open up or down are vertical parabolas and parabolas that open left or right are horizontal parabolas. Horizontal parabolas do not represent functions.

2f x ax bx c

0.a





We will discuss quadratic functions in each of the following five forms where a, b, c, h and k are constants:

2y ax2y ax k

2y a x h

2y a x h k

2y ax bx c




If IfThe graph lies Below the x-axis Above the x-axis

Vertex This is the only point where the graph touches the x-axis.

Domain (All real numbers)Range

Quadratic Functions

Functions of the Form 2y ax

0a 0a

0,0

0y 0y




Quadratic Functions

Domain: {x|x is any real number}Range: {y|y ≥ 0}In interval notation: Domain: , Range: , 0,

Domain: {x|x is any real number}Range: {y|y ≤ 0}In interval notation: Domain: , Range: ,0 ,




Quadratic Functions

The graphs on the previous slide illustrate the following characteristics of quadratic equations of the form :a.If , the parabola “opens upward.”b.If , the parabola “opens downward.”c.The bigger is, the narrower the opening.d.The smaller is, the wider the opening.e.The line (the y-axis) is the line of symmetry.

2y ax0a 0a

a

a

0x




If IfThe y-value Decreases by Increases by

Shift Down units Up unitsVertex Line of

Symmetry

Quadratic Functions

Functions of the Form 2y ax k

0k 0k

k

k k

k

0x

0,k

2y ax




Quadratic Functions

Domain: {x|x is any real number}Range: {y|y ≥ k}In interval notation: Domain: , Range: , ,k

Domain: {x|x is any real number}Range: {y|y ≤ k}In interval notation: Domain: , Range: ,k ,




If IfThe graph lies Below or on the

x-axisAbove or on the

x-axisVertex

The graph opens Down Up

Quadratic Functions

Functions of the Form 2y a x h

,0h

0a 0a




Quadratic Functions

Domain: {x|x is any real number}Range: {y|y ≥ 0}In interval notation: Domain: , Range: , 0,

Domain: {x|x is any real number}Range: {y|y ≤ 0}In interval notation: Domain: , Range: ,0 ,




If IfThe x-value Decreases by Increases by

Shift Left units Right unitsLine of

Symmetry

Quadratic Functions

For the graph of , 2-y a x h

0h 0h

h h

x h

h h2y ax




Graph Shift of Vertex Line of Symmetry

Does not applyUp or DownLeft or Right

Quadratic Functions

In summary,

2y ax

2y ax k 2

y a x h

2y ax 0,0

x h ,0h

0,k 0 x 0 x




Quadratic Functions

Ex: Graph the quadratic function . Find the line of symmetry and the vertex, and state the domain and range of each function.Solution:Line of symmetry is .

The parabola opens upward.

Vertex:

22 3y x

0x Any function of the form has x = 0 for its line of symmetry.

2y ax k

a = 2, so a is positive.

0, 3 , so the vertex is .3k 0, 3





Quadratic Functions

Domain: orRange: or

| is any real numberx x True of all functions of the form .2y ax k ,

| 3y y 3, Because .3k




Example 1: Quadratic Functions

Graph the quadratic function . Find the line

of symmetry and the vertex, and state the domain and range of each function.

Solution:Line of symmetry is .

The parabola opens .

Vertex:

25

2y x

5

2x A function of the form

has x = h for its line of symmetry. 2

y a x h

a = G 1, so a is negative.

5,0

2

, so the vertex is .5

2h 5

,02


downward




Quadratic Functions

Domain: orRange: or

| is any real numberx x True of all functions of the form . 2

y a x h , | 0y y ,0

Because .1a

( )




Quadratic Functions

Functions of the form combine both the vertical shift of units and the horizontal shift of units. The vertex is at . For example, the graph of the function is a shift of the graph of up 5 units and to the right 3 units and has its vertex at .

The graph of is the same as the graph

of but is shifted left unit and down 2 units.

The vertex is at .

2y a x h k

k h ,h k

22 3 5y x

22y x 3,5

21

22

y x

2y x1

2

1/ 2, 2




Quadratic Functions




Quadratic Functions

To easily find the vertex, line of symmetry, range, and to graph the parabola, we want to change the general form into the form . This can be accomplished by “completing the square” using the following technique.

2y ax bx c 2y a x h k

2f x ax bx c

2 ba x x c

a

2 2

22

24 4

bba x

a a ac

bx

Write the function.

Factor a from the first two terms.

Complete the square.





Quadratic Functions2

22

2

44

b ba

b

ax x c

a a

2 24

2 4

b ac ba x

a a

24,

2 4

b ac b

a a

Note: the vertex must lie at

. ,2 2

b bf

a a

Move the negative factor out of the parentheses, first multiplying by a.

Write the square of the binomial and simplify.

In terms of the coefficients a, b and c,

is the line of symmetry

and is the vertex.

2

bx

a




Quadratic Functions

Note!Rather than memorize the formulas for the coordinates of the vertex, you should just remember that the

x-coordinate of the vertex is . Substituting this

value for x in the function will give the y-value for the vertex.

2

bx

a




Zeros of a Quadratic Equation

The points where a parabola crosses the x-axis, if any, are the x-intercepts. This is where . These points are also called the zeros of the function. We find these points by substituting 0 for y and solving the resulting quadratic equation.

If the solutions are non-real complex numbers, then the graph does not cross the x-axis.

0y

2ax x cy b 20 ax bx c

quadratic functionquadratic equation




Zeros of a Quadratic Function

Ex: Find the zeros of the function, the line of symmetry, the vertex, the domain, the range and graph the parabola.Solution:

2 6 1 0x x

26 6 4

2x

6 32

2x

6 4 2

2x

3 2 2x

Quadratic Formula

The zeros of the function.Continued on the next slide…




Zeros of a Quadratic Equation

Change the form of the function for easier graphing.

2 6 1y x x

2 9 96 1x x

2 9 96 1x x

33 8x

Add inside the parenthesis.0 9 9




Zeros of a Quadratic Function

SummaryZeros:Line of Symmetry:Vertex:Domain:

Range:

3 2 23x

3, 8 , ;

8, ; | 8y y | is any real numberx x




Example 2: Zeros of a Quadratic Function

Find the zeros of the function, the line of symmetry, the vertex, the domain, the range and graph the parabola.

Solution:

2 4 2 0x x

24 4 4 1 2

2 1x

4 24

2x

4 2 6

2x

2 6x

Quadratic Formula.

The zeros of the function.Continued on the next slide…




Example 2: Zeros of a Quadratic Function (Cont.)

Change the form of the function. 2 4 2y x x

2 4 2x x

2 44 24x x

22 6x

Add inside the parenthesis.0 4 4

2 44 24x x

Factor from the first two terms.1

Put outside of the parenthesis. 1 4






SummaryZeros:Line of Symmetry:Vertex:Domain:

Range:

2 6 2x

2,6 , ;

,6 ; | 6y y | is any real numberx x




Example 3: Zeros of a Quadratic Function

Find the zeros of the function, the line of symmetry, the vertex, the domain, the range and graph the parabola.

Solution:

22 6 5 0x x

26 6 4 2 5

2 2x

6 4

4x

Quadratic Formula.


There are no real zeros because the discriminant

is negative. The graph will not cross the x-axis.





To find the vertex, use another approach. First find the x-value.2

bx

a

23 3 3

2 6 52 2 2

f

99 5

2 Simplify.

1

2

Plug the x-value into the original equation.


3 1,

2 2

So we have the following vertex.

6 3

2 2 2





Insert graph p. 597 example 2 c.

Summary

Zeros:

Line of Symmetry:

Vertex:

Domain:

Range:

No real zeros3

2x

3 1,

2 2 , ;

1 1, ; |

2 2y y

| is any real numberx x




Applications with Maximum/Minimum Values

Note:The vertex of a vertical parabola is either the lowest point or the highest point on the graph of the parabola. Therefore, the vertex can be used to determine the maximum or minimum value of a quadratic function.




Applications with Minimum/Maximum Values

Minimum and Maximum ValuesFor a parabola with its equation in the form

1. If , then the parabola opens upward and is the lowest point and is called the minimum value of the function.

2. If , then the parabola opens downward and is the highest point and is called the maximum value of the function.

2,y a x h k

0a

0a ,h ky k

,h k y k




Application and Minimum/Maximum Values

If the function is in the general quadratic form , then the maximum or minimum value

can be found by letting and solving for y.

The concepts of minimum and maximum values of a function help not only in graphing but also in solving many types of applications.

2y ax bx c

2

bx

a




Example 3: Maximization/Minimization Problems

A movie theater sells regular adult tickets for $7.50 each. On average, they sell 5,000 tickets per day. The company estimates that each time they raise ticket prices by 50¢, they will sell 1,000 fewer tickets. What price should they charge to maximize their revenue (income) per day? What will be the maximum revenue?Let x = number of 50¢ increases in price.Then the price per ticket =and the number of tickets sold =Revenue = (price per unit)(number of units sold)

7.50 .50x5,000 1 .,000x





Example 3: Maximization/Minimization Problems (Cont.)

So,

The revenue represented by a quadratic function and the maximum revenue occurs at the point where

For , price per ticket =And the maximum revenue =

7.50 .50 5,000 1,000R x x 237,500 7500 2500 500x x x

2500 5000 37,500x x

2

bx

a

5000

2 500

5

5x 7.50 . 550

$50,000 5,000 1, $000 55 $5




Example 4: Maximization/Minimization Problems

A farmer plans to use 2500 feet of spare fencing material to form a rectangular area for cows to graze alongside a river, using the river as one side of the rectangular area. How should he split up the fencing among the other three sides in order to maximize the rectangular area?

If we let x represent the length of one side of the rectangular area then the dimensions of the rectangular area are x feet by 2500-2x feet (see image above). We will let A be the name of our function that we wish to maximize in this problem, so we want to find the maximum possible value of . 2500 2A x x x

MOO!


2500 2x

x

x




Note: If we multiply out the formula for A, we get a quadratic function.We know this function is a parabola opening down. We also know that the vertex is the maximum point on this graph. Remember, the vertex is the point

.

So, plugging in the values, we get the vertexTherefore, the maximum possible area is A(625):

Example 4: Maximization/Minimization Problems (Cont.)

2500 2A x x x 22 2500x x

24, ,

2 4 2o

2r

b ac b b bf

a a a a

625, 5 .62A

(625) 781250 square t. feeA





Section 7.6a: Solving Quadratic Inequalities




Objectives

o Solve quadratic inequalities.

o Solve higher degree inequalities.

o Graph the solutions for inequalities on real number lines.




Quadratic Inequalities

For values of x on either side of a number a, the sign for an expression of the form changes. x a

If , then is positive.

If , then is negative.

x a x a

x a x a

a

0x a 0x a

any x a any x a




To Solve Quadratic or Higher Degree Inequalities

1. Arrange the terms so that one side of the inequality is 0.

2. Factor the algebraic expression, if possible, and find the points where each factor is 0. (Use the quadratic formula, if necessary, to find the points where a quadratic expression is 0.)

3. Mark these points on a number line. Consider these points as endpoints of intervals





To Solve Quadratic or Higher Degree Inequalities (Cont.)

4. Test one point from each interval to determine the sign of the expression for that interval.

5. The solution consists of those intervals where the test points satisfy the original inequality.




Example 1: Quadratic Inequalities

Solve the following inequality by factoring and using a number line. Then, graph the solution set on a number line.

2 2 15x x

Step 1: Add -15 to both sides so that the right-hand side is 0.

2 15 11 52 5x x

Step 2: Factor the left-hand side.

2 2 15 0x x

5 3 0x x




Example 1: Quadratic Inequalities (Cont.)

5 3 0x x Step 3: Set each factor equal to

0 and solve to find the endpoints.

5 0x 3 0x 5x 3x

Step 4: Mark these points on a number line and test one point from each of the intervals.

5 3





05 37 5Test one point from each interval.

5 3 0x x

37 75 Test 7x

2 10 20 0 5 3 0 5 if x x x

Test 0x

0 5 30 5 3 15 0 5 3 0 5 3 if x x x

Test 5x

5 5 35 10 2 20 0 5 3 0 3 if x x x

Step 5: , 5

5,3 3,





We have found that:

2 2 15x x

5 3 0 5 if x x x

5 3 0 5 3 if x x x

5 3 0 3 if x x x

5 ., 3, is in x

5 3

Therefore, the solution is:

) (

5x 3x or

or




Example 2: Cubic Inequality

Solve the following inequality by using a number line and graph each solution set on a number line.

3 28 9 0x x x

Factor the left-hand side. 2 8 9 0x x x

1 9 0x x x

Set each factor equal to 0 and solve to find the endpoints.

1 0x 9 0x

1x 9x

0x

0x




Example 2: Cubic Inequality (Cont.)

Mark the endpoints on a number line and test one point from each of the intervals found.

Endpoints: 1x 9x 0x

1 9021

2

3 10

Test 2x

1 9 0x x x

12 92 2 2 1 11 22 0

Test 3x 3 3 1 93 3 4 6 72 0

, 1 1,0 0,9 9,




Example 2: Cubic Inequality (Cont.)

1 9021

2

3 10

1 9 0x x x

Test1

2x

1 1 1

2 2 21 9

1 1 19

2 2 2

190

8

Test 10x

11 10 90 10 10 11 1 110 0




Example 2: Cubic Inequality (Cont.)3 28 9 0x x x

We have found that:

1 9 0 1 if x x x x

1 9 0 0 9 if x x x x

1 9 0 9 if x x x x

1 9 0 1 0 if x x x x

Therefore, the solution is , 1 0,9 . is in x

1 90




Example 3: Quadratic Inequality

Solve the following inequality by using a number line and graph each solution set on a number line.

2 6 7 0x x Solution: Since this quadratic equation will not factor with integer coefficients, use the quadratic formula to find the endpoints for the intervals. The test points can be integers.

2 4

2

b b acx

a

26 6 4 1 7

2 1x

6 36 28

2x




Example 3: Quadratic Inequality (Cont.)

6 36 28

2x

6 8

2x

6 2 2

2x

3 2x

3 2 3 20





3 2 3 20 3 7

0x Test

2 6 7 0x x

20 6 70 0 0 7 7 0

Test 3x

23 6 73 9 18 7 2 0

Test 7x

27 6 77 49 42 7 14 0





2 6 7 0x x We have

23 2 6 7 0If then x x x

23 2 6 7 0If then x x x

3 2 3 2

The solution is: ,3 2 3 2, . is in x




Hawkes Learning Systems:Intermediate AlgebraSection 7.6b: Solving Inequalities with

Rational Expressions




Objectives

o Solve inequalities containing rational expressions.

o Graph the solutions for inequalities containing rational expressions.




Solving Inequalities with Rational Expressions

o For linear inequalities, the solution always contains the points to one side or the other of the point where the inequality has the value 0.

o A rational inequality may involve the product or quotient of several first-degree expressions. For example,

involves the two first-degree expressions and . o For ( ), if , then is negative; if ,

is positive.

40

3

x

x

x a x a x a x a x a

4x 3x





Procedure for Solving Polynomial Inequalities with Rational Expressions

1. Simplify the inequality so that one side is 0 and the other side has a single fraction with both the numerator and denominator in factored form.

2. Find the points that cause the factors in the numerator or in the denominator to be 0.

3. Mark each of these points on a number line. These are interval endpoints.






4. Test one point from each interval to determine the sign of the polynomial expression for all points in that interval.

5. The solution consists of those intervals where the test points satisfy the original inequality.

6. Mark a bracket for an endpoint that is included and a parenthesis for an endpoint that is not included. Remember that no denominator can be 0.





The steps are as follows:

a. Find the points where each linear factor has the v value 0.

b. Mark each of these points on a number line. C Consider these points as endpoints of intervals.

4 0x 3 0x 4x 3x

4 30

The three intervals formed are , 4 , 4,3 , 3, .

40

3

x

x





c. Choose any number from each interval as a test value to determine the sign of the expression for all values in that interval. Remember, we are not interested in the value of the expression, only whether it is positive or negative.

4 36 0 4

We have chosen the test values , , and . 6 0 4

, 4 4,3 3,





Substituting the values , , and into the original inequality gives:

Results Explanation

The expression is positive for all x in (ng inf,neg4)

The expression is negative for all x in (-4,3)

The expression is positive for all x in (3,posinf)

6 4 2

6 33

2

9

4

9

x

x

0 4 4

0 3

4

3

4

33

x

x

4 4 8

4

4

3 38

1

x

x

, 4 .

4,3 .

3, .

4 3

6 0 4





d. The solution to the inequality consists of all the intervals that indicate the desired sign: + (for ) or – (for ). The solution for the above expression is:

40

3

x

x

, 4 3, .all in or x

In algebraic notation: In interval notation:

Graphically:

4 3. or x x , 4 3, .x

4 3

0x 0x




Example 1: Solving and Graphing Inequalities

Solve and graph the solution for the following inequality:

Solution: Set each linear expression equal to 0 to find the interval endpoints.

Test a value from each of the intervals:

10

5

x

x

1 0x 5 0x 1x 5x

, 1 1,5 5,, , and

1 53 0 6




Example 1: Solving and Graphing Inequalities (Cont.)

Substituting the values , , and into the original inequality gives:

Results Explanation

3

5

1 1

35

x

x

0 1

0 5

1

5

x

x

6 1

6 5

1

5

x

x

3 0 6

1 5

2

8

1

4 The expression is positive

for all x in (neginf,-1). , 1 .

1

5

1

5 The expression is negative

for all x in (-1,5). 1,5 .

7

1 7 The expression is positive

for all x in (5, inf). 5, .





We can conclude that the solution for the above inequality in interval notation is:

10

5

x

x

1,5 .

In algebraic notation: 1 5.x

Graphically:

1 5( )




Example 2: Solving and Graphing Inequalities

Solve and graph the solution for the following inequality:6

1 02

x

x

Solution: Simplify to get one fraction.

6 20

2 2

x x

x x

2 40

2

x

x





Set each linear expression equal to 0 to find the interval endpoints.

2 4 0x 2 0x 2x 2x

Test a value from each of the intervals:

2 24 0 4

, 2 2,2 2,, , and





Substituting the values -4, 0, and 4 into the original inequality gives:

Results Explanation

2 4 4

4

2 4

2 2

x

x

4

6

2

3

2 0 4

0 2

2 4

2

x

x

4

2

2

2 4 4

4 2

2 4

2

x

x

12

2 6

The expression is positive for all x in (neginf, -4) , 2 .

The expression is positive for all x in (2, inf). 2, .

The expression is negative for all x in (-2,2). 2,2 .

2 2





61 0

2

x

x

We can conclude that the solution for the above inequality in interval notation is: , 2 2, .

In algebraic notation:

Graphically:

2 2. or x x

Note: 2 from the denominator is not included in the solution set because the quotient is undefined there.

2 2(

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