Algebra II Summer Packet 2014 - Abraham Clark High Schoolachs.roselleschools.org/UserFiles/Servers/Server_3154010/File/News... · When solving math problems with more than one operation,

ACHS – ENTERING ALGEBRA 2 – SUMMER 2015 WORK Page 1 of 22

Name:_______________________________________________________

Algebra II Summer Assignment

Due Date: First Monday of school year 2015-2016

To maintain a high quality program, students entering Algebra II are expected to remember the basics of the mathematics taught in their Basic Algebra course. In order to review the basic concepts prior to taking Algebra II, the mathematics department has prepared this review packet. For each algebra topic addressed, this packet contains several review examples followed by online tutorials and some practice questions for the student to do.

Since this material is designed as review, you are responsible for completing this packet on your own. This assignment will be collected and graded as your first test, the first Monday of the school year 2015-2016. Be sure to show all appropriate work to support your answers. In addition, a Pre-Assessment will be given to assess the student’s knowledge of the covered topics during the first month in the new school year. All questions must be completed with the correct work. You must return in September knowing how to do all the material in this packet. Also, the material should be done without the use of any graphing or scientific calculator, as the new PAARC test will have a portion of test that you can’t use a calculator for.

For assistant with the packet you may contact Mrs. Fischer at [email protected], or Mr. Naem at [email protected]. Emails may take a few days for responses. Please be specific in your email what you need assistance with, include the section and the question number as well. Sample online websites are giving at the end of the packet. More websites could be found by searching the outline topic for each section.

Late packets will not be accepted.

Good Luck

See you in September 2015!


I. PROPORTION

A proportion is an equation showing two ratios are equal.

Extreme is another way of saying at the ends.

Mean is another way of saying in the middle.

Note: This is not the same mean as used in taking an average.

Cross product: The product of the extremes is equal to the product of the means.

EXAMPLE: Check to see if the proportion is a real proportion. Cross multiply to find the products.

34

= 1216

4(12) = 3(16) 48 = 48

Since both sides are equal, it is a proportion.

A proportion can be used to find missing information.

! Set up the proportion using a variable for the missing information. ! Make sure the set up is correct. If the proportion is set up incorrectly, the result will be

incorrect. ! Cross multiply.

EXAMPLE: On a map, 1 inch represents 5 miles. Find the distance two cities are from each other if they are 4 inches apart on the map.

inchesmiles

15

= 4x

1(x) = 5(4) x = 20 miles

TUTORIAL: http://www.ck12.org/arithmetic/Proportions/

PRACTICE: Find the proportion for each of the questions.

1. 812

= x15

ANSWER: ________

2. 65x

= 54

ANSWER: ________

cd

extreme

extreme

ab =

mean

mean


3. A certain factory produces 80 good motors for every 3 defective motors. At this defect rate,how many defective motors will result with a production of 480 good motors?

ANSWER: ________

4. The scale on a map of Europe states: 1 cm = 62 miles. If Rome is 11.2 cm from Paris on thismap, find the distance in miles between these two cities.

ANSWER: ________

5. A community college has a student to faculty ratio of 21 to 2. If this college has 234 facultymembers, how many students are enrolled there?

ANSWER: ________

II. PERCENT

A percent is out of 100. It is part of 100. The symbol used is %.

Percent is a ratio or fraction where the denominator is 100.

General percent equation: y is x% of z, where: y = the part of the whole x = the percent z = the base.

This can also be set up as a proportion.

EXAMPLE: (find the percent)

What percent of 10 is 7? 7 is x% of 10

10x = 100(7) 10x10

= 70010

x = 70%

(Using proportion) x100

= 710

10x = 100(7) 10x10

= 70010

x = 70%

yz

percent

base

x100

=

part

whole


EXAMPLE: (find the part)

What is 70% of 10? y is 70% of 10

10(70) = 100y 70010

= 100y100

y = 7

(Using proportion) 70100

= y10

10(70) = 100y 70010

= 100y100

y = 7

EXAMPLE: (find the base)

7 is 70% of what number? 7 is 70% of z

70z = 100(7) 70z70

= 70070

z = 10

(Using proportion) 70100

= 7z

70z = 100(7) 70z70

= 70070

z = 10

TUTORIAL: http://www.ck12.org/arithmetic/Percents/

PRACTICE: Find the missing information using either method.

1. What percent of 129 is 86? ANSWER: ________

2. 28 is what percent of 280? ANSWER: ________

3. What is 6% of 75? ANSWER: ________

4. Find 5.5% of 128. ANSWER: ________

5. 18 is 25% of what number? ANSWER: ________

6. 349 is 106% of what number? ANSWER: ________


III. PERCENT – DECIMAL – FRACTION CONVERSIONS

FRACTION TO…

Convert from fraction to decimal:

! Divide the denominator into the numerator.

EXAMPLE: Convert 18

to a decimal. 1 ÷ 8 = 0.125

Convert from fraction to percent:

! Divide the denominator into the numerator. Then multiply by 100.

EXAMPLE: Convert 18

to a percent. (1 ÷ 8)100 = 0.125(100) = 12.5%

DECIMAL TO…

Convert from decimal to fraction:

! Determine the place value of the last number on the right that is not a zero. ! Remove the decimal point. ! Use the numbers from the decimal as the numerator. ! Use the place value found as the denominator. ! Simplify if possible.

EXAMPLE: Convert 0.12 to a fraction. The place value of the 5 is the thousandths place. 12100

= 325

Convert from decimal to percent:

! Move the decimal place 2 units to the right and put a percent sign after it. EXAMPLE: Convert 0.125 to a percent. 0.125 = 12.5%

PERCENT TO…

Convert from percent to decimal:

! Remove the percent sign and move the decimal place 2 units to the left.

EXAMPLE: Convert 12.5% to a decimal. 12.5% = 0.125

Convert from percent to a fraction:

! Percent is over 100. ! Remove the percent sign and place over 100. Do not remove the decimal yet. ! Look at how many places there are after the decimal. Remove the decimal and put add a

zero to the denominator for each decimal place. ! Simplify if possible.


EXAMPLE: Convert 12.5% to a fraction.

12.5100

= 1251000

= 18

PRACTICE: Fill in the table.

IV. ORDER OF OPERATIONS

When solving math problems with more than one operation, you must follow the correct order of operations.

Go through the operations in the following order. Parenthesis (in order from inner to outer) Exponents (including radicals) Multiplication/Division (in order from left to right) Addition/Subtraction (in order from left to right)

Ignoring these rules will give different answers. ! Notice the difference the parenthesis makes in the two EXAMPLEs below.

8 – 7 + 3 = 1 + 3 = 4 8 – (7 + 3) = 8 – 10 = –2

! Notice the difference the order of operations makes in the two EXAMPLEs below. 16 ÷ 2 • 4 = 8 • 4 = 32 16 • 2 ÷ 4 = 32 ÷ 4 = 8

EXAMPLE: Simplify 3(2 + 5)2 ÷ 7 + 9

Step 1: Parenthesis 2 + 5 = 7 3(7)2 ÷ 7 + 9

Step 2: Exponents (radicals) 72 = 49 9 = 3 3(49) ÷ 7 + 3

Step 3: Multiplication/Division 3(49) = 147 147 ÷ 7 = 21 21 + 3

Step 6: Addition 21 + 3 = 24 NOTE: When you have a fraction, simplify the numerator and denominator first, then simplify the fraction.

Tutorial: http://www.regentsprep.org/Regents/Math/orderop/Lorder.htm http://www.math.com/school/subject2/lessons/S2U1L2GL.html http://www.ck12.org/algebra/PEMDAS/

Fraction Decimal Percent 0.17

62% 1.2

7/20 80%

1/3


PRACTICE: Use order of operations to simplify the following problems. Show each step.

1. 6 + 2 x 8 – 12 + 9 ÷ 3 2. 25 – (23 + 5 x 2 – 3)

3. −2 • (−30) + 0.5 •2042 − 6

4. 15 − [8 − (2 + 5)]18 − 52

V. LAWS OF EXPONENTS

PRODUCT RULE: am · an = am + n

(When multiplying like bases, add the powers)

EXAMPLE: 1. x4 · x5 = x4 + 5 = x9 2. a7 · a · a12 = a7 + 1 + 12 = a20

3. (3x6)(2x4) = (3)(2)x6 + 4 = 6x10

POWER RULE: (ambm)n = amnbmn (When taking a monomial to a power, multiply the powers including the coefficient)

EXAMPLES: 1. (a4b3)2 = a8b6 2. (3m2n5)4 = 34m8n20 = 81m8n20

3.9 6 2

4 2 5(6a b )(–c d )

= 2 18 12

5 20 106 a b(–1) c d

= 18 12

20 1036a b–1c d

QUOTIENT RULE: am

an= am−n

(When dividing with like bases, subtract the powers) Note: it is always the numerator's power minus the denominator's power.

EXAMPLES: 1.6

4xx

= x6 – 4 = x2 2. 5 7

4 2m nm n

= m5 – 4n7 – 2 = mn5

ZERO POWER RULE: a0 = 1 (Any term to the zero power is one)

EXAMPLES: 1. (m5 n7)0 = 1 2. (4m8n2)(–2mn4)0 = (4m8n2)(1) = 4m8n2

NEGATIVE POWER RULE: a–n = n1a

and –n1a

= an

(Take the reciprocal of the variable to the negative power) NOTE: Apply the negative power rule to only negative POWERS.

EXAMPLES: 1. 3x–4 = 3 41x

⎛ ⎞⎜ ⎟⎝ ⎠

= 43x

2. –8 2

10 –5n

y–5mx

= 2 5

8 10n yx

–5m


TUTORIALS: http://www.purplemath.com/modules/exponent.htm http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_ExponentsRules.xml http://www.mathsisfun.com/algebra/exponent-laws.html http://www.algebralab.org/practice/practice.aspx

PRACTICES: Simplify each expression using the laws of exponents. Write the answers as positive exponents.

1. 15–4(158) = 2. a7(a8)(a) = 3. (3m4n6)(2m2n) =

4. 6 –3 5

11 –5 5–28a b c7a b c

= 5. (–x5y6)10 =

VI. SIMPLIFYING RADICALS

An expression under a radical sign is in simplest radical form when: 1) There is no integer under the radical sign with a perfect square factor,2) There are no fractions under the radical sign,3) There are no radicals in the denominator

1. n a n b = n ab Example: 2 8 = (2)(8) = 16 = 4

2. an

bn=

ab

n Example: 543

23=

542

3 = 273 = 3

3. amn = anm Example: 6432 = 646 = 2

4. amn = ( an )m Example: 1632 = ( 162 )3 = 43 = 64

5. am/n = amn Example: 163/2 = 1632 = ( 162 )3 = 43 = 64

TUTORIALS: http://www.regentsprep.org/Regents/Math/radicals/Lsimplify.htm http://www.freemathhelp.com/Lessons/Algebra_1_Simplifying_Radicals_BB.htm

PRACTICE: Express the following in simplest radical form.

1) 50 2) 24 3) 169 4) 147

5) 1349

6) 6

277)

3

6


VII. Evaluating Algebraic ExpressionsTo evaluate an algebraic expression: 1. Substitute the given value(s) of the variable(s).2. Use order of operations to find the value of the resulting numerical expression.TUTORIALS: http://www.math.com/school/subject2/lessons/S2U2L3GL.html http://www.purplemath.com/modules/evaluate.htm

PRACTICE: Evaluate.

1. x y2+3z 2

!

"##

$

%&& −2x if x =

12, y = 4, z = −2 2. 12a – 4a2 + 7a3 if a = -3

3. −b + b2 − 4ac2a

if a = 1, b = −4, c = −21 4. 1.2(3)x if x = 3

5. 3(x + y ) −2(x − y )5x + y

if x = 3 and y = 4

VIII. SOLVING LINEAR EQUATIONS

To solve linear equations, first simplify both sides of the equation. If the equation contains fractions, multiply the equation by the LCD to clear the equation of fractions. Use the addition and subtraction properties of equality to get variables on one side and constants on the other side of the equal sign. Use the multiplication and division properties of equality to solve for the variable. Express all answers as fractions in lowest terms.

EXAMPLES:

1. 3(x + 5) + 4(x + 2) = 213x + 15 + 4x + 8 = 21

7x + 23= 217x = -2

x = - 27

2. 2(5x - 4) - lOx = 6x + 3(2x - 5)lOx - 8 - lOx = 6x + 6x - 15

−8 = 12x - 157 = 12x712

= x

3. 23x +5 = 6x −

34

12 23x +5 = 6x −

34

"

#$$

%

&''

8x +60 = 72x −969 = 64x6964

= x

http://www.ck12.org/user:sccmath101/section/Literal-Equations/ ACHS – ENTERING ALGEBRA 2 – SUMMER 2015 WORK Page 10 of 22

TUTORIALS: Solving Linear Equations: http://www.purplemath.com/modules/solvelin.htm Solving Equations: http://www.regentsprep.org/Regents/Math/solveq/LSolvEq.htm

PRACTICE: Solve for the indicated variable:

1. 3n + 1 = 7n – 5 2. 2[x + 3(x – 1)] = 18

3. 5 + 2(k + 4) = 5(k - 3) + 10 4. 6 + 2x(x – 3) = 2x2

5. 23x − 18 = x

66. x −2

3=2x + 1

4

IX. LITERAL EQUATIONS

Formulas are equations that state a fact or rule relating two or more variables. You can solve a formula for any of its variables using the rules for solving equations.

! Some common formulas: Temperature F = 9

5C + 32 Distance d = rt

Circumference of a circle C = πd Area of a triangle A = 12

bh

Density d = mv

! At times it is necessary to move the variables around in the formula without solving the formula. Use the rules of algebra to do this.

EXAMPLE: Formula: A = 12 bh Solve for b

Multiply both sides by 2 2A = bhDivide both sides by h 2A

h = b

Note: When solving literal equations, the result will be an algebraic expression, not a number.

EXAMPLE: Formula: a2 + b2 = c2 Solve for a Subtract b2 from both sides a2 = c2 – b2 Take the square root of both sides a = 2 2c – b

TUTORIAL: http://www.ck12.org/algebra/Linear-Equations/enrichment/Solving-Literal-Equations---Overview/


PRACTICE: Solve each equation for the variable required. Show all work.

1. Simple interest formula: I = 120

pt Evaluate for p p = _____________

2. Perimeter of a rectangle formula: P = 2L + 2W Evaluate for L. L = _____________

3. Volume of a cylinder: V = πr2h Evaluate for r. r = _____________

X. SLOPE OF A LINE

The slope of a line measures the steepness of the line. This is usually referred to as “rise over run.”

Run means how far left or right you move from point to point. On the graph, that would mean a change of x values.

Rise means how many units you move up or down from point to point. On the graph that would be a change in the y values.

The direction of the line will refer to the type of slope it has. Look at the diagram below.

Slope Formula Given Two Points

Given two points (x1, y1) and (x2, y2)

m = riserun

= change in ychangeinx

= 2 1

2 1

y – yx – x

EXAMPLE: Find the slope of the line that passes through the points (–5, 2) and (4, –7).

m = –7 – 24 – (–5)

= –94 + 5

= –99

= –1

TUTORIAL: http://www.ck12.org/algebra/Slope-of-a-Line-Using-Two-Points/

Positive Slope Negative Slope

Zero Slope Numerator = 0

Undefined Slope Denominator = 0


PRACTICE: Find the slope of the line through the given points.

1. (–7, 10 ) and (1, 10) m = ____________

2. (3, 7) and (3, –8) m = ____________

3. (100, 10) and (–3, 50) m = ____________

4. (4, 6) and (7, 7) m = ____________

XI. EQUATION OF LINES

The slope intercept form of a line is y = mx + b where m is the slope and b is the y-intercept. The point slope form of a line is )( 11 xxmyy −=− where m is the slope and )( 1,1 yx is a point on the line. You should be very comfortable using the point slope form of the line. Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes.

EXAMPLE: Find the equations of (a) line parallel and (b) perpendicular to y = −13x + 5 that contains

the point (-2,1)

Solution: Part a (using slope from EXAMPLE above)

1= −13(−2)+ b Using the slope-intercept form with m =

−13 and point (-2,1)

1= 23+ b Multiply -1and -2

1− 23= b Subtract 2

3from both sides

33−23= b Get a common denominator of 3

13= b Combine like terms

y = −13x + 13 This is the equation of the line parallel to the given line that contains (-2,1)

Part b (using slope from EXAMPLE above) b+−= )2(31 Using the slope-intercept form with 3=m and point (-2,1)

b+−= 61 Multiply 3 and -2 b=−− 61 Subtract -6 from both sides

b=7 Subtract 73 += xy This is the equation of the line perpendicular to the given line that contains (-2,1)


EXAMPLE: Find the slope and y-intercept of 6x − 5y =15

Solution: First you must get the line in slope-intercept form.

xy 6155 −=− Subtract 6x form both sides y = 15− 6x

−5Divide by -5

y = 65x −3 Simplify

The slope is m= 65

and the y-intercept is -3

EXAMPLE 3: Find the equation of the line that passes through the point (1,-2) and has slope m= -3.

Solution: Since we are given a point and slope it is easier to use the point slope form of a line. y−−2 = −3(x −1) Substitute into point slope form for (x1,y1) and m y+ 2 = −3x +1 Minus a negative makes + and distribute -3 y = −3x −1 Subtract 2 from both sides

EXAMPLE 4: Find the equation of the line that passes through (-1,3) and (4,5).

Solution: You will need to find slope using m =y2 − y1x2 − x1

m =5−34−−1

=25

Choose one point to substitute back into either the point slope or slope-intercept form of a line.

5= 25(4)+ b Using the slope-intercept form with m =

25

and point (4,5)

5= 85+ b Multiply 2 and 4

5− 85= b Subtract 8

5from both sides

255−85= b Get a common denominator of 5

175= b Combine like terms

y = 25x −17

5 Equation of the line in slope intercepts form.

TUTORIALS: Using the slope and y-intercept to graph lines: http://www.purplemath.com/modules/slopgrph.htm Straight-line equations (slope-intercept form): http://www.purplemath.com/modules/strtlneq.htm Slopes and Equations of Lines: http://www.regentsprep.org/Regents/math/ALGEBRA/AC1/indexAC1.htm Write the equation of a line given two points http://www.ck12.org/algebra/Standard-Form-of-Linear-Equations/


PRACTICES:

Write an equation of a line through the point (a) parallel to the given line and (b) perpendicular to the given line:

1. Point: (1,5) line: 6x-2y=8 2. Point: (-2,2) line: 3x+5y=8

Find the slope and y-intercept of the line:

3. 2x - 3y = 12 4. 4x + y=1

5. y = 2 6. x = -5

Find the equation of a line in slope intercept form:

7. Contains (1,2) and (-2,4) 8. Contains (2,1) and m=4

9. Contains (1,7) and m=0 10. Contains (6,5) and m is undefined

Write the following equations in point slope form

11. Contains (-1,4) and (3,8) 12. Passes through (6,2) and had a y-intercept of 5


XII. GRAPHING LINEAR EQUATIONS

EXAMPLE: Graph the following equation: xy =+ 33

Solution: First you must get the equation in slope-intercept form: 3y = x −3 y = 13x −1

The slope =1/3 and the y-intercept = -1

Plot the point (0,-1) From the first point go Now connect the two points up 1 and over 3 to the right to get the line to get a second point

TUTORIAL: http://www.ck12.org/algebra/Graphs-Using-Slope-Intercept-Form/

PRACTICES: Sketch a graph of the equation:

1. y = - 3x+ 2 2. x = 4 – y 3. y – 1 = 3x + 12


XIII. INTERCEPTS

The x-intercept is where the graph crosses the x-axis. You can find the x-intercept by setting y=0. The y-intercept is where the graph crosses the y-axis. You can find the y-intercept by setting x=0.

EXAMPLE: Find the intercepts for 4)3( 2 −+= xy Solution: X-intercept Y-intercept

4)3(0 2 −+= x Set y=0 4)30( 2 −+=y Set x=0 2)3(4 += x Add 4 to both sides 432 −=y Add 0+3 )3(2 +=± x Take square root of both sides 49 −=y Square 3

)3(2 +=− x or )3(2 += x Write as 2 equations 5=y Add 4 to both sides x=− 5 or x=−1 Subtract 3 from both side

TUTORIAL: http://www.khanacademy.org/math/algebra/linear-equations-and-inequalitie/graphing_with_intercepts/v/x--and-y-intercepts http://www.mathsisfun.com/algebra/finding-intercepts-equation.html

PRACTICE: Find the intercepts for each of the following. 1. 3 2y x= − + 2. y = x3 + 2

3. 2

2

3(3 1)x xyx+

=+

4. 2 3 4y x x= −

XIV. OPERATIONS WITH POLYNOMIALS

To add or subtract polynomials, just combine like terms. To multiply polynomials, multiply the numerical coefficients and apply the rules for exponents.

EXAMPLE:

1. (x2 + 3x - 2) - (3x2 - x + 5) 2. 4(5x2 + 3x - 4) + 3(-2x2 - 2x + 3)x2 + 3x - 2 - 3x2 + x -5 20x2 + 12x - 16 - 6x2 - 6x + 9 -2x2 + 4x – 7 14x2 + 6x - 7

3. 3x(2x + 5)2 4. (4x - 5)(3x + 7)3x(4x2 + 20x + 25) 12x2 + 28x - 15x - 3512x3 + 60x2 + 75x 12x2 + 13x - 35


TUTORIALS: Polynomials (adding & subtracting): http://www.purplemath.com/modules/polyadd.htm,

http://www.regentsprep.org/Regents/math/ALGEBRA/AV2/indexAV2.htm Polynomials (multiplying): http://www.purplemath.com/modules/polymult.htm,

http://www.regentsprep.org/Regents/math/ALGEBRA/AV3/indexAV3.htm

PRACTICE: Perform the indicated operations and simplify:

1) (7x2 + 4x - 3) - (-5x2 - 3x + 2) 2) (7x - 3)(3x + 7)

3) (4x + 5)(5x + 4) 4) (n2 + 5n + 3) + (2n2 + 8n + 8)

5) (5x2 - 4) – 2(3x2 + 8x + 4) 6) -2x(5x + 11)

7) (2m + 6)(2m + 6) 8) (5x – 6)2


XV. FACTORING POLYNOMIALS

EXAMPLE:

Factoring out the GCF Difference of Squares Perfect Square Trinomial a) 6x2 + 21x b) x2 - 64 c) x2 - 10x + 25

3x(2x + 7) (x - 8)(x + 8) (x-5)(x-5) or (x – 5)2

Trinomial d) 3x2 + 7x + 2

Trinomial e) 2x2 - 13x + 15

Trinomial f) 6x2 + x – 1

(3x + l)(x + 2) (2x - 3)(x - 5) (3x - 1)(2x + 1)

TUTORIALS: Factoring Trinomials (skip substitution method): http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut28_facttri.htm Factoring Polynomials (video): http://www.youtube.com/watch?v=uoEoWzHXaJ8 Factoring a Trinomial: http://www.algebrahelp.com/lessons/factoring/trinomial/ Factoring: http://www.regentsprep.org/Regents/Math/math-topic.cfm?TopicCode=factor

PRACTICES:

Factor Completely.

1. 16y2 + 8y 2. 18x2 - 12x 3. 6m2 - 60m + 10

4. 6y2 - 13y – 5 5. 20x2 + 31x - 7 6. 12x2 + 23x + 10

7. x2 - 2x – 63 8. 8x2 - 6x - 9 9. x2 – 121


XVI. SOVING SYSTEMS OF LINEAR EQUATIONS

EXAMPLE: Solve for x and y:

x = 2y + 5 3x + 7y = 2

Using substitution method:

3(2y + 5) + 7y = 2 6y + 15 + 7y = 2

13y = -13 y = -1

x = 2(-1) + 5 x=3

Solution: (3, -1)

Solve for x and y: 3x + 5y = 1 2x + 3y = 0

Using linear combination (addition/ subtraction) method:

3(3x + 5y = 1) -5(2x + 3y = 0)

9x + 15y = 3 -l0x - 15y = 0

-1x = 3 x = -3

2(-3) + 3y = 0 y=2

Solution: (-3, 2)

Solve each system of equations by either the substitution method or the linear combination (addition/ subtraction) method. Write your answer as an ordered pair.

TUTORIALS: Solve systems of linear equations: http://www.regentsprep.org/regents/math/math-topic.cfm?TopicCode=syslin Solve systems of equations (video): http://www.youtube.com/watch?v=qxHCEwrpMw0 Systems of Linear Equations: http://www.purplemath.com/modules/systlin1.htm

PRACTICE: Solve the following systems of linear equations:

1. y = 2x + 4 2. 2x + 3y = 6-3x + y = - 9 -3x + 2y = 17


XVII. GRAPHING INEQUALITIES

EXAMPLE: Graph the solution to y ≤ 2x + 3.

Just as for number-line inequalities, the first step is to find the "equals" part. For two-variable linear inequalities, the "equals" part is the graph of the straight line; in this case, that means the "equals" part is the line y = 2x + 3:

The illustration on the right has the "or equal to" part (it's just the line) graphed; now the "y less than" part needs to be graphed. In other words, this is where it is shaded on one side of the line or the other. Less than is below the line and greater than is above the line. So shade in below the line for this one:

And that's all there is to it: the side shaded is the "solution region" they want.

EXAMPLE: Graph the solution to 2x – 3y < 6.

First, solve for y: 2x – 3y < 6 –3y < –2x + 6y > ( 2/3 )x – 2 (Note the flipped inequality sign in the last line. Do not forget to flip the inequality if multiplying or dividing through by a negative!) Now, find the "equals" part, which is the line y = ( 2/3 )x – 2. It looks like this:

But this exercise is what is called a "strict" inequality. That is, it isn't an "or equals to" inequality; it's only "y greater than". In the case of these linear inequalities, the notation for a strict inequality is a dashed line. So the border of the solution region actually looks like this:

By using a dashed line, the border isn't included in the solution. Since this is a "y greater than" inequality, I want to shade above the line, so the solution looks like this:

Here the straight line means the equations is using the ≤ or ≥

The graph is shaded down because the equation is using the less than or equal to sign ( ≤ )

This line is dotted because you are strictly using just the greater than or just the less than sign ( >,< )

This graph is shaded upwards because the equation is using the greater than or equal to sign ( ≥ )


TUTORIALS: http://www.webmath.com/plotineq.html http://www.coolmath.com/algebra/08-lines/index.html

PRACTICES:

1. x + y ≤ 5 2. 8x + 12y ≥ 24

3. 12x + 12y ≥ 36 4. 2x + y ≥ 4


XVIII. QUADRATIC EQUATIONS

An equation in the form ax2 + bx + c = 0 where a ≠ 0 is called a quadratic equation. Its related quadratic function is y = ax2 + bx + c. If you graph the related quadratic function, the solutions of the quadratic equation are x-values where the graph crosses the x-axis. A linear equation can have only one solution. However, a quadratic equation can have 2, 1, or 0 real-number solutions.

EXAMPLE:

The related function of −x2 + 4 = 0 is y = −x2 + 4. The graph of y = −x2 + 4 is shown below.

The related function of x2 − 2x + 1 = 0 is y = x2 − 2x + 1. The graph of y = x2 − 2x + 1 is shown below.

The related function of x2 − x + 2 = 0 is y = x2 − x + 2. The graph of y = x2 − x + 2 is shown below.

TUTORIAL: http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html http://www.purplemath.com/modules/solvquad5.htm

PRACTICE: Solve each equation by graphing the related function. If the equation has no real-number solution, write no solution.

1. x2 + 3 = 0 2. x2 + 4x + 4 = 0 3. x2 + x − 2 = 0

Documents

Algebra II Summer Packet 2014 - Abraham Clark High Schoolachs.roselleschools.org/UserFiles/Servers/Server_3154010/File/News... · When solving math problems with more than one operation,