algebra 2 cpe summer packet - rihrhsd · algebra 2 cpe summer packet 2 name!_____! to the students:...

Algebra 2 CPE Summer Packet 1

RAMAPO-‐INDIAN HILLS SCHOOL DISTRICT Dear Ramapo-‐Indian Hills Student: Please find attached the summer packet for your upcoming math course. The purpose of the summer packet is to provide you with an opportunity to review prerequisite skills and concepts in preparation for your next year’s mathematics course. While you may find some problems in this packet to be easy, you may also find others to be more difficult; therefore, you are not necessarily expected to answer every question correctly. Rather, the expectation is for students to put forth their best effort, and work diligently through each problem. To that end, you may wish to review notes from prior courses or on-‐line videos (www.KhanAcademy.com, www.glencoe.com, www.youtube.com) to refresh your memory on how to complete these problems. We recommend you circle any problems that cause you difficulty, and ask your teachers to review the respective questions when you return to school in September. Again, given that math builds on prior concepts, the purpose of this packet is to help prepare you for your upcoming math course by reviewing these prerequisite skills; therefore, the greater effort you put forth on this packet, the greater it will benefit you when you return to school. Please bring your packet and completed work to the first day of class in September. Teachers will plan to review concepts from the summer packets in class and will also be available to answer questions during their extra help hours after school. Teachers may assess on the material in these summer packets after reviewing with the class. If there are any questions, please do not hesitate to contact the Math Supervisors at the numbers noted below. Enjoy your summer! Ramapo High School Michael Kaplan [email protected] 201-‐891-‐1500 x2255 Indian Hills High School Amanda Zielenkievicz [email protected] 201-‐337-‐0100 x3355


NAME __________________________________________________________________________ To the students: The following set of review problems were designed to prepare you for your Algebra 2 CP/CPE course. You can either print out the problems or complete them on a separate piece of paper. Please bring the packet and your completed work on the first day of school in September.

Thank you.

Practice 1-‐4: Solve each equation. 1. |x + 12| = 9 2. |4b+1|-‐8 = 1 3. |8 + y| = 2y – 3

v Section 1-‐4 Solving Absolute Value Equations Example 1: |x+5|-‐3 =8 |x+5| = 11 x+ 5 = 11 or x + 5 = -‐11 x = 6 or x = -‐16 Example 2: |3+x| = 2x – 5 3+x = 2x – 5 or 3+ x = -‐(2x-‐5) 3 = x-‐5 3+x = -‐2x + 5

x= 8 3x = 2 x = 2/3 – check answers, 2/3 is an extraneous solution!


Practice 1-‐5: Solve each inequality. Graph the solution on a number line.

1. 3. 4𝑛 − 5 𝑛 − 3 > 3 𝑛 + 1 − 4

€

4x − 32

≥ −3.5

€

9(2r − 5) − 3 < 7r − 4

v Section 1-‐5: Solving Inequalities Inequalities: statements comparing two quantities. For any two real numbers, a and b, exactly one of the following statements is true:

𝑎 < 𝑏 𝑎 = 𝑏 𝑎 > 𝑏

The solution to an inequality is the set of numbers that make the inequality true. The procedures for solving an inequality are the same as those for solving an equation, except any time the inequality is multiplied or divided by a negative number, the inequality sign must be flipped. When graphing an inequality on a number line, the starting point is noted with either or . The open circle is used when the value is strictly less than or greater than (< or >). The closed circle is used when the value is less than or equal to or greater than or equal to ( ).


Practice 2-‐1: State the domain and range of each relation. Then determine whether each relation is a function. If it is a function, determine if it is one-‐to-‐one, onto, both, or neither. 1. {(-‐6, -‐1), (-‐5, -‐9), (-‐3,-‐7), (-‐1, 7), (6,-‐9)} 2. {(2, -‐2), (-‐1, -‐1), (-‐2, 0), (-‐1, 0), (2, 2)} Find each value, if 3. 𝑓(−3) 4. 𝑔(5) 5. 𝑔(2) ∙ 𝑔(3) 6. g(2a) 7. f(x+1)

€

f (x) = −2x + 4, and g(x) = x 3 − x

v Section 2-‐1: Relations and Functions Function: a relation in which each element of the domain is paired with exactly one element in the

range. One-‐to-‐one function: Each element of the domain pairs to exactly one unique element of the range.

Ex. (1, 2), (2, 3), (3, 4) Onto function: Each element of the range corresponds to an element of the domain. Ex. (1,2), (2,

2), (3, 3), (4, 6) One-‐to-‐one and Onto: Each element of the domain is paired to exactly one element of the range, and

each element of the range element corresponds to a unique element of the domain. Example: State the domain and range of the relation {(-‐4, -‐2), (-‐3, 1), (0, -‐2), (1, 2), (3, 3)}. Then determine whether each relation is a function. If it is a function, determine if it is one-‐to-‐one, onto, both, or neither. ANSWER: Domain: Range: function, one-‐to-‐one Equations that represent functions are often written in function notation. For example, y=2x+3 can be written as f(x) = 2x + 3. Function notation emphasizes the fact that the y values, the dependent variables, depend on the values of x, the independent variable. Example: Given the function f(x) = x2 + 2 , find f(5) and f(9). ANSWERS: ,


Practice 2-‐3: Find the slope of the line that passes through each pair of points 1. (5, 10) and ( -‐1, -‐2) 2. (6, 4) and ( 3, 4) 3. (1, 9) and ( 0 , 6) Determine if the 2 lines are parallel, perpendicular are neither 4) line going through points (4,3) and (1, -‐3) and line through points (1, 2) and (-‐1, 3) 5) line through points (1, 5) and (3, 7) and line through (-‐1, -‐4) and (1, -‐2)

v Section 2-‐3: Rate of Change and Slope Slope: the ratio of the change in y-‐ coordinates to the corresponding change in x-‐ coordinates. The slope of a line is the same as its rate of change. Suppose a line passes through and , then slope=

.

Facts about slope:

Vertical lines have Undefined slope or No Slope Horizontal lines have 0 slope.

Parallel lines have the same slope. Perpendicular lines have slopes that are opposite reciprocals.


Practice 2-‐4: Write an equation in slope-‐intercept form of the equation given the following information. 1. Slope=3, and passes through (0, -‐6) 2. Passes through (-‐2, 5) and (3, 1) 3. Passes through (-‐1, -‐2) and (-‐3, 1) 4. x-‐intercept = 2, y-‐intercept = 5

v Section 2-‐4: Writing Linear Equations Slope Intercept form: , where m is the slope and b is the y-‐intercept.

Example1: Write an equation in Example 2 : Write an equation in slope-‐intercept form for the line slope-‐intercept form of the line that has slope -‐3/2 and passes through (-‐1, 4) and (-‐4, 5). through (-‐4, 1)


v v 2-‐8 Graphing Linear and Absolute Value Inequalities

To graph a linear inequality, graph the linear equation associated with the inequality. You may need to convert to y=mx+b form. If < or > draw line with a dotted line. If ≤ 𝑜𝑟 ≥ then draw a solid line. Then shade the appropriate area. Example: Graph 𝑥 − 2𝑦 < 4

ANSWER. Graph y > 12x − 2 with a dotted line. Shade above (see graph).

To graph an absolute value inequality follow the same rules for linear equality. Graph the absolute value, using the appropriate dotted or solid line. Then shade appropriately. Example: Graph. 𝑦 ≥ 𝑥 − 2 ANSWER: Graph y = x − 2 , solid line, shade above (see graph). Remember that the shaded region represents the values that make the statement true. The shaded region is your solution set. Practice 2-‐8: Graph each inequality. 1. y ≥ 2x-‐3 2. 3.

€

x − 3y < 6

€

y + 3 ≥ x +1


Practice 4.1: For each quadratic function complete the following: a. State the direction of the opening (up or down) b. Find the equation for the axis of symmetry c. Find the coordinates of the vertex

1. 𝑦 = −𝑥! − 4𝑥 + 5 2. 𝑦 = 3𝑥! − 12𝑥 − 2

v Section 4.1 – Quadratics A quadratic function in standard form is given by the equation 𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙+ 𝒄, where a, b, and c are real numbers and 𝑎 ≠ 0. Remember when graphing a quadratic equation it creates a parabola (the ‘U’ shape)

Important facts:

• The value of 𝑎 determines whether the parabola will be opening up or down. If 𝑎 is a positive number the parabola will be opening up and if 𝑎 is a negative number the parabola will be opening down.

• The axis of symmetry can be found using the equation 𝑥 = − !!!. Remember the axis of

symmetry is a line that cuts the parabola exactly in half.

• The vertex is the parabola’s maximum or minimum point depending on which way the parabola is opening. The value you found for the axis of symmetry is the x-‐coordinate of the vertex, to find the y-‐coordinate you plug in the x – value into the equation and evaluate. !− !

!!, 𝑓 !− !

!!!!


Solve each equation by factoring.

1. 20𝑥! + 15𝑥 = 0 2. 6𝑥! + 18𝑥! = 0

3. 𝑥! − 16𝑥 + 64 = 0 4. 𝑥! − 4𝑥 − 21 = 0

5. 𝑥! − 7𝑥 + 12 = 0 6. 𝑥! − 25 = 0

7. 2𝑥! − 5𝑥 + 2 = 0

v 4.2 – Solving Quadratic equations by factoring

Factoring is used to represent quadratic equations in the factored form of a(x – p)(x – q) = 0, and solve this equation.

Factoring GCF In a quadratic equation you may factor out the Greatest Common Factor. Ex 1: 16𝑥! + 8𝑥 = 0. GCF = 8x

8x(2x + 1) = 0 Zero Product Rule

8x = 0 or 2x +1 = 0

x= 0 or x = -‐1/2

𝑎𝑥! + 𝑏𝑥 + 𝑐

Factoring where a = 1

𝐄𝐱 𝟐: 𝑥! + 9𝑥 + 20 = 0 To Factor we want to find two numbers that multiply to 20 and add to 9.

(x+ 5)(𝑥 + 4) = 0 5 +4 = 0 and 5 *4 = 20

(𝑥 + 5) = 0 𝑜𝑟 (𝑥 + 4) = 0 Zero Product Rule

𝑥 = −5 𝑜𝑟 𝑥 = −4


Practice 3.1:

1. 𝑦 = !!𝑥 + 2

4𝑥 − 10𝑦 = −20

2. 6𝑥 − 4𝑦 = 16 2𝑦 = −2𝑥 + 2

v 3.1: Solving Systems by Graphing Example: Solve the system of equations by graphing. 𝑦 = 2𝑥 − 1 𝑥 + 𝑦 = 5 Step 1: Graph each line. The second equation here needs to be changed to 𝑦 = 𝑚𝑥 + 𝑏. 𝑥 + 𝑦 = 5 −𝑥 − 𝑥 𝑦 = −𝑥 + 5 Step 2: Find the point of intersection. ‘ If the lines overlap (same line) – infinitely many solutions If lines are parallel – no solution Solution: (2, 3)


Solve by substitution. 1. −5𝑥 + 3𝑦 = 12 2. 𝑥 – 4𝑦 = 22 𝑥 + 2𝑦 = 8 2𝑥 + 5𝑦 = −21

v 3.1: Solving Systems by Substitution Example: Solve the system of equations by substitution. 𝑦 − 3𝑥 = −3 −2𝑥 − 4𝑦 = 26 Step 1: Solve for a variable for either equation. (It is ideal to pick the variable with a coefficient of 1) 𝑦 − 3𝑥 = −3 +3𝑥 + 3𝑥 𝑦 = 3𝑥 − 3 Step 2: Plug the expression 3𝑥 − 3 in for 𝑦 of the OTHER equation. −2𝑥 − 4𝑦 = 26 −2𝑥 − 4(3𝑥 − 3) = 26 Step 3: Solve for 𝑥. −2𝑥 − 4(3𝑥 − 3) = 26 −2𝑥 − 12𝑥 + 12 = 26 −14𝑥 + 12 = 26 −14𝑥 = 14 𝑥 = −1 Step 4: Plug in x for either equation to solve for y. 𝑦 = 3𝑥 − 3 𝑦 = 3(−1) − 3 𝑦 = −6 Final Solution: (−1,−6)


Solve using elimination:

1. 2x +3y = 56x + 9y =15

2. 5x +3y = 5215x + 9y = 54

v 3.1: Solving Systems by Elimination Example: Solve the system of equations by elimination 4𝑥 − 3𝑦 = 25 −3𝑥 + 8𝑦 = 10 Step 1: Decide which variable you want to eliminate and find the LCM of the two coefficients for that variable. Eliminate 𝑥 à 4 and -‐3 have an LCM of 12 Step 2: Multiply each equation by the number that will make the x-‐terms have a coefficient of 12. One must be negative and the other must be positive. 3(4𝑥 − 3𝑦 = 25) à 12𝑥 − 9𝑦 = 75 4(−3𝑥 + 8𝑦 = 10) −12𝑥 + 32𝑦 = 40 Step 3: Add the columns of like terms. 12𝑥 − 9𝑦 = 75 −12𝑥 + 32𝑦 = 40 23𝑦 = 115 𝑦 = 5 Step 4: Plug in 𝑦 for either equation to solve for 𝑥. 4𝑥 − 3𝑦 = 25 4𝑥 − 3(5) = 25 𝑥 = 10 Final Solution: (10,5)


Square Roots and Simplifying Radicals Simplify the following radicals. Remember, no radicals can be left in the denominator.

1. 23

2. 32

3. 50 ⋅ 10 4. 16 ⋅ 25 5. 98x3y6 6. 56a2b4c5

7. 8149

8. 10p3

27

9. 45+ 2 3

10. 3 52− 2

● Product Property for two numbers, ,

● Quotient Property for any numbers a and b, where ≥ 0,

Ex. 1 Simplify √45 = √9 ⋅ 5 = √3 ⋅ 3 ⋅ 5= 3√5

Ex. 2 Simplify

Ex. 3 Simplify !2516= √25

√16= 54

Ex. 4 Simplify =


Exponents

Simplify the following. Remember there should be no negative exponents.

1. −4x5y−2 2. 35

33

3. a−2b3

c−4d−1 4. 7a3b−1( )

0

5. 6x3( )2 6. 1

2−4

7. 2x3y2!

"#

$

%&

3

8. 52x6

13x−7

Rules of Exponents

● ●

● ●

●

●

Ex. 1

Ex. 2

Ex. 3

algebra 2 cpe summer packet - rihrhsd · algebra 2 cpe summer packet 2 name!_____! to the students:...

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