Name________________________

CP1 Math 2 Cumulative Exam Review Deductive Geometry (Ch. 6) • Writinggeometricproofs• Trianglecongruenceanditsuses• Parallellinesandtheiruses• Quadrilateralsandtheirproperties Exponents & Radicals (Ch. 1) Arithmetic with Radicals

Adding, subtracting, multiplying, dividing – includes rationalizing the denominator Simplifying radicals Solve simple equations ( 3 + 48 = k 3 ) Classify numbers (real, rational, irrational, integer, non-real, counting)

Properties of Exponents (use in both directions) Solving exponential equations (change to same base and solve) Quadratics & Polynomials (Ch. 2 & 3) Solving Quadratic Equations 3 methods to solve quadratics equations:

• Factoring o Types of quadratics we factored: difference of 2 perfect squares, monics and non-

monics (with and without GCF’s) o Methods: factor out GCF; sums and products, splitting the middle, z-substitution

• Completing the square • Quadratic formula

Connections between equations and Graphs of Quadratics

• x-intercepts, roots, zeros are solutions to the quadratic when equal to 0 • y-intercept à point on y-axis (i.e. f(0)) • line of symmetry à average of the roots • vertex à maximum or minimum of parabola, (avg roots, f(avg roots)) or complete the

square • other points à use symmetry of graph to find additional point(s)

Formats for quadratic functions & ability to move between forms:

• Standard Form: y = ax2 + bx + c • Factored Form: y = a(x-r1)(x – r2) • Vertex Form: y = a(x – h)2 + k

Standard à vertex (complete square); Standard à factored (factor) factored or vertex à standard (multiply and simplify)

Applications of quadratics Types: Projectile Motion & Maximizing area

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Part I – Deductive Geometry (Chapter 6) 1. ∠𝐸 is a right angle. Find ∠𝐶. 2. Find the value of 𝑥:

3. ΔABC is isosceles with 𝐴𝐵 ≅ 𝐴𝐶. ∠𝐴 = 50°. The bisectors of ∠𝐵 and ∠𝐶 meet at D. Find the measure of ∠𝐵𝐷𝐶. 4. Given: 𝐴𝑋 ≅ 𝐵𝑋; ∠𝐴𝐶𝑋 ≅ ∠𝐵𝐷𝑋 Prove: 𝐴𝐶 ∥ 𝐷𝐵 5. Given: 𝐵𝐶𝐷𝐸 is an isosceles trapezoid with legs 𝐷𝐸 and 𝐶𝐵. Prove: ∠1 ≅ ∠2 6. Given: ∠3 ≅ ∠4; ∠1 ≅ ∠2 Prove: 𝐴𝑋 ≅ 𝐵𝑌

BC

D

X

A

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7. Given: 𝐴𝐷 is an altitude of ∆𝐴𝐵𝐶 A 𝐴𝐷 bisects ∠𝐵𝐴𝐶 Prove: 𝐴𝐷 is the median to 𝐵𝐶 B C D 8. Given: Parallelogram 𝐴𝐵𝐶𝐷 with diagonals intersecting at point 𝑂. Prove: 𝑋𝑂 ≅ 𝑌𝑂

9. Ronda and Stephanie are working to prove that 𝐴𝐶 is parallel to 𝐵𝐷 in the diagram to the right. Ronda’s argument is:

𝐴𝐶 ∥ 𝐵𝐷 because: ∡ 1 = ∡2+ ∡3 and ∡1 = ∡4+ ∡5 by the exterior angle sum theorem. Therefore ∡3 = ∡4. Since these are alternate interior angles and they are congruent we know by AIP that the lines are parallel.

Where is the incorrect step in Ronda’s proof? 10. Given: 𝐽𝑈𝑀𝑃 is a parallelogram; 𝑆,𝑇,𝐴,𝑎𝑛𝑑 𝑅 are

midpoints.

a. Using geometric relationships explain why 𝑆𝑇𝐴𝑅 is not necessarily a rhombus. b. What must be true in order for 𝑆𝑇𝐴𝑅 to be a rhombus?

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Answers 1. ∠𝐶 = 26° 2a. 𝑥 = 100° b. 𝑥 = 64° 3. ∠𝐵𝐷𝐶 = 115° 4.

Given Prove 1. 𝐴𝑋 ≅ 𝐵𝑋; 𝐶𝑋 ≅ 𝐷𝑋 1. Given 2. . ∠𝐴𝑋𝐶 ≅ ∠𝐵𝑋𝐷 2.VAT 3. . Δ𝐴𝑋𝐶 ≅ Δ𝐵𝑋𝐷 3. SAS 4. ∠𝑋𝐴𝐶 ≅ ∠𝑋𝐵𝐷 4. CPCTC 5. 𝐴𝐶 ∥ 𝐷𝐵 5. AIP 5.

Given Prove 1. 𝐵𝐶𝐷𝐸 is an isosceles trapezoid with legs 𝐷𝐸 and 𝐶𝐵

1. Given

2. 𝐴𝐷 ∥ 𝐸𝐵 2. Definition of a trapezoid 3. ∠1 ≅ ∠𝐶𝐵𝐸 3. The base angles of an isosceles

trapezoid are congruent 4. ∠𝐶𝐵𝐸 ≅ ∠2 4. PAI 5. ∠1 ≅ ∠2 5. Transitive property 6.

Given Prove 1. ∠3 ≅ ∠4; ∠1 ≅ ∠2 1. Given 2. ∠𝐴𝑋𝑌 ≅ ∠𝐵𝑌𝑋 2. Addition Property 3. 𝑋𝑌 ≅ 𝑌𝑋 3. Reflexive Property 4. Δ𝐴𝑋𝑌 ≅ Δ𝐵𝑌𝑋 4. ASA 5. 𝐴𝑋 ≅ 𝐵𝑌 5. CPCTC 7.

Given Prove 1. 𝐴𝐷 is an altitude of ∆𝐴𝐵𝐶 𝐴𝐷 bisects ∠𝐵𝐴𝐶

1. Given

2. 𝐴𝐷 ⊥ 𝐵𝐶 2. Definition of altitude 3. ∠𝐴𝐷𝐵 𝑎𝑛𝑑 ∠𝐴𝐷𝐶 are right angle 3. Perpendicular lines form right angles 4. ∠𝐴𝐷𝐵 ≅ ∠𝐴𝐷𝐶 4. All right angles are congruent 5. ∠𝐵𝐴𝐷 ≅ ∠𝐶𝐴𝐷 5. Definition of angle bisector 6. Δ𝐴𝐵𝐷 ≅ Δ𝐴𝐶𝐷 6. ASA 7. 𝐵𝐷 ≅ 𝐶𝐷 7. CPCTC 8. D is the midpoint of 𝐵𝐶 8. Definition of midpoint 9. 𝐴𝐷 is the median to 𝐵𝐶 9. Definition of median

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8. Given Prove

1. Parallelogram 𝐴𝐵𝐶𝐷 with diagonals intersecting at point 𝑂.

1. Given

2. 𝐴𝐶 and 𝐷𝐵 bisect each other 2. The diagonals of a parallelogram bisect each other

3. 𝐴𝑂 ≅ 𝐶𝑂 3. Definition of segment bisector 4. ∠𝐴𝑂𝑋 ≅ ∠𝐶𝑂𝑌 4. VAT 5. 𝐴𝐵 ∥ 𝐷𝐶 5. Opposite sides of a parallelogram are

parallel (Definition of parallelogram) 6. ∠𝑌𝐶𝑂 ≅ ∠𝑋𝐴𝑂 6. PAI 7. Δ𝐴𝑂𝑋 ≅ Δ𝐶𝑂𝑌 7. ASA 8. 𝑋𝑂 ≅ 𝑌𝑂 8. CPCTC 9. The step that is incorrect is: Therefore ∡3 = ∡4. Though it is true that ∡ 1 = ∡2+ ∡3 and ∡1 = ∡4+ ∡5, making ∡4+ ∡5 = ∡2+ ∡3 by the transitive property. It is not true that any of the individual angles need to be equal to each other. For example 20 + 70 = 40 + 50 is true. Both sums equal 90, but none of the numbers in the equation are equal. 10. a. The opposite sides and angles are equal in a parallelogram, so halves of equal sides would also be equal. This would make Δ𝑆𝑈𝑇 ≅ Δ𝐴𝑃𝑅 by SAS so that 𝑆𝑇 ≅ 𝐴𝑅 by CPCTC. Also, Δ𝑆𝐽𝑅 ≅ Δ𝐴𝑀𝑇 by SAS so that 𝑆𝑅 ≅ 𝐴𝑇 by CPCTC. This does not mean that 𝑆𝑅 ≅ 𝐴𝑇 ≅ 𝑆𝑇 ≅𝐴𝑅 which would have to be true to make the STAR a rhombus. b. In order for STAR to be a rhombus, ∡𝑈 ≅ ∡𝑃 ≅ ∡𝐽 ≅ ∡𝑀. In other words, JUMP must be a rectangle. Part II – Exponents (Chapter 1) 1. For each equation, find the value of k that satisfies the equation.

a. 98 = 𝑘 2

b. 125 = 𝑘 5

c. 5+ 125 = 𝑘 5

d. 2 6+ 150 = 𝑘 6

2. Write each expression as a single power of x. Simplify numerical exponents when the

exponent is 4 or less.

a. 𝑥! ! !! b. 𝑥!" ∙ !!!

! c. 𝑥

! ∙ 𝑥! !

d. 3𝑥!𝑦! ! e. 3 ! ∙ (5)! f. !!!!!

!!!!

3. If 𝑔(𝑥) = 𝑥, find 𝑔(32) and 𝑔(45). Simplify your answers as much as possible.

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4. Write each radical in simplified or standard form.

a. 121 b. !! c. !

! d. !

! !

5. If 𝑓(𝑥) = 9!, find 𝑓(− !!)

6. Evaluate each expression.

a. 4

!! ∙ 4

!!

b. 25

!!!

c. 27!! ∙ 27

!! d. 8

!!!!!

7. Determine whether each of the following numbers is rational or irrational. Explain your

answer.

a. 16.33

b. 52− 3

c. !.!"#!

d. 13

Answers 1. a. k = 7 b. k = 5 c. k = 6 d. k = 7 2. a. 𝑥!! b. 𝑥!! c. 𝑥!" d. 27𝑥!"𝑦! e.15! f.3𝑥!𝑧! 3. 𝑔(32) = 4 2 𝑔(45) = 3 5 4. a. 11 b. !

! c. ! !

! d. !

!

5. 𝑓(− !

!) = !

!"

6. a. 4 b. 5 c. 3 d. !

!

7. a. rational, can be written as !"##

!"" b. rational, can be written as !

!

c. rational, can be written as !!"

d. irrational, non-perfect square under radical.

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Part III - Quadratics (Chapters 2 & 3) Directions – Problems 1-5 should be completed WITHOUT a calculator. 1. Here is a quadratic in vertex form: 𝑦 = 4(𝑥 − 2)! − 100 Sketch a graph clearly labeling: The vertex The zeros The axis of symmetry The y-intercept The other point on the graph with the same y-value as the y-intercept. 2. Here is a quadratic in factored form:𝑦 = −4(𝑥 − 2.5)(𝑥 + 8.5) Sketch a graph clearly labeling: The vertex The zeros The axis of symmetry The y-intercept The other point on the graph with the same y-value as the y-intercept. 3. A parabola has x-intercepts at 4 and −3. The y-intercept is at (0, 6). Write an equation in factored form 𝑦 = 𝑎(𝑥 − 𝑟)(𝑥 − 𝑠) to represent the function described.

Hint for finding a: You know the zeros so plug them in, and then figure out the a value by temporarily plugging in the given y-intercept.

4. Solve by any method you choose.

a. 3𝑥! − 48 = 0 b. 𝑥! + 4𝑥 = −4 c. −𝑥! = −4𝑥 d. 5𝑥! − 20𝑥 = 60 e. 3(𝑥 + 5)! = 9

5. Write an equation in normal form for a quadratic function with roots of 3+ 2 and 3− 2 Directions – You may use a calculator to complete problems 6 – 8. 6. Solve by completing the square: 2𝑥! − 20𝑥 + 14 = 0 7. Put in vertex form and identify the vertex: a. 𝑦 = 3𝑥! + 18𝑥 − 11 b. 𝑦 = 5𝑥! + 12𝑥 − 8

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8. A ball is thrown upward from a flat surface. Its height in feet as a function of time since it was thrown (in seconds) is given by the equation ℎ(𝑡) = −16𝑡! + 34𝑡 + 25.

a. Evaluate ℎ(1), and explain what it means in the context of this problem. b. What was the highest the ball got, and when did it reach that height? c. At what time does the ball land?

9. The diagram at the right is a graph of f(x) = –2x2 + 20x – 32. ∆ABC has vertices at the x intercepts and vertex of the parabola.

a. Find the area of ∆ABC. b. Find the location of point D that makes quadrilateral ABCD a rhombus. c. Find the quadratic function whose vertex, point E, would make quadrilateral ABCE a kite.

Answers: 1. Vertex: (2,−100) Zeros: −3 and 7 Line of symmetry: 𝑥 = 2 y-intercept: (0,−84) Other point: (4,−84) 2. Vertex: (−3, 121) Zeros: −8.5 and 2.5 Line of symmetry: 𝑥 = −3 y-intercept: (0, 85) Other point: (−6, 85) 3. 𝑦 = −0.5(𝑥 − 4)(𝑥 + 3) 4. a. 4, -4 Solving directly is the easiest way b. -2 This is a perfect square already. c. 0, 4 Common factor factoring works d. 6, -2 Factor out a GCF and then use sum/product e. -5 Solve directly by taking the square root of both sides. 5. 𝑓(𝑥) = 𝑥! + 6𝑥 + 7 6. 𝑥 = 5± 3 2 7. a.. 𝑦 = 3(𝑥 + 3)! − 38 Vertex at (−3,−38)

b. 𝑦 = 5(𝑥 + 1.2)! − 15.2 Vertex at (−1.2,−15.2) 8. a. ℎ(1) = 43 , this means that after 1 second the ball is 43 feet high b. 3.4 seconds, 82.8 feet c. 7.47 seconds 9. a. 54 u2 b. (5, –18) c. y = –2(x – 5)2 – 3 (y-value of vertex may vary)