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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)