WCCUSD Algebra 2 Benchmark 2 Study Guide...WCCUSD Algebra 2 Benchmark 2 Study Guide Page 7 of 14...

WCCUSD Algebra 2 Benchmark 2 Study Guide

Page 1 of 14 MCC@WCCUSD 01/17/13

1 a) Simplify

Method 1: Expansion Method 2: Exponent Properties The Power of a power rule for exponents says that

(b x ) y = bx • y

b) Simplify Method 1: Expansion Method 2: Exponent Properties

3.0/A.APR.1

1´ You try:

a) Simplify b) Simplify

Expand Commute like factors

Expand and find factors of one

Multiply like terms



2 What expression is equivalent to 1634 ?

Method 1: Method 2:

12.0/A.APR.1

2´ You try:

What expression is equivalent to 843 ?



3 What is the solution to log4 x = −3 ? Method 1: Change to exponential form

The logarithm of a number x with base b is the solution y for the equation by = x

logb x = y such that by = x Method 2: Exponentiation Method 3: Base change

We will use the base change formula for logarithms which states

14.0/A.SSE.3f

3´ You try: What is the solution to log2 y = 4 ?



4 Express as a single logarithm logx 3+ logx 21

Method 1: Product Identity

The product identity for logarithmic functions states that logb(x • y) = logb(x) + logb(y) Method 2: Exponentiation

Let N be the value of logx 3 + logx 21 N is such that N = logx 3 + logx 21 ∴ logx 3 + logx 21 = logx 63

14.0/A.SSE.3e

4´ You try: Express as a single logarithm

log410+ log4 6



5 What are the solutions to the equation x2 = 2x +15 ?

Rewrite in standard form: (ax2 + bx + c =0)

Method 1: Factoring From standard form x2 – 2x – 15 = 0

a = 1, b = – 2, c = – 15 We want two numbers whose product is a•c = – 15 and whose sum is b = – 2

The solutions are x = –3, 5

Method 2: Completing the square Find the value of c that makes x2 – 2x + c a perfect square trinomial.

8.0/A.SSE.3a

5´ You try: What are the solutions to the equation

x2 – 9x = – 14 ?

Those two numbers are – 5 and 3

– 15

– 5 3 – 2

(x – 5)(x + 3) = 0



6 What are the solutions to x2 + 4x – 1 = 0 ? The equation is already in standard form.

a = 1, b = 4, and c = – 1 Method 1: Factoring

We want 2 numbers whose product is – 1 and sum is 4. Method 2: Quadratic formula

The quadratic formula is:

In this case, a = 1, b = 4, and c = – 1

8.0/N.CN.7

6´ You try: What are the solutions to x2 – 6x + 3 = 0 ?

Product Sum (1)•(–1) = -1 YES 1 + (– 1) = 0 NO (–1)•(1) = -1 YES –1 + 1 = 0 NO

There are no rational numbers such that their product is –1 and their sum is 4, therefore this quadratic is not factorable.



7 a) Find the vertex of this parabola y = 3x2 − 6x + 2

The graph of y = ax2 + bx + c is a parabola that has a

vertex with an x–coordinate of −b

2a

To find the y–coordinate, evaluate y = 3x2 – 6x + 2 for x = 1

Vertex is (1, – 1)

b) Match y = 3x2 − 6x + 2 to its correct graph. So the graph that matches y = 3x2 – 6x + 2 is graph C.

10.0/G.GPE.3

7´ You try: a) Find the vertex of this parabola

y = − 12x2 + 2x +3

b) Match y = − 12x2 + 2x +3 to its correct

graph.

A. B.

C. D.

• Opens up if a > 0, down if a < 0 • Has a y –intercept of (0, c) • Is narrower than graph of y = x2 if |a| >1 and wider if |a|<1

• a = 3, 3>0 so it opens up [can not be B or D] • c = 2 so the y-int is (0, 2) [can not be A] • a = 3, |3| > 1 so is narrow [must be C]

FACTS ABOUT PARABOLAS for y = 3x2 – 6x + 2

A. B.

C. D.



8 a) Describe the graphical change from y = x2

to y = 12(x − 2)2 +3

b) Match each equation with its graph.

9.0/G.GPE.3.2

8´ You try: a) Describe the graphical change from y = x2

to y = −3(x + 2)2 b) Match each equation with its graph.

• (h, k) is the vertex

• If a > 0, the parabola opens up. If a < 0, it opens down. • If |a| >1 it is narrower than graph of y = x2 and wider if |a| < 1

• (2, 3) is the vertex The graph of y = x2 will shift 2 units to the right and 3 up.

• a = , >0 so the parabola

will open up No change from y = x2

• a = , < 1 so it is a

wider graph The graph will be wider than y = x2

VERTEX FORM y = a(x – h)2 + k

1.

2.

3. 4.

A. B.

C. D.

1. B The vertex is at (–3, 2); a > 0 so it opens up; |a| = 1 so it is neither wider nor narrower than y = x2 2. C The vertex is at (3, –2); a < 0 so it opens down; |a| = 1 so it is neither wider nor narrower than y = x2 3. A The vertex is at (3, 2); a > 0 so it opens up; |a| > 1 so it is narrower than y = x2 4. D The vertex is at (3, 2); a > 0 so it opens up; |a| < 1 so it is wider than y = x2

1.

2.

3. 4.

A. B.

C. D.



9 Match each equation to its graph.

17.0/G.GPE.3.2

9´ You try: Match each equation to its graph.

End of Study Guide

1.

2. 3.

4.

A.

B.

C.

D.

1. B This equation is a hyperbola.

• (h, k) is the center [(2, 1)] • ±a: the distance of rectangle from center in x-direction [±3] ±b: the distance of rectangle from center in y-direction [±4] • asymptotes go through vertices of rectangle • because y is first, the hyperbola opens in the y-direction

2. D This equation is an ellipse.

• (h, k) is the center [(2, 1)] • ±a: the distance of rectangle from center in x-direction [±4] ±b: the distance of rectangle from center in y-direction [±3] 3. C This equation is a circle. • (h, k) is the center [(2, 1)] • r is the radius [r=3] 4. A This equation is a parabola. • (h, k) is the vertex [(2, 1)] • a < 0 so it opens down • |a| > 1 so it is narrower than y = x2

1.

2. 3. 4.

A.

B.

C.

D.



You Try Solutions: 1´ You try:

a) Simplify

Method 1: Expansion Method 2: Exponent Properties The Power of a power rule for exponents says that

(b x ) y = bx • y

b) Simplify Method 1: Expansion Method 2: Exponent Properties

2´ You try:

What expression is equivalent to 843 ?

Method 1: Method 2:

Write 1 coefficient Expand and

commute factors

Expand and find factors of one



3´ You try: What is the solution to log2 y = 4 ? Method 1: Change to exponential form The logarithm of a number x with base b is the solution y for the equation by = x

logb x = y such that by = x Method 2: Exponentiation Method 3: Base change We will use the base change formula for logarithms which states

4´ You try: Express as a single logarithm

log410+ log4 6 Method 1: Product Identity

The product identity for logarithmic functions states that logb(x • y) = logb(x) + logb(y) Method 2: Exponentiation

Let N be the value of log4 10 + log4 6 N is such that N = log4 10 + log4 6 ∴ log4 10 + log4 6= log4 60



5´ You try: What are the solutions to the equation

x2 – 9x = – 14 ? Rewrite in standard form: (ax2 + bx + c =0) Method 1: Factoring

From standard form x2 – 9x + 14 = 0 a = 1, b = – 9, c = 14

We want two numbers whose product is a•c = 14 and whose sum is b = – 9 The solutions are x = 7, 2 Method 2: Completing the square Find the value of c that makes x2 – 9x + c a perfect square trinomial.

6´ You try: What are the solutions to x2 – 6x + 3 = 0 ? The equation is already in standard form.

a = 1, b = – 6, and c = 3 Method 1: Factoring

We want 2 numbers whose product is 3 and sum is – 6. Method 2: Quadratic formula

The quadratic formula is:

In this case, a = 1, b = – 6, and c = 3

Those two numbers are – 7 and –2

14

– 7 – 2

– 9 (x – 7)(x – 2) = 0

Product Sum 1 • 3 = 3 YES 1 + 3 = 4 NO –1• –3 = 3 YES –1 + –3 = – 4 NO

There are no rational numbers such that their product is 3 and their sum is – 6, therefore this quadratic is not factorable.



7´ You try: Vertex is (2, 5)

8´ You try: a) Describe the graphical change from y = x2

to y = −3(x + 2)2 b) Match each equation with its graph

a) Find the vertex of this parabola

The graph of y = ax2 + bx + c is a parabola that has a

vertex with an x–coordinate of

b) Match to its correct graph.

A. B.

C. D.

• Opens up if a > 0, down if a < 0 • Has a y –intercept of (0, c) • Is narrower than graph of y = x2 if |a| >1 and wider if |a|<1

• a = – ½, – ½<0 so down [can not be A or C] • c = 3 so the y-int is (0, 3) [can not be B] • a = - ½, |- ½| < 1 so wide [must be D]

FACTS ABOUT PARABOLAS for y = - ½x2 + 2x + 3

So the graph that matches y = - ½x2 + 2x + 3 is graph D.

• (h, k) is the vertex

• If a > 0, the parabola opens up. If a < 0, it opens down. • If |a| >1 it is narrower than graph of y = x2 and wider if |a| < 1

• (– 2, 0) is the vertex The graph of y = x2 will shift 2 units to the left. • a = – 3, –3 < 0 so the parabola will open down Opens the in the opposite direction of y = x2 • a = – 3, |–3| > 1 so it is a narrower graph The graph will be narrower than y = x2

VERTEX FORM y = a(x – h)2 + k

1.

2.

3. 4.

A. B.

C. D.

1. A The vertex is at (2, – 1); a > 0 so it opens up; |a| > 1 so it is narrower than y = x2 2. D The vertex is at (2, 1); a > 0 so it opens up; |a| = 1 so it is neither wider nor narrower than y = x2 3. B The vertex is at (2, 1); a < 0 so it opens down; |a| = 1 so it is neither wider nor narrower than y = x2 4. C The vertex is at (2, – 1); a > 0 so it opens up; |a| < 1 so it is wider than y = x2



9´ You try: ANSWERS:

1. C 2. B 3. A 4. D

JUSTIFICATION is in the next column.

JUSTIFICATION:

Match each equation to its graph.

1.

2. 3. 4.

A.

B.

C.

D.

1. C This equation is a parabola. • (h, k) is the vertex [(–1, 3)] • a > 0 so it opens up • |a| < 1 so it is wider than y = x2

2. B This equation is a hyperbola.

• (h, k) is the center [(–1, 3)] • ±a: the distance of rectangle from center in x-direction [±2] ±b: the distance of rectangle from center in y-direction [±5] • asymptotes go through vertices of rectangle • because x is first, the hyperbola opens in the x-direction

3. A This equation is an ellipse.

• (h, k) is the center [(–1, 3)] • ±a: the distance of rectangle from center in x-direction [±2] ±b: the distance of rectangle from center in y-direction [±5] 4. D This equation is a circle. • (h, k) is the center [(–1, 3)] • r is the radius [r=5]

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WCCUSD Algebra 2 Benchmark 2 Study Guide...WCCUSD Algebra 2 Benchmark 2 Study Guide Page 7 of 14...

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