ALGEBRA II SEMESTER EXAM REVIEW

ALGEBRA II

SEMESTER EXAM

REVIEW

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KNOWT LINK:

What to know: 

• Everything Domain and Range/End Behavior

• Transformations of Parabolic and Absolute Value Functions

• Vertex and Standard form of a quadratic function

• Solve a quadratic function by completing the square, factoring, or using the quadratic formula

• Solve quadratic equations with non-real solutions

• Solve quadratic inequalities algebraically

• Write the equation of a quadratic function given conic attributes. (Directrix and Focus)

• Solving systems of linear and quadratic equations

• Solve 3 variable systems

• Add, Subtract, Multiply, and Divide Polynomials

• Factor polynomial expressions with a degree of 3 or 4

Domain and Range/End Behavior

Domain and Range Display Methods

1. Inequalities:
   a ≤ x ≤ b, a < x < b

2. Set Notation:
   {x | a ≤ x ≤ b}, {x | a < x < b}

3. Interval Notation:
   [a, b], (a, b), [a, b), (a, b]
   (Brackets mean the point is included; parentheses mean not included)

For all methods, use x for domain and y for range.

• End behavior

  • Explains what the y values do as x approaches Infinity

  • As x->∞ y->_

  • As x->-∞ y->_

Domain: All real numbers, (-∞,∞)

Range: y≥1, [1,∞), or {y|y≥1}

1. End Behavior: 

   1. as x → ∞, y → ∞

   2. as x → -∞, y → ∞

Domain: -3 < x ≤ 1, (-3, 1], or {x | -3 < x ≤ 1}

Range: -4 ≤ y ≤ 0, [-4, 0], or {y | -4 ≤ y ≤ 0}

No end behavior because it has endpoints

Transformations

• Quadratic

  • f(x)= a(x - h)² + k

    • a: Vertical stretch/compression

      • a > 1: Stretch

      • 0 < a < 1: Compression

      • a < 0: Reflection over x-axis

    • h: Horizontal shift

      • Positive h: Left shift

      • Negative h: Right shift

    • k: Vertical shift

      • k > 0: Up shift

      • k < 0: Down shift

  • Parent function is f(x)=x²

• Absolute value

  • f(x) = a |bx - h| + k (same as quadratic but using absolute value)

    • a: Vertical stretch/compression and reflection

      • ( a > 1 ): Vertical stretch (makes the V-shape taller)

      • ( 0 < a < 1 ): Vertical compression (flattens the V-shape)

      • ( a < 0 ): Reflection over the x-axis (flips the V-shape upside down)

    • h: Horizontal shift

      • Positive ( h ): Shifts the graph to the right by ( h ) units

      • Negative ( h ): Shifts the graph to the left by ( h ) units

    • k: Vertical shift

      • ( k > 0 ): Shifts the graph upward by ( k ) units

      • ( k < 0 ): Shifts the graph downward by ( k ) units

  • Parent function is f(x)=|x|

Vertex/Standard Form

• Vertex form is y=a(x-h)² + k

  • vertex=(h,k)

• Standard form is y = ax² + bx + c

• You can convert vertex to standard by simplifying the equation

Solving Quadratic Equations

• Completing the square

  • Move all constants to the other side.

  • Add (½ b)² to both sides.

  • Take the square root of both sides.

• Factoring ((No Coefficients for x²))

  • Find two numbers that add to b and multiply to c.

  • Form: y = (x - a)(x - b).

  • If there’s a coefficient for x², divide by that coefficient.

  • Express in the form y = a(x² + bx + c).

  • You can also solve by factoring the normal way, but instead it needs to multiply to a*c and add to b

• Quadratic function

  • x= ( -b ± √( b^2 - 4ac ) ) / ( 2a )

  • Input a,b, and c values into the formula and simplify

Non-real Solutions (i)

Definition:

Nonreal solutions arise when the discriminant of a quadratic equation is negative, indicating no real solutions.

Quadratic Equations:

• Standard Form:
  ( ax^2 + bx + c = 0 )

• Discriminant Formula:
  ( D = b^2 - 4ac ) 

  • If ( D < 0 ), the equation has nonreal (complex) solutions.

Complex Numbers:

• Form:
  ( a + bi )

  • Where:

    • ( a ) = real part

    • ( b ) = imaginary part

  • Pure imaginary numbers occur when ( a = 0 ) (e.g., ( 0 + bi )).

Finding Imaginary Solutions:

• Use the Quadratic Formula:
  ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} )

  • If ( b^2 - 4ac < 0 ), then solutions will involve imaginary numbers.

Example:

• Equation:
  ( x^2 - 2x + 2 = 0 )

• Discriminant:
  ( (-2)^2 - 4(1)(2) = 4 - 8 = -4 )

• Solutions:
  x = ( -b ± √( b^2 - 4ac ) ) / ( 2a )

  • Results in two imaginary solutions:
    ( 1 + i ) and ( 1 - i ).

Key Concepts:

• Imaginary Unit (i): Represents ( \sqrt{-1} ).

• Complex Conjugates: For any complex number ( a + bi ), its conjugate is ( a - bi ).

• Addition/Subtraction: Complex numbers can be added or subtracted by combining real and imaginary parts separately.

Quadratic Inequalities

• You can just use Desmos on the final, but it is useful to know how to solve them algebraically

Step 1: Rearrange the Inequality

Move all terms to one side of the inequality so that one side equals zero. The inequality should be in the form:

• [

• ax^2 + bx + c \leq 0 \quad \text{or} \quad ax^2 + bx + c \geq 0

]

Step 2: Factor the Quadratic Expression

If possible, factor the quadratic expression on one side of the inequality to obtain:

• [

• (rx + s)(tx + v) \leq 0 \quad \text{or} \quad (rx + s)(tx + v) \geq 0

]

Step 3: Find the Zeros (Roots)

Set each factor equal to zero to find the critical points (zeros) of the quadratic. For example, if you have:

• [

• (rx + s)(tx + v) = 0

]

You would solve for the zeros.

Step 4: Create a Number Line

Plot the zeros on a number line. These points will divide the number line into intervals.

Step 5: Test Each Interval

Choose a test point from each interval created by the zeros and substitute it back into the original inequality to determine whether that interval satisfies the inequality. Note if the expression is positive or negative in that interval.

Step 6: Determine the Solution Set

Based on your test results, identify which intervals satisfy the original inequality. Write your solution in either inequality notation or interval notation.

Example

Problem

Solve the inequality:

Step 1: Rearrange the Inequality

The inequality is already in standard form:

x−x−12>0

x

2

−x−12>0

Step 2: Factor the Quadratic Expression

We can factor the quadratic expression:

(x−4)(x+3)>0

(x−4)(x+3)>0

Step 3: Find the Zeros (Roots)

Set each factor equal to zero to find the critical points:

1. x−4=0

2. x−4=0

   • Solution: 

   • x=4

   • x=4

4. x+3=0

5. x+3=0

   • Solution: 

   • x=−3

   • x=−3

Step 4: Create a Number Line

Plot the critical points 

x=−3

x=−3 and 

x=4

x=4 on a number line. These points divide the number line into three intervals:

• (−∞,−3)

• (−∞,−3)

• (−3,4)

• (−3,4)

• (4,∞)

• (4,∞)

Step 5: Test Each Interval

Choose a test point from each interval and substitute it back into the factored inequality:

1. Interval 

2. (−∞,−3)

3. (−∞,−3): Test with 

4. x=−4

5. x=−4

6. (−)(−)>0 True 

7. (−)(−)>0 True 

9. Interval 

10. (−3,4)

11. (−3,4): Test with 

12. x=0

13. x=0

14. (−)(+)<0 False 

15. (−)(+)<0 False 

17. Interval 

18. (4,∞)

19. (4,∞): Test with 

20. x=5

21. x=5

22. (+)(+)>0 True 

23. (+)(+)>0 True 

Step 6: Determine the Solution Set

Based on the test results, we find that the intervals satisfying the original inequality are:

• From interval 

• (−∞,−3)

• (−∞,−3): True

• From interval 

• (−3,4)

• (−3,4): False

• From interval 

• (4,∞)

• (4,∞): True

Thus, the solution set is:

(−∞,−3)∪(4,∞)

(−∞,−3)∪(4,∞)

CONICS!!!!!!!!!!!!

Parabola:

• Equation: y = a(x-h)² + k or x = a(y-k)² + h

• Directrix: y = k - 1/(4a) or x = h - 1/(4a)

• Focus: (h, k + 1/(4a)) or (h + 1/(4a), k)

• Vertex: (h, k)

Circle:

• Equation: (x-h)² + (y-k)² = r²

• Center: (h, k)

• Radius: r

Ellipse:

• Equation: (x-h)²/a² + (y-k)²/b² = 1

• Center: (h, k)

• Foci: (h ± c, k) or (h, k ± c), where c² = a² - b²

Hyperbola:

• Equation: (x-h)²/a² - (y-k)²/b² = 1 or (y-k)²/a² - (x-h)²/b² = 1

• Center: (h, k)

• Vertices: (h ± a, k) or (h, k ± a)

• Foci: (h ± c, k) or (h, k ± c), where c² = a² + b²

SOLVING SYSTEMS OF LINEAR AND QUADRATIC EQUATIONS

Methods for Solving Systems of Equations

1. Substitution Method

This method involves solving one equation for one variable and substituting that expression into the other equation.

Steps:

1. Solve one equation for one variable (e.g., y in terms of x).

2. Substitute this expression into the other equation.

3. Simplify and solve for the remaining variable.

4. Substitute the solved value back into one of the original equations to find the other variable.

5. Write the solution as an ordered pair (x,y).

Example:

Solve:

y = x² + 1

y = 2x - 1

Solution: The system has complex solutions (1+i, y) and (1-i, y).

2. Elimination Method

This method eliminates one variable by adding or subtracting equations.

Steps:

1. Align both equations so like terms are in columns.

2. Multiply one or both equations (if necessary) to make the coefficients of one variable opposites.

3. Add or subtract the equations to eliminate that variable.

4. Solve for the remaining variable.

5. Substitute this value into one of the original equations to find the other variable.

Example:

Solve:

3y + 2x = 6

5y - 2x = 10

Solution: The solution is (x,y) = (0,2).

3. Graphing Method

This method involves graphing both equations on a coordinate plane and finding their point of intersection.

Steps:

1. Rewrite each equation in slope-intercept form (y = mx + b).

2. Plot both lines on a graph.

3. Identify where the lines intersect; this point is the solution.

4. If lines are parallel, there is no solution. If they overlap, there are infinitely many solutions.

Example:

Solve:

y = 0.5x + 2

y = x - 2

Solution: The lines intersect at (8,6).

Summary Table

SOLVING 3 VARIABLE SYSTEMS

Substitution Method

1. Solve one equation for one variable in terms of the others

2. Substitute this expression into the other two equations

3. Solve the resulting 2-variable system

4. Substitute back to find the third variable

Elimination Method

1. Use addition or subtraction to eliminate one variable from two equations

2. Repeat to get a 2-variable system

3. Solve the 2-variable system

4. Substitute back to find the third variable

Matrix Method

1. Write the system as an augmented matrix

2. Use row operations to transform the matrix into row echelon form

3. Back-substitute to solve for the variables

Examples

1. Substitution Method

Solve the system:

x + y + z = 6

2x - y + z = 4

x + 2y - z = 3

1. Solve the third equation for z:
   z = x + 2y - 3

2. Substitute this expression into the first two equations:
   x + y + (x + 2y - 3) = 6
   2x - y + (x + 2y - 3) = 4

3. Simplify:
   2x + 3y = 9
   3x + y = 7

4. Solve this 2-variable system:
   x = 12/7, y = 13/7

5. Substitute back to find z:
   z = 12/7 + 2(13/7) - 3 = 17/7

Solution: (12/7, 13/7, 17/7)

2. Elimination Method

Solve the system:

x + y + z = 6

2x - y + z = 4

x + 2y - z = 3

1. Eliminate z by subtracting equation 1 from equation 2:
   x - 2y = -2

2. Eliminate z by adding equation 1 and equation 3:
   2x + 3y = 9

3. Solve the resulting 2-variable system:
   x = 12/7, y = 13/7

4. Substitute back to find z:
   z = 6 - 12/7 - 13/7 = 17/7

Solution: (12/7, 13/7, 17/7)

3. Matrix Method

Solve the system:

x + y - z = -2

2x - y + z = 5

-x + 2y + 2z = 1

1. Write the augmented matrix:
   [1 1 -1 | -2]
   [2 -1 1 | 5]
   [-1 2 2 | 1]

2. Perform row operations to get row echelon form:
   [1 1 -1 | -2]
   [0 -3 3 | 9]
   [0 0 4 | 8]

3. Back-substitute:
   4z = 8, so z = 2
   -3y + 3(2) = 9, so y = -1
   x + (-1) - 2 = -2, so x = 1

Solution: (1, -1, 2)

Add, Subtract, Multiply, and Divide Polynomials

Adding Polynomials

Steps:

1. Align like terms vertically or horizontally

2. Combine like terms by adding their coefficients

3. Write the result, keeping all terms

Example:

Add (3x² + 2x - 5) and (2x² - 4x + 7)

Solution:

(3x² + 2x - 5) + (2x² - 4x + 7)

= (3x² + 2x²) + (2x - 4x) + (-5 + 7)

= 5x² - 2x + 2

Subtracting Polynomials

Steps:

1. Change the sign of each term in the polynomial being subtracted

2. Add the polynomials using the addition method

Example:

Subtract (4x³ - 2x + 1) from (6x³ + 3x² - 5x + 8)

Solution:

(6x³ + 3x² - 5x + 8) - (4x³ - 2x + 1)

= 6x³ + 3x² - 5x + 8 + (-4x³ + 2x - 1)

= 2x³ + 3x² - 3x + 7

Multiplying Polynomials

Steps:

1. Multiply each term of one polynomial by every term of the other

2. Combine like terms in the result

Example:

Multiply (2x + 3) and (x² - 4x + 5)

Solution:

(2x + 3)(x² - 4x + 5)

= 2x(x²) + 2x(-4x) + 2x(5) + 3(x²) + 3(-4x) + 3(5)

= 2x³ - 8x² + 10x + 3x² - 12x + 15

= 2x³ - 5x² - 2x + 15

Dividing Polynomials

Steps (Long Division):

1. Organize the polynomial:
   • Arrange both the dividend and divisor in descending order of degree
   • Ensure all terms are present, using zero coefficients if necessary

2. Initiate the division:
   • Divide the leading term of the dividend by the leading term of the divisor
   • This becomes the first term of the quotient

3. Perform subtraction:
   • Multiply the term obtained in step 2 by the entire divisor
   • Subtract this result from the dividend
   • This forms a new polynomial of lower degree

4. Continue the process:
   • Bring down the next term of the dividend if available
   • Repeat steps 2-4 with this new polynomial as the dividend
   • Continue until the degree of the remainder is less than the divisor's degree

5. Conclude the division:
   • The terms collected in the quotient form the polynomial quotient
   • The final remainder is the polynomial remainder

Steps (Synthetic Division):

1. Prepare the problem:
   • Arrange the polynomial in descending order of exponents
   • Insert 0 coefficients for any missing terms
   • Note the constant from the divisor (x - a), changing its sign

2. Construct the division structure:
   • List the coefficients of the polynomial in a row
   • Draw a horizontal line beneath this row
   • Place the modified divisor constant to the left of the line

3. Execute the division process:
   • Transfer the first coefficient downward
   • Multiply this coefficient by the divisor constant
   • Record the product beneath the next coefficient
   • Sum the numbers in this column
   • Repeat this process for all remaining columns

4. Interpret the results:
   • The bottom row represents the quotient coefficients
   • The final number in this row is the remainder

5. Factor verification:
   • A remainder of 0 indicates that the divisor is a factor of the polynomial

Example:

Divide (x³ - 2x² - 4x + 8) by (x - 2)

Solution: x²-4

Therefore, (x³ - 2x² - 4x + 8) ÷ (x - 2) = x² - 2x + 4

Factor polynomial expressions with a degree of 3 or 4

Steps

1. Identify the GCF:
   Factor out any GCF to simplify.

2. Determine the Number of Terms:

   • Four Terms: Use grouping.

   • Three Terms (Trinomials): Use trial and error or the AC method.

3. Apply Specific Factoring Techniques:

Factoring a Polynomial with Four Terms (Using Grouping)

1. Group the Terms:

   • Example: ( x³ + x² - x - 1 → (x³ + x²) + (-x - 1) )

2. Factor Out GCF from Each Group:

   • From ( x³ + x²: , x²(x + 1) )

   • From ( -x - 1: , -1(x + 1) )

3. Combine Results:
   ( x²(x + 1) - 1(x + 1) )

4. Factor Out Common Factors:
   Result: ( (x + 1)(x² - 1) )

5. Factor Remaining Expressions:
   Result: ( (x + 1)²(x - 1) )

Factoring a Polynomial with Three Terms

1. Identify Coefficients:

   • ( a, b, c ) from ( ax² + bx + c )

   • Example: For ( 2x² + 5x + 3: , a=2, b=5, c=3 )

2. Use the AC Method:

   • Multiply ( a ) and ( c ); find two numbers that multiply to ( ac ) and add to ( b ).

   • For ( a = 2, b = 5, c = 3: ) Multiply: ( ac = 6; ) Find numbers: ( (2,3). )

3. Rewrite Middle Term:
   Example: Rewrite as ( 2x² + 2x + 3x + 3 )

4. Factor by Grouping:
   Result: ( (2x(x + 1) + 3(x + 1)) = (x + 1)(2x + 3) )

Example of Factoring a Cubic Polynomial

Consider the cubic polynomial ( 2x³ - 3x² + 18x - 27 ): 

1. Group Terms:
   ( (2x³ - 3x²) + (18x - 27) )

2. Factor Each Group:
   ( x²(2x - 3) + 9(2x - 3) )

3. Combine Factors:
   Result: ( (2x - 3)(x² + 9) )

To find the roots of the polynomial, we can set each factor equal to zero. This gives us two equations to solve:

1. 2x - 3 = 0

2. x² + 9 = 0

Solving these equations results in:

1. x = 3/2 (real root)

2. x² + 9 = 0 does not yield real roots, as it leads to x = ±3i (complex roots).