1. Snapshot of the School Year Algebra 2 Honors Class Alex Asemota Mrs. Vittese Period 4 May 3, 2012

2. Chapter 2  To solve this problem: 2x-3 11  Solve the first inequality:  2x-3<5 (Add 3 on both sides)  2x<8 (Now divide by 2 on both sides)  x<4  Then solve the second inequality:  x+14>11 (Subtract 14 from both sides  x>-3  The solution is:  x>-3 and x<4  This means that all numbers between -3 and 4 are solutions.

3. Chapter 3  Find function composition given f(x) = 3x and g(x) = x + 2.  Question: f(g(8))  Step 1: Plug in given for x and solve.  g(8) = (8) + 2 = 10  Step 2 : Plug in g(4) for into f function and solve.  f(g(8)) = f(10) = 3(10) = 30

4. Chapter 4  Solve 4x + 3y = 19 and 5x + 7y = 40  Step 1: Multiply equations to make the coefficients of the x terms additive inverses.  5 (4x + 3y) = 5 19  −4 (5x + 7y) = −4 40  Step 2: Simplify,  20x + 15y = 95  −20x − 28y = −160  Step 3: Then you have to divide to get y  −13y = −65  y = 5  Step 4: Substitute for y in the first equation to get x  4x + 3(5) = 19  4x = 4  x = 1  The solution is (1, 5).

5. Chapter 5  Solve: x 2 + 5x + 7x + 35  Step 1: Put related terms next to each other  x(x + 5) + 7(x + 5) =  Step 2: Simplify  (x + 7)(x + 5)

6. Chapter 6  Solve: x + 9/x = 10  x(x + 9/x)=10  Step 1:Multiply through by x to clear the fractions.  x 2 + 9 = 10x  Step 2: Subtract 10x to make it all equal 0.  x 2 − 10x + 9 = 0  Step 3: Factor  (x − 1)(x − 9) = 0  Step 4: Solve for 0  x = 1 OR x = 9  Step 5: Check answer  Both 1 and 9 check in the original equation. The solutions are 1 and 9.

7. Chapter 7  Solve:´√ 3x+3 + √ x+2 = 5  Step 1: Subtract √ x+2: √ 3x+3=- 5 − √ x+2  ( √ 3x+3) 2 = (5 − √ x+2) 2  Step 2: Square the terms,  3x + 3 = 25 − √ x+2 + x + 2  Step 3: Simplify  10 √ x+2 = 24 − 2x  Step 4: Factor and Divide  5 √ x+2 = (12 − x)  Step 5: Square to to cancel square roots  (5 √ x+2) 2 = (12 − x) 2  Step 6-9: Simplify and Solve  25(x + 2) = 144 − 24x + x2  25x + 50 = 144 − 24x + x2  0 = 94 − 49x + x2  0 = (x − 2)(x − 47)  x = 2 OR x = 47  Step 10: Check  2 checks in the original equation; 47 does not check. The solution is 2.

8. Chapter 8  Solve: 5x2 + 6x + 1 = 0  Step 1 : Plug information for as in ax 2 + bx + c=0  Step 2: Simplify Answer: x = {-1/5, -1}

9. Chapter 9  Step 1: Start with the given equation  Step 2: Subtract from both sides  Step 3:Factor out the leading coefficient  Step 4: Take half of the x coefficient  Step 5:Now square  Step 6: Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of does not change the equation  Step 7: Distribute  Step 8: Multiply  Step 9: Now add to both sides to isolate y  Step 10: Combine like terms  Vertex (-1,6)  AoS x= -1  Minimum is the vertex (-1,6)

10. Chapter 11  Using the Remainder Theorem, find the value of f (–5), for f (x) = 3x4 + 2x3 + 4x.  I need to do the synthetic division, remembering to put zeroes in for the powers of x that are not included in the polynomial:  Since the remainder is 1605, then, thanks to the Remainder Theorem, I know that:  f (–5) = 1605.