for the following, find a) the axis of symmetry, b) the vertex, and c) the y intercept.
4. ( y = x^2 - 2x + 1 )
6. ( y = x^2 - 6x + 5 )
5. ( y = x^2 - 4x + 3 )
7. ( y = -2x^2 - 8x + 10 )
factor
8. ( x^2 - 7x - 8 )
10. ( x^2 + 4x - 12 )
9. ( x^2 - 5x + 6 )
11. ( x^2 + 6x + 5 )

Question

Accepted Answer

### Problem 4: $ y = x^2 - 2x + 1 $
#### a) Axis of Symmetry
# Explanation:
## Step1: Recall the formula for the axis of symmetry of a quadratic function $ y = ax^2 + bx + c $, which is $ x = -\frac{b}{2a} $.
For $ y = x^2 - 2x + 1 $, $ a = 1 $, $ b = -2 $.
## Step2: Substitute $ a $ and $ b $ into the formula.
$ x = -\frac{-2}{2(1)} = \frac{2}{2} = 1 $
# Answer: $ x = 1 $

#### b) Vertex
# Explanation:
## Step1: The x-coordinate of the vertex is the axis of symmetry, so $ x = 1 $.
## Step2: Substitute $ x = 1 $ into the function to find the y-coordinate.
$ y = (1)^2 - 2(1) + 1 = 1 - 2 + 1 = 0 $
# Answer: $ (1, 0) $

#### c) y-intercept
# Explanation:
## Step1: The y-intercept occurs where $ x = 0 $. Substitute $ x = 0 $ into the function.
$ y = (0)^2 - 2(0) + 1 = 0 - 0 + 1 = 1 $
# Answer: $ (0, 1) $

### Problem 5: $ y = x^2 - 4x + 3 $
#### a) Axis of Symmetry
# Explanation:
## Step1: Use the formula $ x = -\frac{b}{2a} $ with $ a = 1 $, $ b = -4 $.
## Step2: Substitute values: $ x = -\frac{-4}{2(1)} = \frac{4}{2} = 2 $
# Answer: $ x = 2 $

#### b) Vertex
# Explanation:
## Step1: x-coordinate is 2 (from axis of symmetry).
## Step2: Substitute $ x = 2 $ into the function: $ y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1 $
# Answer: $ (2, -1) $

#### c) y-intercept
# Explanation:
## Step1: Substitute $ x = 0 $ into the function.
$ y = (0)^2 - 4(0) + 3 = 0 - 0 + 3 = 3 $
# Answer: $ (0, 3) $

### Problem 6: $ y = x^2 - 6x + 5 $
#### a) Axis of Symmetry
# Explanation:
## Step1: Use $ x = -\frac{b}{2a} $ with $ a = 1 $, $ b = -6 $.
## Step2: Substitute: $ x = -\frac{-6}{2(1)} = \frac{6}{2} = 3 $
# Answer: $ x = 3 $

#### b) Vertex
# Explanation:
## Step1: x-coordinate is 3.
## Step2: Substitute $ x = 3 $ into the function: $ y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4 $
# Answer: $ (3, -4) $

#### c) y-intercept
# Explanation:
## Step1: Substitute $ x = 0 $: $ y = (0)^2 - 6(0) + 5 = 0 - 0 + 5 = 5 $
# Answer: $ (0, 5) $

### Problem 7: $ y = -2x^2 - 8x + 10 $
#### a) Axis of Symmetry
# Explanation:
## Step1: Use $ x = -\frac{b}{2a} $ with $ a = -2 $, $ b = -8 $.
## Step2: Substitute: $ x = -\frac{-8}{2(-2)} = \frac{8}{-4} = -2 $
# Answer: $ x = -2 $

#### b) Vertex
# Explanation:
## Step1: x-coordinate is -2.
## Step2: Substitute $ x = -2 $ into the function: $ y = -2(-2)^2 - 8(-2) + 10 = -2(4) + 16 + 10 = -8 + 16 + 10 = 18 $
# Answer: $ (-2, 18) $

#### c) y-intercept
# Explanation:
## Step1: Substitute $ x = 0 $: $ y = -2(0)^2 - 8(0) + 10 = 0 - 0 + 10 = 10 $
# Answer: $ (0, 10) $

### Problem 8: Factor $ x^2 - 7x - 8 $
# Explanation:
## Step1: We need two numbers that multiply to $ -8 $ and add to $ -7 $.
The numbers are $ -8 $ and $ 1 $ (since $ -8 \times 1 = -8 $ and $ -8 + 1 = -7 $).
## Step2: Rewrite the middle term and factor by grouping.
$ x^2 - 8x + x - 8 = x(x - 8) + 1(x - 8) = (x - 8)(x + 1) $
# Answer: $ (x - 8)(x + 1) $

### Problem 9: Factor $ x^2 - 5x + 6 $
# Explanation:
## Step1: Find two numbers that multiply to $ 6 $ and add to $ -5 $.
The numbers are $ -2 $ and $ -3 $ (since $ -2 \times -3 = 6 $ and $ -2 + (-3) = -5 $).
## Step2: Rewrite the middle term and factor by grouping.
$ x^2 - 2x - 3x + 6 = x(x - 2) - 3(x - 2) = (x - 2)(x - 3) $
# Answer: $ (x - 2)(x - 3) $

### Problem 10: Factor $ x^2 + 4x - 12 $
# Explanation:
## Step1: Find two numbers that multiply to $ -12 $ and add to $ 4 $.
The numbers are $ 6 $ and $ -2 $ (since $ 6 \times -2 = -12 $ and $ 6 + (-2) = 4 $).
## Step2: Rewrite the middle term and factor by grouping.
$ x^2 + 6x - 2x - 12 = x(x + 6) - 2(x + 6) = (x + 6)(x - 2) $
# Answer: $ (x + 6)(x - 2) $

### Problem 11: Factor $ x^2 + 6x + 5 $
# Explanation:
## Step1: Find two numbers that multiply to $ 5 $ and add to $ 6 $.
The numbers are $ 1 $ and $ 5 $ (since $ 1 \times 5 = 5 $ and $ 1 + 5 = 6 $).
## Step2: Rewrite the middle term and factor by grouping.
$ x^2 + x + 5x + 5 = x(x + 1) + 5(x + 1) = (x + 1)(x + 5) $
# Answer: $ (x + 1)(x + 5) $

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QUESTION IMAGE

Question

Explanation:

Problem 4: \( y = x^2 - 2x + 1 \)

a) Axis of Symmetry

Step1: Recall the formula for the axis of symmetry of a quadratic function \( y = ax^2 + bx + c \), which is \( x = -\frac{b}{2a} \).

Step2: Substitute \( a \) and \( b \) into the formula.

Step1: The x-coordinate of the vertex is the axis of symmetry, so \( x = 1 \).

Step2: Substitute \( x = 1 \) into the function to find the y-coordinate.

Step1: The y-intercept occurs where \( x = 0 \). Substitute \( x = 0 \) into the function.

Answer:

b) Vertex