Instructions: For the following quadratic functions, write the function in factored form and then find the x𝑥-intercepts, axis of symmetry, vertex, and domain and range.y=x2−6x+8𝑦=𝑥2−6𝑥+8Factored Form: y=(x𝑦=(𝑥 Answer 1 Question 5 Answer 2 Question 5 )(x)(𝑥 Answer 3 Question 5 Answer 4 Question 5 )) (Type least to greatest.)x𝑥-Intercepts: (( Answer 5 Question 5 ,, Answer 6 Question 5 )) and (( Answer 7 Question 5 ,, Answer 8 Question 5 )) (Type least to greatest.)Axis of Symmetry: x=𝑥= Answer 9 Question 5Vertex: (( Answer 10 Question 5 ,, Answer 11 Question 5 ))Domain: Answer 12 Question 5Range: y𝑦 Answer 13 Question 5 Answer 14 Question 5

The quadratic function given is y = x^2 - 6x + 8.

1. Factored Form: To write the function in factored form, we need to factorize the quadratic equation. The factored form of the equation is y = (x - 2)(x - 4).

2. x-intercepts: The x-intercepts of the function are the values of x for which y = 0. Setting y = 0 in the factored form of the equation gives us x - 2 = 0 and x - 4 = 0. Solving these equations gives us x = 2 and x = 4. So, the x-intercepts are (2, 0) and (4, 0).

3. Axis of Symmetry: The axis of symmetry of a parabola y = ax^2 + bx + c is given by the formula x = -b/2a. Here, a = 1 and b = -6. So, the axis of symmetry is x = -(-6)/2*1 = 3.

4. Vertex: The vertex of the parabola is the point (h, k) where h is the axis of symmetry and k is the value of the function at h. So, h = 3 and k = (3)^2 - 6*(3) + 8 = -1. So, the vertex is (3, -1).

5. Domain: The domain of a quadratic function is all real numbers. So, the domain is (-∞, ∞).

6. Range: Since the coefficient of x^2 is positive, the parabola opens upwards. So, the minimum value of the function is the y-coordinate of the vertex and the maximum value is ∞. So, the range is [-1, ∞).