Answers needed.

Question

Consider the function f(x) = x^2-6x+8 .

1. What is the vertex of the graph of this function? _______  

1. The maximum/minimum value is  _______ .  

The axis of symmetry is  _______  .

Answer 1

x^2-6x + 8 = 0

complete the square, take half of 6, square it and add it to both sides to get

x^2 -6x + (6/2)^2 = -8 + (6/2)^2

(x-3)^2 = 1

vertex is (3,-1) which is also the minimum point

axis of symmetry is a vertical line through the vertex or x=3

or

x^2-6x +8 = 0

factor

(x-4)(x-2)= 0, the x intercepts are 2 and 4

the x coordinate of the vertex is the average or midpoint between 2 and 4, (2+4)/2 = 3

plug in 3 into the function to get the y coordinate of the vertex 3^2 -6(2) + 8 = 9-18+8 = -1

It's an upward opening parabola with x-intercepts 2 and 4, and y intercept 8 or (0,8)