Question

How do you write `y=x^2-6x+8` in vertex form and identify the vertex, y intercept and x intercept?

Answer 1

`y = x^2 - 6x + 8`

x of vertex: `x = (-b/2a) = 6/2 = 3`
y of vertex: y = f(3) = 9 - 18 + 8 = -1

Vertex form:`y = (x - 3)^2 - 1`

y-intercept when x = 0 --> y = 8

x-intercept when y = 0, solve `x^2 - 6x + 8 = 0.`

Compose factor pairs of 8 -> (1, 8)(2, 4). This sum is 6 = -b. Then the 2 real roots (x-intercepts) are 2 and 4