Question

How do you find the axis of symmetry, vertex and x intercepts for `y=x^2-6x+8`?

Answer 1

Vertex (3, -1)
x-intercepts: 2 and 4

Explanation:

`y = x^2 - 6x + 8`
x-coordinate of axis of symmetry, and of vertex:
`x = - b/(2a) = 6/2 = 3`
y-coordinate of vertex:
`y(3) = 9 - 6(3) + 8 = -9 + 8 = - 1`
Vertex (3, -1)
To find the 2 x-intercepts, make y = 0 and solve:
`y = x^2 - 6x + 8 = 0`
Find 2 real roots (x- intercepts) knowing the sum (-b = 6) and the product (c = 8). They are 2 and 4.