Question

What is the axis of symmetry and vertex for the graph `y = x^2 -4x + 8`?

Answer 1

Axis of symmetry: `x=2`, vertex: `(2, 4)`

Explanation:

To find out the axis of symmetry and vertex, the equation must be put in standard form (vertex form): `y=a(x-h)^2+k`
where `a =` constant, axis of symmetry `=h` and vertex `= (h,k)`

Use completing of the square:

1. Combine the x-terms: `y=(x^2-4x)+8`
2. half the x-term coeficient `: 1/2 (-4) = -2` and put it inside the squared factor `(x-2)^2`
3. Subtract the added term: `(- 2)^2 =4`
   `y= (x-2)^2+8-4`
   Note: since `(x-2)^2 = (x-2)(x-2) = x^2-4x+4` the added term is `4`.
4. Simplify: `y = (x-2)^2+4`

From the graph you can also see the axis & vertex:
graph{x^2-4x+8 [-10.17, 9.83, -0.76, 9.24]}