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

Feb 15, 2017

Axis of symmetry: $x = 2$, vertex: $\left(2 , 4\right)$

Explanation:

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

Use completing of the square:

1. Combine the x-terms: $y = \left({x}^{2} - 4 x\right) + 8$
2. half the x-term coeficient $: \frac{1}{2} \left(- 4\right) = - 2$ and put it inside the squared factor ${\left(x - 2\right)}^{2}$
3. Subtract the added term: ${\left(- 2\right)}^{2} = 4$
$y = {\left(x - 2\right)}^{2} + 8 - 4$
Note: since ${\left(x - 2\right)}^{2} = \left(x - 2\right) \left(x - 2\right) = {x}^{2} - 4 x + 4$ the added term is $4$.
4. Simplify: $y = {\left(x - 2\right)}^{2} + 4$

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