What is the axis of symmetry and vertex for the graph y=x^2-6x+4?

Mar 8, 2016

The vertex is located at the point $\left(3 , - 5\right)$. The line of symmetry is through the $x$ value of the vertex, in this case that is $3$. So the equation is $x = 3$

Explanation:

As usual there are several methods you might choose to solve this question. One method is "completing the square". If you are not familiar with the method is may seem to be a bit strange. Here's the solution:

$y = {x}^{2} - 6 x + 4$
$y = \left({x}^{2} - 6 x\right) + 4$ completing the square inside the brackets I get:
$y = \left({x}^{2} - 6 x + 9 - 9\right) + 4$
Notice that ${x}^{2} - 6 x + 9$ is the perfect square ${\left(x - 3\right)}^{2}$ so I can rewrite the equation as :
$y = {\left(x - 3\right)}^{2} - 9 + 4$. Now simplify to get :
$y = {\left(x - 3\right)}^{2} - 5$.
When a quadratic equation is written in this form it is telling me that the vertex is located at $\left(3 , - 5\right)$ The line of symmetry is always the line through the $x$ value of the vertex. Therefore the equation is $x = 3$