# How do you find the axis of symmetry, and the maximum or minimum value of the function y=x^2-6x+2?

Jun 7, 2017

The axis of symmetry is $x = 3$.

The minimum (vertex) is $\left(3 , - 7\right)$.

#### Explanation:

$y = {x}^{2} - 6 x + 2$ is a quadratic equation in the form: $a {x}^{2} - b x + c$, where $a = 1$, $b = - 6$, and $c = 2$.

The axis of symmetry:

$x = \frac{- b}{2 a}$

$x = \frac{- \left(- 6\right)}{2}$

Simplify.

$x = \frac{6}{2} = 3$

$x = 3$

The vertex is the maximum or minimum of the parabola.

If $a > 0$, the parabola will open upwards, and the vertex will be a minimum.

If $a < 0$, the parabola will open downward, and the vertex will be a maximum.

In the current equation $> 0$ so the parabola so the vertex is the minimum and the parabola will open upward.

With a standard equation, substitute the value for $x$ and solve for $y$. The point $\left(x , y\right)$ is the vertex.

$y = {x}^{2} - 6 x + 2$

$y = {3}^{2} - 6 \left(3\right) + 2$

Simplify.

$y = 9 - 18 + 2$

$y = - 7$

The vertex is $\left(3 , - 7\right)$, which is also the minimum point.
graph{y=x^2-6x+2 [-16.02, 16.01, -8.01, 8.01]}