# How do you find the axis of symmetry, and the maximum or minimum value of the function y=x^{2}-6x+4?

Apr 4, 2017

The minimum value is at $\left(3 , - 5\right)$

#### Explanation:

The equation of this parabola is in the form $y = a {x}^{2} + b x + c$

$y = {x}^{2} - 6 x + 4 \text{ "rarr" "a=1," } b = - 6 \mathmr{and} c = 4$

The axis of symmetry is found from $x = \frac{- b}{2 a}$

$x = \frac{- \left(- 6\right)}{2 \times 1} = \frac{6}{2} = 3$

The Turning Point is the maximum or minimum value and it lies on the axis of symmetry, so you have the $x$-coordinate

If $x = 3$, substitute to find $y$.

$y = {\left(3\right)}^{2} - 6 \left(3\right) + 4 = 9 - 18 + 4$

$y = - 5$

In this case the parabola opens upwards ($a > 0$, ie positive), therefore the Turning Point is a minimum value.

$\left(3 , - 5\right)$