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

Question

2460 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-07T00:31:58

The axis of symmetry is $x=3$ .

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

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

The axis of symmetry:

$x=(-b)/(2a)$

$x=(-(-6))/(2)$

Simplify.

$x=6/2=3$

$x=3$

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 $(x,y)$ is the vertex.

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

$y=3^2-6(3)+2$

Simplify.

$y=9-18+2$

$y=-7$

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

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