Question

What are the important points needed to graph `y=x^2- 6x+2`?

Answer 1

`y = x^2-6x+2` represents a parabola. Axis of symmetry is x = 3. Vertex is `V(3, -7)`. Parameter `a=1/4`. Focus is `S(3, -27/4)`. Cuts x-axis at `(3+-sqrt7, 0)`. Directrix equation: `y=-29/4`. .

Explanation:

Standardize the form to `y+7=(x-3)^2`.
Parameter a is given 4a = coefficient of `x^2` = 1.
Vertex is `V(3, -7)`.
The parabola cuts x-axis y = 0 at `(3+-sqrt7, 0)`.
The axis of symmetry is x = 3, parallel to y-axis, in the positive direction, from the vertex

Focus is S(3, -7-1.4)#, on the axis x = 3, at a distance a =1/4, above the focus.

Directrix is perpendicular to the axis, below the vertex, at a distance a = 1/4, V bisects the altitude from S on the directrix.