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

1 Answer
A. S. Adikesavan
Apr 14, 2016

#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.

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
1243 views around the world
You can reuse this answer
Creative Commons License