How do you find the axis of symmetry and the vertex when dealing with quadratics?

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Author Question: How do you find the axis of symmetry and the vertex when dealing with quadratics? (Read 1615 times)

coco

Hero Member
Posts: 739

on: Jun 18, 2013

Ex. y= x^2 -6x + 5

Millan

Sr. Member
Posts: 368

Reply #1 on: Jun 18, 2013

You have to turn it into vertex form: y=(x-h)^2+k where h and k are the vertex.

To do that, you need to "complete the square" - A fast way to factor a quadratic

y=(x^2-6x+___)+5-___
To find the spot in the blank, do (b/2)^2
(-6/2)^2=+9
y=(x^2-6x+9)+5-9
y=(x^2-6x+9)-4
The parenthesis area can be factored into (x-3)(x-3) which = (x-3)^2

Thus, you vertex form is y=(x-3)^2-4
Your axis of symmetry is +3, vertex is (+3, -4)

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