Monday, March 15, 2010

Devin's Reflection

Parabolas:

how to find the axis of symmetry, vertex, focus, & directrx??

1.) to find the axis of symmetry: x = -b/2a

2.) for the vertex: (-b/2a, f(-b/2a)) or use complete the square:

y = (x+a)^2 + b.....a & b are numbers and (-a,b) = vertex

3.) to find the focus: 1/4p= the coefficient of x^2 and then add p

Note:

*If opens up, add to y value from vertex, if opens down, subtract

*If opens right, add to x value to vertex, if opens left, subtract)

4.) directrix: is p units behind the vertex

Note:

*If opens up, subtract; if opens down, add from y-value of vertex.
*If opens right, subtract x-value
*If opens left, add x-value

Example: x^2 + 1

~vertex:

x = -b/2a

x = 0/2(1) = 0

0^2 + 1 = 1

(0,1)

~Focus:

1/4p = 1

4p = 1

p = 1/4

(0, 1 + 1/4)

(0, 5/4)

~directrix:

y = 1 - 1/4

y = 3/4

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