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November 29, 2014

November 29, 2014

Posted by **sophie** on Thursday, October 25, 2012 at 12:13pm.

a.vertex

b.maximum or minimum

c.axis of symmetry

d.y-intercept

e.x-intercept(zeros)

f.graph

- geometery -
**Reiny**, Thursday, October 25, 2012 at 2:19pmcomplete the square ....

y = x^2 - 4x**+ 4 - 4**+ 5

= (x-2)^2 + 1

vertex is (2,1)

opens up, so minimum

axis of symmetry: x = 2

y-intercepts: let x = 0, y = 5

x-intercepts:

let y = 0

x^2 - 4x+ 5 = 0

(x-5)(x+1) = 0

x = 5, x = -1 , notice the x of the vertex is midway between these.

graph: you do it.

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