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October 1, 2014

October 1, 2014

Posted by **Anonymous** on Sunday, December 16, 2012 at 8:47pm.

- Intermediate Algebra -
**Reiny**, Sunday, December 16, 2012 at 10:40pmy=(x-3)^2-2

or

y=+1(x-3)^2-2 vs y = a(x-p)^2 + q

since a=1 , a> 0 , so it opens upwards

y=intercept, let x = 0

y = (-3)^2 - 2 = 7

x-intercept, let y = 0

(x-3)^2 - 2 = 0

(x-3)^2 = 2

x-3 = ± √2

x = 3 ± √2

"points of the vertex " ????

the vertex IS a point, it does not have points, it has coordinates

they are (3,-2) , (from (p,q) )

For D) just pick any suitable values for x, evaluate the corresponding y's and plot.

(pick values of x immediately above and below +3 )

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