Wednesday

March 4, 2015

March 4, 2015

Posted by **Anonymous** on Sunday, December 16, 2012 at 3:29pm.

- Intermediate Algebra -
**Henry**, Monday, December 17, 2012 at 9:48pmVertex Form: Y = a(x-h)^2 + k.

A. The parabola opens upward because "a"

is positive.

B. Y = (x-3)^2 - 2.

Let X = o and solve for Y:

Y = (0-3)^2 - 2 = 7. = Y-int.

C. Let Y = 0 and solve forX:

Y = (x-3)^2-2 = 0.

(x-3)^2 = 2

Take sgrt of both sides:

x-3 = +-sqrt2

X = +-1.414 + 3 = 4.14, and 1.59. = X-intercepts.

D. V(h,k) = V(3,-2).

E. Select values of X below and above

h and calculate the corresponding value of Y.

(x,y). Y = (x-3)^2 - 2.

(-2,23)

(-1,14)

(0,7)

(1,2)

(2,-1)

V(3,-2)

(4,-1)

(5,2)

(6,7)

(7,14)

(8,23).

- Intermediate Algebra -
**meri**, Friday, December 19, 2014 at 7:18pmfor the quadratic equation y=(x-3)^2-2 determine A) weather the graph opens up or down B) the y- intercept; C) the x intercept D) the points of the vertex and E ) graph the equation with at least 10 pairs of points

**Answer this Question**

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