Posted by **Jon** on Sunday, March 2, 2008 at 5:13pm.

Identify the vertex, axis of symmetry, and direction of opening for y=2(x+3)^2-5

y = |2x+3|^2-5

h = 3

k = -5

vertex = (3,-5)

axis of symmetry = x = 3

direction = up b/c a = 2 and 2>0

- Algebra II -
**Damon**, Sunday, March 2, 2008 at 5:24pm
y = 2 (x+3)^2 - 5

(x+3)^2 = (1/2)(y+5)

(x+3)^2 = 4 (1/8) (y+5)

vertex (-3,-5)

symmetric about x = -3

indeed opens up because a = +1/8, check as x becomes very positive or very negative, y gets big positive

- Algebra II(can you explain) -
**Jon**, Sunday, March 2, 2008 at 5:41pm
I don't understand what you did after the 1st line.

- Algebra II -
**Damon**, Sunday, March 2, 2008 at 7:16pm
Hmmm

I know, and I bet it is in your book, that

a formula of type:

(x-h)^2 = 4 a (y-k)

is a parabola with vertex at (h,k)

a is the distance from vertex to focus and from vertex to directrix

if a is +, opens up. if a is -, opens down

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