Posted by **Jon** on Sunday, March 2, 2008 at 12:02pm.

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

vertex = 1/2

axis = 1/2

direction = up

- Algebra II -
**Reiny**, Sunday, March 2, 2008 at 12:18pm
your text should have the following:

for y = a(x-p)^2 + q

vertex is (p,q)

axis of symmetry: x=p

opening: up if a>0, down if a<0

please redo your answers and let me know what you got.

- Algebra II -
**Jon**, Sunday, March 2, 2008 at 12:33pm
y=2x+3^2-5

y=2x+9-5

x=-b/2a

x=-9/2(-9)

x=1/2

I dont know if I have the value for a right

- Algebra II(I think I got it) -
**Jon**, Sunday, March 2, 2008 at 3:07pm
vertex= (3,-5)

axis of symmetry= x=2

direction= up b/c 2>0

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