SOLUTION: I'm not sure if this is the right section. I have two question's with no textbook. 1. identify the vertex, axis of symmetry, and direction of opening for y=2(x+3)^2-5. 2. Writ

Question 166497: I'm not sure if this is the right section. I have two question's with no textbook.
1. identify the vertex, axis of symmetry, and direction of opening for y=2(x+3)^2-5.
2. Write y=-4x^2+8x-1 in vertex form.
I must show my work.
Please HELP!!!
Thanks
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
# 1

Start with the given equation

Rewrite as

Now the equation above is in vertex form where (h,k) is the vertex and 'a' determines which direction it opens up. In this case, , and .

So because , this tells us that the parabola opens up. Also, since and , the vertex is (-3,-5). Finally, the axis of symmetry is simply the x-coordinate of the vertex. So the axis of symmetry is

=======================
Answer:

Direction of opening: up
Vertex: (-3,-5)
Axis of Symmetry:

# 2

Start with the right side of the given equation.

Factor out the coefficient . This step is very important: the coefficient MUST be equal to 1.

Take half of the coefficient to get . In other words, .

Now square to get . In other words,

Now add and subtract inside the parenthesis. Make sure to place this after the "x" term. Notice how . So the expression is not changed.

Group the first three terms.

Factor to get .

Combine like terms.

Distribute.

Multiply.

So after completing the square, transforms to . So .

================================
Answer:

So the answer is which is now in vertex form where , , and
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