Write each quadratic function in vertex form, if not already in it, identify the vertex, axis of symmetry?

Question

and also their direction of opening?

y = x^2 - 2x - 5

y = -3^2 + 24x

y = x^2 - 6


please help

Answer 1

1.y=(x-1)^2-6
vertex=(1,6)
axis of symmetry=x=1

2.y=-3(x^2-8x+16)+48
y=-3(x-4)^2+48
vertex=(4,-48)
axis of symmetryx=4

3.x=x^2-6
vertex=)0,6)
axis of symmetry x=0 or the y axis

Answer 2

Vertex kind ability there is not any remoted x time period: y = a(x-b)^2 + c. that's called this because you may study off the vertex as (b, c). y = -x^2 + 4: already in vertex kind. The vertex is (0, 4), the axis of symmetry is x = 0 and it opens downwards. y = -5x^2 + 9: already in vertex kind. The vertex is (0, 9), the axis of symmetry is x = 0 and it opens downwards. y = x^2 + 6x + 2 = (x + 3)^2 - 3^2 + 2 = (x + 3)^2 - 7. The vertex is (-3, -7), the axis of symmetry is x = -3 and it opens upwards.

Answer 3

Vertex form looks like
y = a(x-h)^2 + k
Your vertex would be (h,k)
Your axis of symmetry is x=h
If a is positive, then it opens up. If not, it opens down.

For y=x^2 - 2x - 5, you would have to complete the square.
y = (x^2 - 2x) - 5
Take (2/2)^2 = 1
y = (x^2 - 2x + 1 - 1) - 5
y = (x^2-2x+1) - 6
y = (x-1)^2 - 6
This is in vertex form.
Your vertex is just going to be (h,k) = (1,-6)
Axis of symmetry is x = 1.
Opens up.

I think you have a typo in the second question. It is most likely
y = -3x^2 +24x
Once again, complete the square.
y = -3(x^2 - 8x)
Take (8/2)^2 = 16
y = -3(x^2 - 8x+16-16)
y = -3(x-4)^2 + 48
Your vertex is (4,48). Your axis of symmetry would be x=4.
Opens down since -3 is negative.

Your third question is already in vertex form since it can be written as
y = (x-0)^2 - 6
So your vertex is (0,-6)
Your axis of symmetry is x=0.
Opens up.

Answer 4

These are all parabolas. Vertex form for a parabola with center (h,k) is:

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

So complete the square for the x term and see what constant amount you have left over for k.

Here is an example for the first one.

y = x^2 - 2x - 5
y = (x^2 - 2x +1) - 1 - 5
y = (x - 1)^2 - 6
y + 6 = (x - 1)^2

So the vertex is at (h,k) = (1,-6).
The coefficient for the x^2 term is positive so the parabola opens up.

You can do the rest.