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#### MammaOrnelas

##### New member

- May 2, 2012

- 6

- Thread starter MammaOrnelas
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- #1

- May 2, 2012

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- Feb 1, 2012

- 20

That form of a quadratic equation is called the completed square form.

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- #3

- May 2, 2012

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- Feb 1, 2012

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I'd be very happy to help you as much as I can here, should you have any problems.

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- #5

- May 2, 2012

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---------- Post added at 23:09 ---------- Previous post was at 23:07 ----------

the problem is f(x)= -3(x-6)^2-4 I know I have to set it equal to zero. It wants the vertex, y intercept and axis of symmetry. I don't know how to do this.

I had to miss three days of class last week due to a death of a family member in another state. I am extremely lost..............

- Feb 1, 2012

- 20

Okay, from what I gave you earlier, can you see that:

a = -3

b = 6

c = -4

?

a = -3

b = 6

c = -4

?

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- #7

- May 2, 2012

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- Feb 1, 2012

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That one general form is: [tex]y = ax^2 + bx +c[/tex]

Your example: [tex]y = x^2 + 4x + 8[/tex]

Which makes: a = 1, b = 4 and c = 8.

In the completed square form, we have:

[tex]y = a(x-b)^2 + c[/tex]

And in your problem: [tex]y= -3(x-6)^2-4[/tex]

In the same way, we have: a = -3, b = 6 and c = 4.

Does that make things clearer?

And subsequently, remember what I said in my first post here:

Vertex = (b, c)

Axis of symmetry = x = b

y-intercept is obtained by putting x = 0.

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- #9

- May 2, 2012

- 6

---------- Post added at 23:30 ---------- Previous post was at 23:26 ----------

I got all of it right but it said that the y-intercept is (0,-112).....how did they get that

- Feb 1, 2012

- 20

[tex]y = -3(x-6)^2-4[/tex]

By putting x = 0, we get:

[tex]y = -3(0-6)^2-4[/tex]

[tex]y = -3(36)-4[/tex]

[tex]y = -108-4[/tex]

[tex]y = -117[/tex]

Wasn't that easy?

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- #11

- May 2, 2012

- 6