vertex standard

I know how to convert from vertex to standard, but I dunno how to do standard to vertex...?

Help? : ) Google isn't being very friendly.
Okay, here's a specific sample:

y = x² – 6x + 11

Question Number 1 :
For this equation x^2 - 6*x + 11 = 0 , answer the following questions :
A. Use completing the square to find the root of the equation !

Answer Number 1 :
The equation x^2 - 6*x + 11 = 0 is already in a*x^2+b*x+c=0 form.
As the value is already arranged in a*x^2+b*x+c=0 form, we get the value of a = 1, b = -6, c = 11.

1A. Use completing the square to find the root of the equation !
x^2 - 6*x + 11 = 0 ,divide both side with 1
So we get x^2 - 6*x + 11 = 0 ,
Which means that the coefficient of x is -6
We have to use the fact that ( x + q )^2 = x^2 + 2*q*x + q^2 , and assume that q = -6/2 = -3
Next, we have to separate the constant to form x^2 - 6*x + 9 + 2 = 0


So we will get ( x - 3 )^2 + 2 = 0 -> this is the vertex form

2009-06-18 21:42:54 by EddieLightning

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I know how to write it in vertex form, but how do I get it to standard form?

y=(x+a)^2+b

y= (x+a)(x+b)

Or something like that rofl

2008-03-29 22:17:20 by Cirelli

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This equation is in vertex form y=-0.083(x-0)^2+10, how do you convert it to standard form, please show all your work .

The quadratic term (x-0)^2 makes this a very simple problem to convert.

Step 1:
FOIL the quadratic term, and place the expanded quadratic in the parentheses.
(x-0)^2 = (x^2 +0x +0) = (x^2)

After step 1 we get:
y = -0.083 (x^2) + 10

Step 2:
Distribute the coefficient outside of the parentheses to the quadratic expression inside the parentheses.
-0.083 (x^2) = -0.083x^2

After Step 2 we get:
y = -0.083x^2 + 10

Step 3:
Combine like terms, if necessary.
In this case, this step is not necessary.

Our final quadratic equation is:

y = -0.083x^2 + 10

(If we had to plug these numbers into the quadratic formula, a=-0.083, b=0, c=10)

2009-06-11 16:41:24 by Karen

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How can I change any standard form equation to vertex form,and vise versa? Also how would I change Factored Form to standard form and vertex form. Can you please show examples. Thanks!!

are you talking about conic sections?

to get to the vertex form... simply complete the square.. if there are any quadratic expressions...

say
x^2 + 2y^2 + 4x + 4y - 3 = 0

(x^2+ 4x) + (2y^2 + 4y) = 3
(x^2+4x+4) + 2(y^2+2y+1) = 3 + 4 + 2
(x+2)^2 + 2(y+1)^2 = 9

... although the point discussed here is the center not the vertex, completing the square where only one variable is quadratic will lead to a parabola and the point is really the vertex...

§

2007-10-04 18:19:48 by seniorgirl

first the vertex form:
a(x-k)^2+c

The conversion:

a(x^2-2xk+k^2)+c
ax^2-2axk+ak^2+c

Does this make sense, am I doing it right?

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2009-07-17 20:00:30 by GCal