Write equation of parabola?

How do I write the equation of the parabola in standard form:

y=x^2-6x-1

2009-06-24T10:17:22Z

Hm. No one responded with the multiple choice I received.

A) Y+10= (X-3)^2
B) y-1=(x-3)^2
C)y+6=(x-3)^2
D)y-3=(x-3)^2

?2009-06-24T10:09:13Z

Favorite Answer

you need to do the method known as completing the square.


y = x^2 -6x -1
y = (x^2 -6x +9 - 9) -1
y = (x^2 -6x +9) -10
y = (x - 3)^2 -10


Steps:

1. If the value in front of 'x' isn't 1 then factor it out of the first two terms

2. Divide the second term by 2 and then square it

3. Add and then subtract this value

4. Take the fourth thing in the brackets out

5. Combine like terms and factor

Anonymous2009-06-24T10:05:09Z

y = x^2 - 6x - 1
==> y = (x^2 - 6x + 9) - 1 - 9
==> y = (x - 3)^2 - 10

EDIT:

Depending on the study, the standard forms of parabolas do vary. In this case, the answer is A.

Hope this helps!

Anonymous2016-04-11T06:11:16Z

Standard equation: y= a(x-h)^2 + k {(h,k) is the vertex} Re-written as: (x – 7)^2 = 4p(y – 5) {4p = 1/a} Assuming Directrix Y=1, Distance between vertex and directrix = absolute value (1-5) = +4 = p (x-7)^2 = 4(4)(y-5) y-5 = [(x-6)^2]/16 y= [(x-6)^2]/16 + 5

thailine2009-06-24T10:05:21Z

That is standard form for a parabola.

a = 1, b = -6, c = -1

The previous answer is in vertex form

musicalmissy2009-06-24T10:07:14Z

i think it would be -y= 7x+1

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