Parabola:
Rewrite 3y^2 + 6y + 108x - 969=0 in standard form, show work.
3(y^2 + 2y + ...) - 969 = -108x
3(y^2 + 2y + 1) - 969 - 3 = -108x
3(y+1)^2 - 972 = -108x
(-1/36) (y+1)^2 + 9 = x
I will assume you realize that this is a "horizontal" parabola and that you know how to read the vertex from that form.
Just realized that Steve had already answered this question before you reposted it
http://www.jiskha.com/display.cgi?id=1324054249
Have patience and always check if your question has been answered before re-posting it.
It saves unnecessary work on our part.
To rewrite the given equation in standard form for a parabola, we need to complete the square for both the x and y terms.
Step 1: Group the x and y terms separately.
3y^2 + 6y = -108x + 969
Step 2: Factor out the coefficient of y^2 (which is 3) from the y terms.
3(y^2 + 2y) = -108x + 969
Step 3: Complete the square for the y terms. Add the square of half the coefficient of y to both sides of the equation.
3(y^2 + 2y + 1) = -108x + 969 + 3(1)
Simplifying:
3(y + 1)^2 = -108x + 972
Step 4: Divide both sides of the equation by the coefficient of (y + 1)^2 to make it equal to 1.
(y + 1)^2 = (-108/3)x + 324/3
Simplifying:
(y + 1)^2 = -36x + 108
Step 5: Rewrite the equation in standard form, which is in the form (x - h)^2 = 4p(y - k).
(x - h)^2 = 4p(y - k)
Comparing with the equation we obtained:
(y + 1)^2 = -36x + 108
We can see that h = 0, k = -1, and 4p = -36.
Dividing the equation by -36:
(y + 1)^2 / -36 = x - 3
Finally, rearranging the terms:
x = -(y + 1)^2 / 36 + 3
So, the equation 3y^2 + 6y + 108x - 969 = 0 can be rewritten in standard form as x = -(y + 1)^2 / 36 + 3.