Pls help
What type of parabola and the value of "P" in 24y² - 36y = 0?
12y(2y - 3y) = 0
Solutions:
y = 0
y = 3/2
24y^2 -36y
x=24y^2-36y
x= 24 (y^2 - (36/24) y)
x = 24 (y^2 - (3/2) y )
x = 24 (y^2 -2(3/2)(1/2) y)
x = 24 (y^2 -(2)(3/4) y + (3/4)^2 -(3/4)^2)
x = 24 ( (y^2 -(2)(3/4)y +(9/16) ) - 9/16)
x = 24 (y-3/4)^2) - 216/16
x+216/16 = 24( y-3/4)^2
x + 27/2 = 24(y-3/4)^2
(y-3/4)^2 = (1/24) (x+27/2)
x- h = 4p(y-k)
h=-27/2 ; k= 3/4
(h,k) = (-27/2,3/4) --vertex
4p=1/24
p=1/96
If the equation were 24y^2 - 36y = x, then it would be a parabola that opens to the right.
24y^2 - 36y = 0 is not a parabola, but two points on the number line, namely
24y^2 - 36y = 0
2y^2- 3y = 0
y(2y - 3) = 0
y = 0 or y = 3/2
I can't tell you the value of P because you haven't defined P.
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Answers & Comments
Pls help
What type of parabola and the value of "P" in 24y² - 36y = 0?
12y(2y - 3y) = 0
Solutions:
y = 0
y = 3/2
24y^2 -36y
x=24y^2-36y
x= 24 (y^2 - (36/24) y)
x = 24 (y^2 - (3/2) y )
x = 24 (y^2 -2(3/2)(1/2) y)
x = 24 (y^2 -(2)(3/4) y + (3/4)^2 -(3/4)^2)
x = 24 ( (y^2 -(2)(3/4)y +(9/16) ) - 9/16)
x = 24 (y-3/4)^2) - 216/16
x+216/16 = 24( y-3/4)^2
x + 27/2 = 24(y-3/4)^2
(y-3/4)^2 = (1/24) (x+27/2)
x- h = 4p(y-k)
h=-27/2 ; k= 3/4
(h,k) = (-27/2,3/4) --vertex
4p=1/24
p=1/96
If the equation were 24y^2 - 36y = x, then it would be a parabola that opens to the right.
24y^2 - 36y = 0 is not a parabola, but two points on the number line, namely
24y^2 - 36y = 0
2y^2- 3y = 0
y(2y - 3) = 0
y = 0 or y = 3/2
I can't tell you the value of P because you haven't defined P.