# Math Help - graphing this parabola

1. ## graphing this parabola

4y^2-4y-4x+5=0

this is what i did:

4y^2-4y=4x-5
4y^2-4y=4x-5+4
4^2-4y+4=4x-5+4
4(y^2-2y+2)=4x-1
4(y-2)^2=4x-1

i don't really know how to move to the next step in order to get something like (y-k)^2=4p(x-h) or (x-h)^2=4p(y-k)

it doesn't make sense cause i have the 4 outside which seems like its part of the 4p but (y-2) is squared.... and i can't complete the square on the right side since no x^2 value, only x.

2. Originally Posted by Frenchie
4y^2-4y-4x+5=0

this is what i did:

4y^2-4y=4x-5
4y^2-4y=4x-5+4
4^2-4y+4=4x-5+4
4(y^2-2y+2)=4x-1
4(y-2)^2=4x-1

i don't really know how to move to the next step in order to get something like (y-k)^2=4p(x-h) or (x-h)^2=4p(y-k)

it doesn't make sense cause i have the 4 outside which seems like its part of the 4p but (y-2) is squared.... and i can't complete the square on the right side since no x^2 value, only x.
Sorry, this is wrong. If you are going to complete the square, the coefficient of the square term must be 1.

$4y^2-4y-4x+5=0$

$4y^2 - 4y + 5 = 4x$

$4\left(y^2 - y + \frac{5}{4}\right) = 4x$

$4\left[y^2 - y + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + \frac{5}{4}\right] = 4x$

$4\left[\left(y-\frac{1}{2}\right)^2 + 1\right] = 4x$

$\left(y - \frac{1}{2}\right)^2 + 1 = x$.

You should be able to graph this now...

3. thank you! I have another question, is that equivalent to this:
(y-1/2)^2=(x-1)

see it doesn't make sense because whats P? there is nothing infront of (x-1) such as 8(x-1) where in that case we could say 4p(x-1), p=2

it doesn't look like this general form (y-k)^2=4p(x-h) =/

4. ah nvm i think its (y-1/2)^2=1/4(x-1)... or is it (y-1/2^2)=4(x-1)... eh w.e good enough

thank you

5. Originally Posted by Frenchie
thank you! I have another question, is that equivalent to this:
(y-1/2)^2=(x-1)

see it doesn't make sense because whats P? there is nothing infront of (x-1) such as 8(x-1) where in that case we could say 4p(x-1), p=2
"no" coefficient means the coefficient is 1. If 4p= 1, then p= 1/4.

it doesn't look like this general form (y-k)^2=4p(x-h) =/
But it is.
$(y- 1/2)^2= x- 1$ is the same as $(y- 1/2)^2= 4(1/4)(x- 1)$. p= 1/4, k= 1/2, and h= 1.

6. ah thank you all this is helpful