write in standard form, is this right?

[gymnastqueen]

This is what I have so far, but I suspect that either my method is wrong or I have made mistakes in my calculations somewhere, so is this correct and if not why not?
write 3y[sup:2c5qkzad]2[/sup:2c5qkzad]-x-12y+14=0 in standard form

3y[sup:2c5qkzad]2[/sup:2c5qkzad]-x-12y+14=0
3y[sup:2c5qkzad]2[/sup:2c5qkzad]-12y-x+14=0
3[(y[sup:2c5qkzad]2[/sup:2c5qkzad]-4y=4)-4]-x+14=0
3(y-2)[sup:2c5qkzad]2[/sup:2c5qkzad]-12-x+14=0
((3(y-2)[sup:2c5qkzad]2[/sup:2c5qkzad])/-2)-(x/2)=(-2/-2)
((3(y-2)[sup:2c5qkzad]2[/sup:2c5qkzad])/-2)-(x/2)=1

What info does this give me to graphing this equation?

 

[wjm11]

This is what I have so far, but I suspect that either my method is wrong or I have made mistakes in my calculations somewhere, so is this correct and if not why not?
write 3y2-x-12y+14=0 in standard form

3y2-x-12y+14=0
3y2-12y-x+14=0
3[(y2-4y=4)-4]-x+14=0
3(y-2)2-12-x+14=0
((3(y-2)2)/-2)-(x/2)=(-2/-2)
((3(y-2)2)/-2)-(x/2)=1

What info does this give me to graphing this equation?

Hello, gymnastqueen,

Your math steps are okay, but let’s analyze the overall problem and see if that’s the way we want to go. Consider your original equation, 3y2-x-12y+14=0. If we move x over to the other side, we have

3y^2 - 12y +14 = x

You might recognize this as a “sideways” parabola with an x intercept of 14. This is written in standard form.

If we complete the square as you did, we’d have

x = 3(y-2)^2 + 2

which is in vertex form, the vertex being the point (2,2).

Now if x and y *both* had squared terms, we’d have either an ellipse or a hyperbola, and your form/approach would have been the correct one for that.

Hope that helps.

 

[Loren]

I believe you are aiming at getting your equation into the form...
\(\displaystyle (y-k)^2 = 4p(x-h)\) where p is a nonzero real number, and the vertex is at (h,k).

 

[gymnastqueen]

I believe you are aiming at getting your equation into the form...
\(\displaystyle (y-k)^2 = 4p(x-h)\) where p is a nonzero real number, and the vertex is at (h,k).

Yes! this is what I'm trying to do, does anybody know what the proper way of achieving this is?

 

[Loren]

3y[sup:22e1o3iv]2[/sup:22e1o3iv]-x-12y+14=0

3y[sup:22e1o3iv]2[/sup:22e1o3iv]-12y=x-14

y[sup:22e1o3iv]2[/sup:22e1o3iv]-4y+___=(1/3)(x-14)+___

Complete the square on the left side will get you started toward having a binomial squared there. I leave the rest of the maneuvering to you.

 

[Mrspi]

I believe you are aiming at getting your equation into the form...
\(\displaystyle (y-k)^2 = 4p(x-h)\) where p is a nonzero real number, and the vertex is at (h,k).

Yes! this is what I'm trying to do, does anybody know what the proper way of achieving this is?

Since the definition of "standard form" might well vary from textbook to textbook, it would have helped us (in the beginning) to know what form you were working towards. Loren has given you a good beginning step. See what you can do with that. If you are still having trouble, please repost showing all of the work you've done.