Question

can someone walk me through how to solve this parabola???
x=1/2y^2 +3y+4

Answer 1

By solve, what exactly do you mean?

Answer 2

Solve for y?

Answer 3

find an equation for it in standard form

Answer 4

Find an equation? I'm confused.

Answer 5

an equation... in standard form...

Answer 6

Do you need it in the form x = a(x - h)^2 + k or x = ax^2 + bx + c?

Answer 7

the first one. it's already in the second form

Answer 8

i just keep messing up and need some help

Answer 9

Oh. ok. THat's called general form where I come from and the second one is standard form.
Ok, we're gonna do something called complete the square. First set x = 0
0=1/2y^2 +3y+4
Then, bring the constant to the other side.
-4 = 1/2y^2 + 3y
Now, remember to never complete the square unless the leading coefficient is 1. To get there, multiply the entire equation by 2.
-8 = y^2 + 6y
Now, you have to recognize that on the right side, it is starting to follow a perfect square trinomial. You have to complete the square by dividing 6 by 2 and then squaring it.
-8 = y^2 + 6y + 9
Since this would change the equation, you also have to add a 9 on the other side to balance the equation.
-8 + 9 = y^2 + 6y + 9
1 = y^2 + 6y + 9
Now it's starting to look like a perfect square trinomial right? You know how to factor it now.
1 = (y + 3)^2
To finish it, bring 1 back to the other side and set it equal to x again.
x = (y + 3)^2 - 1

Answer 10

Sorry that it's so long.

Answer 11

thats ok... and i get what ur saying but my book is telling me a different answer

Answer 12

oh. What's it say?

Answer 13

What does your textbook say is the answer?

Answer 14

x+1/2=1/2(y+3)^2

Answer 15

Oh. I see what they did. Instead of multiplying by 2, they factored out 1/2.
-4 = 1/2y^2 + 3y
-4 = 1/2(y^2 + 6y + 9)
-4 + 9/2 = 1/2(y^2 + 6y + 9)
-4 + 9/2 = 1/2(y + 3)^2
1/2 = 1/2(y + 3)^2
Now, instead of bringing it to the other side, they just brought back x.
x + 1/2 = 1/2(y + 3)^2

Answer 16

Get it?

Answer 17

YES! thank u!!!

Answer 18

You're welcome! :D

Answer 19

and to find the directrix... can u possibly help with that?