Question

I need help for parabola stuff.

Answer 1

here it is https://www.khanacademy.org/math/algebra2/conics_precalc/parabolas_precalc/e/parabola_intuition_2

Answer 2

How do I do this exercise without doing guess and check? @aum

Answer 3

This is me: khanacademy.com/profile/Davhla

Answer 4

From the given equation of the parabola first determine the vertex of the parabola.

Answer 5

Also notice how when the right hand side is expanded, the x^2 term will have a negative coefficient of -1/4. That means the parabola will be upside down or opening downward.

Answer 6

I know how to get the x value thing right, but the other ones don't work well.

Answer 7

@Whitemonsterbunny17 @math215 @mathmale Help plz

Answer 8

@scarymary101 @Dark_Riddles Help please or tag friends here to help!

Answer 9

@Hero Please help me!

Answer 10

y - 4 = -1/4 * (x+3)^2
The vertex of the parabola is (-3, 4).

Due to the negative coefficient, -1/4, of x^2 this is a parabola that opens downward (an inverted parabola). Therefore, the directrix will be above the vertex and focus will be below the vertex. Above or below by how much? By one-fourth of the coefficient of x^2. That is, by \(\large \pm \frac{1}{16}\).

So you have all the information to drag the focus and the directrix to the proper location.

Answer 11

Sorry, but I have NO idea WHATSOEVER what you mean... Can you like make a video and show me, and talk during the video and explain? Thanks.

Answer 12

The link you sent me doesn't play any video. It just shows a dark video screen with no controls to play the video.

Answer 13

oh no...

Answer 14

But instead of giving out all your account details why don't you learn about parabolas and try to do it yourself. Do you know what is the vertex, focus, directrix of a parabola are?

Answer 15

It works perfectly fine for me. Just try refreshing. The problem is probably in your java settings in your control panel.

I have tried that, I have watched the videos 3 times today and I don't understand.

Answer 16

I know what the vertex is, exactly, and perfectly well, If the parabola is like a U, then the vertex is the point at the bottom, the lowest point. If the parabola is like a n, then the vertex is the highest point. I know about the focus and directrix a little bit, but I am not confident in my knowledge of them.

Answer 17

Forget the video. Try this. First click on the directrix and move it up to the horizontal line at y = +4 first a then just a tiny fraction above +4. Do that first and tell me when you are done.

Answer 18

Done!

Answer 19

Click on the focus and drag it near the point (-3,4) such that the vertex (not the focus) of the parabola is at (-3,4). Keep an eye on the changing equation on the right. It should say y - 4 = -(some constant) * (x+3)^2
Keep moving the focus around until you get the -4 part on the left of the equation and the +3 part on the right of the equation next to x. This will happen when the vertex of the parabola is exactly at (-3,4).

Answer 20

Done.

Answer 21

What does the equation on the right say?

Answer 22

y −4 = −1 (x + 3)^2

Answer 23

You have gotten most of it right at this point. To make the -1 into -1/4 you need very precise movement which may not be possible with a mouse and the scale they are using for the x and y axes on the graph. The directrix need to come DOWN by a very tiny fraction and then the focus has to be moved VERTICALLY UP by the same fraction. This is a very small fraction to move and it may not be possible but you can play around until you get it.

Answer 24

In fact I would say it is almost impossible to do because the directrix has to be at the horizontal line \(\large y = 4\frac{1}{16}\). How is one supposed to move one-sixteenth of one division on the graph. The focus has to be one-sixteenth vertically below the vertex. Such high precision may not be possible.

But instead of a fraction if they had a nice round number that is a multiple of 4 in the place of -14 then we can do it precisely.

Answer 25

But instead of a fraction if they had a nice round number that is a multiple of 4 in the place of -1/4 then we can do it precisely.

y −4 = −A (x + 3)^2 where A is a nice round number, preferably a multiple of 4, then we can do it precisely.

Answer 26

There has to be a way to do it.

Answer 27

i looked at the hints and this is what the answer is. i dont understand. still. sorry

Answer 28

If there is a way to magnify the diagram then it may be possible.
But you get the idea on how the focus, directrix, vertex determine the shape and nature of a parabola. After this point it is no longer math leaning but more to do with how precise your mouse control is. Look for the comment section to see if anyone has posted a comment saying they have done it precisely and what device they used to do it.

Answer 29

There is no comment section fro exercises.

Answer 30

There are so many things wrong with the answer you posted. Answer has a positive 1/4 but question has -1/4 which makes the parabola upside down.
y - 4 = -1/4(x+3)^2 and the graph in the answer is:
y-3/2 = 1/4(x+5/2)^2

Answer 31

Nope.... Look closely there is no negative sign.

Answer 32

I am talking about the first problem you posted. is this a different problem?

Answer 33

Thank you so much for putting up with me aum.

This is the problem I've been working on the whole time.

Answer 34

Have you switched to a different problem than the one you posted first?

Answer 35

No. I have only done 2 problems- one long before I came here for help, and this problem.

Answer 36

Maybe the system gave you a different problem.

Answer 37

If I click on the first link you posted I get y - 4 = -1/4(x+3)^2
If I click on the second link you posted I get y - 3/2 = 1/4(x+5/2)^2

completely different problems.

Answer 38

I'm confused now, so can we throw away the old problems and work on a new problem?

Answer 39

Here is the problem.

Answer 40

I am seeing y - 1/2 = -1/14(x - 3/2)^2
Is this what you are seeing?

Answer 41

If you are seeing a different problem, then the site perhaps posts a different problem when the page is accessed from different computers or accounts.

Answer 42

No it is a random problem system.

Answer 43

So no point asking here for help.
But if you write down the problem here I can give you a hint how to get close to it and after that you will have to play with it to tune it precisely.

Answer 44

Yes i gave u the picture of the problem.

Answer 45

Also i don't understand how to do these problems overall.

Answer 46

y - 1/2 = -1/14 * (x - 3/2)^2
The general vertex form of a parabola is: y - k = a(x - h)^2
where (h, k) is the vertex.

Compare the two. h = 3/2 and y = 1/2. And a = -1/14
So the vertex of the parabola is (3/2, 1/2).

Since a = -1/14 is negative this parabola opens downward. If a parabola opens downward, its directrix will be above the vertex (A parabola will never intersect the directrix). And the focus will be below the vertex.

Switch them if a is positive. For positive a, the parabola will open upwards, the directrix will be below the vertex and the focus will be above the vertex.

By how much above or below? By |a / 4|. Here it is by 1 / (14*4) = 1/56.

This should be enough to quickly move the directrix and focus around so the vertex is precisely at (3/2, 1/2). Then you can tweak it a bit finer but I think it would be nearly impossible to get precisely -1/14 using a mouse and with the graph scale they are using.

Answer 47

Can you explain more why you do that? I don't understand and I don't know how to do these problems. @Hero @aum

Answer 48

@ganeshie8 @zepdrix Help math please

Answer 49

@mathmale

Answer 50

You may want to close this one and post a new one where you have a picture of the problem and not the link. Each time a new person comes into this thread and click on the link they are going to have a random problem assigned to them and they will be answering a question that is different from the one you are looking at.

Answer 51

Hi mathmale! I am having trouble with parabola directrex and focus.

Answer 52

Hi, David! I am a bit reluctant to jump in at this point because @aum has been helping you and doing (in my opinion) a very good, very well-informed job of it.

If your math problems are presented as graphs, then I'd suggest you take a screen shot of the graph and share only that screen shot, so that anyone and everyone helping you will see precisely the same graph.

Also, David, I'd suggest you ask questions as specific as you possibly can at this point, rather than saying, "I don't understand."

Answer 53

Which problem would you like to focus on, and what specifically would you like help with?

Answer 54

here is the problem. my trouble is more general. i will attach something better later..

Answer 55

OK, our starting point will be this screen shot (see the attached).
Exactly what would you like to know in regard to this problem?

Answer 56

Aum has correctly explained how to identify the coordinates of the vertex of this parabola. Can you do that?

Answer 57

Yes.

Answer 58

the vertex of the parabola given to us as a formula is (h, k); here, with h and k known, that vertex is ( ? , ? )

Answer 59

I have no idea.

Answer 60

@mathmale help

Answer 61

@Whitemonsterbunny17

Answer 62

help?!!?!?

Answer 63

David, there are several forms of the equations that all represent parabolas.
You are familiar with the quadratic function \[y=ax^2+bx+c\] Another form is

\[y-k = a(x-h)^2\]...where h is the x-coordinate of the vertex of your parabola and k is the y-coordinate. Thus, the vertex is at (h,k).

If you're given y-2=2(x-3)^2, you should be able to tell immediately that the x-coord. of the vertex is 3 and the y-coord. of the vertex is -2.

Please make a note of all this, as you will definitely see these equations again in the future.

Answer 64

going back to the problem at hand, what are the coordinates of the vertex?

Answer 65

ohhh.... i remember....

Answer 66

There is a third form of the equation for a vertical parabola: 4py=x^2, or \[4p(y-k) =(x-h)^2, \]
where h and k are again the x- and y-coords. of the vertex. 'p' represents the distance from the vertex to the focus OR the distance from the directrix to the vertex.

Answer 67

What do you need to know right this minute, David?

Answer 68

How to do these problems: https://www.khanacademy.org/math/algebra2/conics_precalc/parabolas_precalc/e/parabola_intuition_2

Answer 69

look here http://screencast-o-matic.com/watch/c2ib05nHph
no talking in it, just typing

Answer 70

David, i have downloaded the latest version of Java, and have tried several times to run http://screencast-o-matic.com/watch/c2ib05nHph, but am not seeing this video.

Sorry, but at this late hour my creativity is not great enough to help me determine what is wrong with my system so that it will not display this video.

I need to hit the sack.

However, if you can find some alternative way to share the info presented in this video, I'll try my best to be of help. But that'll be tomorrow, not tonight.

Best to you.

Answer 71

Thanks, good night.

Answer 72

You may have to take a screen short or two of the Khan Academy video, or find some other way to share the video, before I can help interpret the meaning of the video for you. Actually, if I see examples of the types of problems you're working on, I could help you with them without seeing the video at all.