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Hi all. I am totally lost here. I have to find the point R. I am given that the range of the parabola is (-2,infinity) So I know the x-coordinate is -2 but how would I find y? As far as I can remember I need to x-intercepts and a max y vallue to do this? I can see that the one point which it passes through is (0,0) I can also see that the axis of symmetry is x=2 so should I times that by 2 and then the negative x-intercept is -4? Thanks a lot in advance
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picture in gif format.S
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hi Deon588
At R, it is the y coordinate that is -2, not x
So Q may not be (-4,0)
I've had a quick go at this and got stuck because of insufficent information
Can you post the wording for the question please?
Got to go to day two of my first aid course. I'll look again after that and post again if no one else has beat me to it.
QUICK EDIT
Definitely more than one answer based on the information seen.
and
Bob
Last edited by Bob (2011-06-13 18:37:29)
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob;
How did the first aid go?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi bob thanks a lot. Can you recomend any software which I can use to practice LaTeX? It would be much faster if could type the math instead of uploading gif images.
Good luck with the first aid!
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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi Bobbym. Thanks but what I meant was a offline editor or something simular. I am currently doing all my work in electronic format. I'm using design science mathtype but it is a bit limited in some ways. Also do you have any ideas on finding R?
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Hi Deon588;
Some people here use an offline program. I do not so I can not help there.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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No worries. Can you please explain how you went about finding the answer? Thanks a lot
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The x value of the parabola's vertex is given by:
So if:
then
You know from the range that y = - 2, so the vertex is located at ( -2,-2). Now you know that Q is at ( -4, 0 ). So you have 3 points (-4,0),(-2,-2) and (0,0). The rest is just curve fitting.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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So for the x-coordinate of R. can I say that because b/2a=2 -b/2a =-2? I find it really amazing that there are people like you on here willing to help at no apparent financial gain? Are you offiliated with mathsisfun.com? Or do you just love the math?
Thanks again
Deon
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here willing to help at no apparent financial gain.
There are other types of gain besides financial.
Are you offiliated with mathsisfun.com?
No, I just hang here. MIF runs the forum and it is his brainchild.
Or do you just love the math?
Yes, that is one of the other types of gain I mentioned.
Glad to help.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi Deon588 and bobbym
I am now re-qualified as a first aider for CPR and stuff like that.
I mostly do my Latex by typing in the commands.
http://www.mathisfunforum.com/viewtopic.php?id=4397
is good to get started and then I peek at other people's Latex to see how they did it. What you do is click on the Latex in someone's post and a window appears with the right code.
You have to 'top' and 'tail' each line of code with square brackets 'math' and '/math' to switch in the Latex interpreter.
Bob
Last edited by Bob (2011-06-15 00:02:20)
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi bob bundy;
Congratulations!
Thanks for pointing out the typo, I fixed it up.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Congratulations Bob!!! I am still doing something wrong as my solution to the equation always is concave down even though I can see from the graph that it's concave up. Sorry I don't usually take this long to understand something but parabolas is something totally new to me.
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hi Deon588
Just read your post. Stay on-line. Answer coming up
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi;
The first method is fine you are not forming your simultaneous linear equations right.
a = 1 / 2 and b = 2.
I did not see your answer Bob. Sorry.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Answer.
The general form for a parabola this way round is
Put in (0,0) and c must be zero.
So now we have
Put in (-4,0)
Put in (-2,-2)
Subtracting (ii) from (i) (watch out for double negatives here!)
so
Hope that helps.
Once again you've pipped me to the post, bobbym!!
Bob
Last edited by Bob (2011-06-14 23:04:32)
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks for the quick responses. Bob thanks a lot for posting that detailed explanation anyhow because I still was a bit unsure but now I know where my mistake lies. I am studying from home at the moment and have a class with a tutor once a week which makes it very difficult when I get stuck somewhere.
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Well you are lucky then because you can ask a question in here as often as you like and someone will answer it.
Once again you've pipped me to the post, bobbym!!
You got there first ( post #16 ), I just did not see it.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi Deon588
Glad to help.
bobbym:
You got there first ( post #16 ), I just did not see it.
Unfortunately, they didn't cover this on the first aid course.
It was a relief to see you had the same values.
Actually, I had it in post #3.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob;
With all the math fear in the world, they may have to start teaching CPR and math together.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi bobbym,
Last time on the Red Cross training they use 'Staying Alive' by the Bee Gees to get the pace.
The St Johns Ambulance have a darker sense of humour. They use 'Another One Bites The Dust' by Queen.
But on the new course we had to 'up the pace' to 'Nellie the Elephant' speed. {For any would-be first aiders reading this, that's about 110 pumps a minute!)
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi;
That does seem kind of brisk. A prolonged pace like that might produce 2 people needing first aid.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi. I am posting this here because it has to do with the same graph. I have to find the point P. As this is one of the points of intersection of g(x) and f(x) I tried solving the system of equations. At the end I got x^2+5x=0 and I didn't know how to proceed. One of my previous questions was find the equation of the line which is paralel to g and passes through the point R Which is (-2,-2) which I got as y=-1/2x-3. Can I solve the system of equations of that line and f(x) and then subtract 3 from the x coordinate and add 3 to the y-coordinate?
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