August 26, 2010 6:45 PM Subscribe

Can someone help me with a maths problem please, how to find y for a given x on a curve?

I am struggling with some probably basic maths. I understand so little maths that I don't know the right terms to use but I'll try and explain.

This image shows a curve. I am trying to find the equation which will let me find y for a given x on a curve like this.

http://imgur.com/hxCWk.png

I want to be able to try different scales on x and y axes and get the resulting numbers for y. I have no idea how to do this. And it's not homework, I am a long time out of school.
posted by zingzangzung to Science & Nature (14 answers total)

I am struggling with some probably basic maths. I understand so little maths that I don't know the right terms to use but I'll try and explain.

This image shows a curve. I am trying to find the equation which will let me find y for a given x on a curve like this.

http://imgur.com/hxCWk.png

I want to be able to try different scales on x and y axes and get the resulting numbers for y. I have no idea how to do this. And it's not homework, I am a long time out of school.

Which of these statements is closest to what you want:

1. Help me find y for a given x on the specific parabola (U-shaped curve) shown at imgur.com/hxCWk.png

2. Help me find y for a given x for any parabola

3. Help me find y for a given x for ANY curve, parabola or not

posted by 23skidoo at 6:55 PM on August 26, 2010

1. Help me find y for a given x on the specific parabola (U-shaped curve) shown at imgur.com/hxCWk.png

2. Help me find y for a given x for any parabola

3. Help me find y for a given x for ANY curve, parabola or not

posted by 23skidoo at 6:55 PM on August 26, 2010

I think he wants to solve for the equation of the curve. (Which I could do in about 3 seconds in high school, but it alludes me now.)

posted by lee at 7:03 PM on August 26, 2010

posted by lee at 7:03 PM on August 26, 2010

23skidoo I think no. 3 is closest to what I want, as I am only really interested in the results between say x = -3 and x=3 and I would like to be able to input different curves so I can get the results for say a flatter curve.

posted by zingzangzung at 7:03 PM on August 26, 2010

posted by zingzangzung at 7:03 PM on August 26, 2010

Well, it looks like a parabola. So we know the equation is something like y = a*x^{2} + b. The question is 'what's a?' We can see that when x = 0, y = 5. So we know b = 5.

So now we only need to know a. We know that when x = 5, y = 0. So that means a*5^{2} + 5 = 0. 5^{2} is 25, so now we know a*25 + 5 = 0.

We can solve that pretty easily.

First divide by 25 and we get a + 1/5 = 0. (5/25 = 1/5 and zero divided by 5 = 0)

Then you can move the one to the other side of the equals sign: a*5 = -1. Whenever you move something across the equals sign, you make it negative.

So now we know a = -1/5.

So the equation is y = -(1/5)*x^{2} + 5

If you put that into wolfram alpha you get the same graph.

---

Hmm, actually I looked a little closer and it doesn't actually cross y at exactly five, but more like 4.5 or something. But anyway that's how you do it, for that particular type of curve.

Playing around with wolfram alpha, using 1/4th instead of 1/5 gets you a little closer. It's also possible that that's not a parabola either.

posted by delmoi at 7:04 PM on August 26, 2010

So now we only need to know a. We know that when x = 5, y = 0. So that means a*5

We can solve that pretty easily.

First divide by 25 and we get a + 1/5 = 0. (5/25 = 1/5 and zero divided by 5 = 0)

Then you can move the one to the other side of the equals sign: a*5 = -1. Whenever you move something across the equals sign, you make it negative.

So now we know a = -1/5.

So the equation is y = -(1/5)*x

If you put that into wolfram alpha you get the same graph.

---

Hmm, actually I looked a little closer and it doesn't actually cross y at exactly five, but more like 4.5 or something. But anyway that's how you do it, for that particular type of curve.

Playing around with wolfram alpha, using 1/4th instead of 1/5 gets you a little closer. It's also possible that that's not a parabola either.

posted by delmoi at 7:04 PM on August 26, 2010

There isn't really any way to specifically find out what kind of curve a curve is. You can only try different curves and measure the error. There is computer software that will do this given a set of points.

posted by delmoi at 7:07 PM on August 26, 2010

Yeah, the key is deciding that the curve is a standard parabola, which is a safe assumption I think.

Given that, you can determine the exact formula that produces that curve by using the spots where it intersects the X and Y axes. You'll have to learn the anatomy of a parabolic curve -- vertex, focus, etc.

posted by intermod at 7:07 PM on August 26, 2010

Given that, you can determine the exact formula that produces that curve by using the spots where it intersects the X and Y axes. You'll have to learn the anatomy of a parabolic curve -- vertex, focus, etc.

posted by intermod at 7:07 PM on August 26, 2010

If you know the vertex of the parabola, as well as another point (such as a root - where it crosses the y-axis) you can easily get the formula for the parabola using the vertex form equation:

y=a(x-h)^2 + k where (h,k) is the vertex. Then just solve for 'a' using another known point.

Vertex form hints: http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php

posted by Diplodocus at 7:14 PM on August 26, 2010

y=a(x-h)^2 + k where (h,k) is the vertex. Then just solve for 'a' using another known point.

Vertex form hints: http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php

posted by Diplodocus at 7:14 PM on August 26, 2010

You're really asking two questions here. Question one is, "Given this graph, how do I find the equation that represents the curve?" Question two is, "Given an equation, how do I calculate y for a given x?"

Question two is much simpler: you take the equation, substitute your known value in place of x, and then solve for y. "Solve for y" means you rearrange terms until you have just y on one side and numbers on the other side.

Question one is a little more involved. You have to know what kind of curve you have a graph of in order to determine its equation, i.e. is it a straight line, parabola, hyperbola, exponential, cubic, quartic, etc. Once you know what kind of curve it is then you need to measure a certain number of points. In the case of a line and a parabola, you need to know two points on the curve to get its equation. delmoi gave you an example of that using (0,5) and (5,0) as the two points because it's generally easiest to measure points where one of the values is zero, since that is where it's crossing an axis and you can just read a value for the other number. But it doesn't have to be at an axis crossing, it can be any two points along the curve.

posted by Rhomboid at 7:49 PM on August 26, 2010

Question two is much simpler: you take the equation, substitute your known value in place of x, and then solve for y. "Solve for y" means you rearrange terms until you have just y on one side and numbers on the other side.

Question one is a little more involved. You have to know what kind of curve you have a graph of in order to determine its equation, i.e. is it a straight line, parabola, hyperbola, exponential, cubic, quartic, etc. Once you know what kind of curve it is then you need to measure a certain number of points. In the case of a line and a parabola, you need to know two points on the curve to get its equation. delmoi gave you an example of that using (0,5) and (5,0) as the two points because it's generally easiest to measure points where one of the values is zero, since that is where it's crossing an axis and you can just read a value for the other number. But it doesn't have to be at an axis crossing, it can be any two points along the curve.

posted by Rhomboid at 7:49 PM on August 26, 2010

ZunZun can find equations to fit curves, if you have data for various x,y points.

posted by novalis_dt at 7:52 PM on August 26, 2010

posted by novalis_dt at 7:52 PM on August 26, 2010

It's a parabola with zeros at x=4.5, x=-4.5. At x=0, y=5 So:

y=(x-4.5)*(x+4.5)*k

Evaluate at x=0:

5 = -20.25*k

k = - (20/81) = -.24691...

So:

y= -.24691*(x-4.5)*(x+4.5)

You can expand the right side if you want but if you just want to plot the line or get values for y, that's sufficient as it is.

posted by chairface at 10:02 PM on August 26, 2010

y=(x-4.5)*(x+4.5)*k

Evaluate at x=0:

5 = -20.25*k

k = - (20/81) = -.24691...

So:

y= -.24691*(x-4.5)*(x+4.5)

You can expand the right side if you want but if you just want to plot the line or get values for y, that's sufficient as it is.

posted by chairface at 10:02 PM on August 26, 2010

Any parabola will be a stretched and shifted version of this graph, so it won't look "flatter". You could get flatter graphs with higher-degree even polynomials like y = ax^4 + b and so on.

posted by scose at 10:47 PM on August 26, 2010

posted by scose at 10:47 PM on August 26, 2010

assuming it is a parabla - it seems to go pretty much directly through the point (3, 3) which would suggest that the curve is y = -(2/9)x^2 + 5

then you just subsititute your required series of values of x and that will give you a series or points that lie on the graph.

posted by mary8nne at 5:00 AM on August 27, 2010

then you just subsititute your required series of values of x and that will give you a series or points that lie on the graph.

posted by mary8nne at 5:00 AM on August 27, 2010

This is going to be impossible to do with pencil and paper. There are easy algorithms out there that you could use to get an approximate answer on a computer in Matlab or some other software.

If you know the form of the curve is a parabola (as most are assuming here -- it may not be, it could be anything; for example it could be a 3rd degree polynomial where the figure has just zoomed in on a portion of it) than you can follow the above instructions.

You say you want to be able to input different curves; will you know the equations of these curves?

posted by bluefly at 6:18 AM on August 27, 2010

This thread is closed to new comments.

posted by lee at 6:55 PM on August 26, 2010