Maths help? :) (1 Viewer)

[theind1996]

Hey guys, I need some help on how to sketch this semi-circle: y=√(4x-x^2 ) Thanks.

 

[findx]

Download Geogebra. Its helpful when you dont know what a graph looks like

 

[Carrotsticks]

EDIT: Forgot to add that you take the positive upper half since it's a positive square root, but that is of course obvious.

 

[nightweaver066]

y^2 = 4x - x^2
x^2 - 4x + y^2 = 0
(x - 2)^2 + y^2 = 4

Since it was originally y = sqrt(4x - x^2), where the range is always positive, you sketch the upper half of the circle getting what findx posted.

 

[theind1996]

Thanks for the help everyone - much appreciated. Repped you all. Carrotsticks especially, damn Latex looks good.

View attachment 24199

EDIT: Forgot to add that you take the positive upper half since it's a positive square root, but that is of course obvious.

Thanks Carrotsticks. However, is there also another method of solving this problem? I swear my teacher showed me another method, but I just can't find it in my notebook.

View attachment 24198

Download Geogebra. Its helpful when you dont know what a graph looks like

Haha thanks , I already had Geogebra. I needed to solve it algebraically, not just using Geogebra, but thanks anyways.

 

[Carrotsticks]

Thanks Carrotsticks. However, is there also another method of solving this problem? I swear my teacher showed me another method, but I just can't find it in my notebook.

You can use a geometric argument (with a bit of logic) to acquire the answer immediately.

1. We recognise that this is a semi-circle

2. This semi-circle obviously has a root at x=0 and x=4 (by observing the square root if we factorise x out of it)

3. Since the 'y' is by itself (in the form y^2, so nothing weird like y^2 + y etc), we can deduce that this circle has its diameter on the x axis (since the circle is not being shifted up or down at all, so its at a 'neutral' position)

4. Since the diameter lies on the x axis, and it has roots at 0 and 4, the circle must obviously have diameter of 4 units, ie: radius is 2 units.

5. Hence, the circle (where the semicircle came from) has radius 2 and centre (2,0)

6. Sketch directly.

The above steps may seem long, but this is just the thought process that should occur in your mind when you see this problem.

It usually gets you the answer immediately.

Is this what your teacher was saying?

 

[theind1996]

You can use a geometric argument (with a bit of logic) to acquire the answer immediately.

1. We recognise that this is a semi-circle

2. This semi-circle obviously has a root at x=0 and x=4 (by observing the square root if we factorise x out of it)

3. Since the 'y' is by itself (in the form y^2, so nothing weird like y^2 + y etc), we can deduce that this circle has its diameter on the x axis (since the circle is not being shifted up or down at all, so its at a 'neutral' position)

4. Since the diameter lies on the x axis, and it has roots at 0 and 4, the circle must obviously have diameter of 4 units, ie: radius is 2 units.

5. Hence, the circle (where the semicircle came from) has radius 2 and centre (2,0)

6. Sketch directly.

The above steps may seem long, but this is just the thought process that should occur in your mind when you see this problem.

It usually gets you the answer immediately.

Is this what your teacher was saying?

Yea, this is kinda what he was saying. To ensure that I have proper understanding, could you please mark my solution to this (pardon my non-usage of Latex):

y= 4 - √(16-x^2)

1. Well I can simply recognise that it is a NEGATIVE semi-circle by seeing the minus sign.
2. The roots are at -4 and 4 by factorising 16-x^2 into (4+x) (4-x).
3. The given equation of the semi-circle is the same as y-4 = - √(16-x^2)
4. I can see that this semi-circle is moved up by 4 units since moving up by 4 units on the y-axis is denoted by (y-h) - where the "h" is 4 as is seen in "y-4".
5. Therefore, this is a NEGATIVE semi-circle of radius 4 with centre at origin (0,0) BUT MOVED UP BY 4 UNITS.
6. Therefore, the graph of the equation of the semi-circle y= 4 - √(16-x^2) should look like this:



And thanks a lot Carrotsticks, I would rep you again if I could.

 

[Carrotsticks]

Your solution is correct.

 

[theind1996]

Your solution is correct.

YAY!

 

[Drongoski]

So we have the upper semi-circle,of a circle, radius 2 and centre (2,0).

 

[theind1996]

So we have the upper semi-circle,of a circle, radius 2 and centre (2,0).

YESS! That's the method my teacher showed me. Thanks a lot. So many kind people. Although I'm a lil' confused as to this 4, how you got it?.. But Repped

 

[Carrotsticks]

YESS! That's the method my teacher showed me. Thanks a lot. So many kind people. Although I'm a lil' confused as to this 4, how you got it?.. But Repped

He did the same 'completing the square' process as I did, but inside the square root instead of outside as I did.

 

[delian]

Yes, I hear Geogebra is a very helpful tool