Curve sketching (1 Viewer)

[Arithela]

Sketch the curves


1. x^2y = 1

2. y = 1/1 + x^2

 

[Arithela]

sorry, 2nd one should be y = 1/(1 + x^2)

 

[sunny]

aka the bell curve.

Graphmatica ftw. www.graphmatica.com

 

[Mark576]

y = 1/x²
dy/dx = -1/x
Therefore this graph is monotonically decreasing for positive values of x, and likewise monotonically increasing for negative values of x. Note that the graph must always be above the x-axis, and that the lim (x-->0^-) = +∞ and that the lim (x-->0⁺) = +∞. We could also work out the limits as x-->±∞ but I think that's obvious enough.

That should help you draw the graph, and yes I've taken note that you can only use 2 unit methods for this graph. This would be much simpler in 4 unit

 

[lacklustre]

hey guys how do you sketch inverse trig function in graphmatica? I can't figure it out. (I tried y = sin^-1 x and other variations)

 

[powerdrive]

hey guys how do you sketch inverse trig function in graphmatica? I can't figure it out. (I tried y = sin^-1 x and other variations)

another way of writing inverse sin/cos/tan x is "arc sin/cos/tan x"

i.e. y=sin^-1 x --> y=arcsin x etc.

in the instructions it says write y=asin x for inverse sin x etc.

 

[Mark576]

Try: y=asin(x) <--the parentheses are important!

 

[lacklustre]

Ah thanks guys. works a treat

Another one: How do you change the axis to radian measure?

 

[Arithela]

heres another question:

3x(x - 2) > 0

solving this, you get:

x < 0, x > 2

but how do we know x is less than 0 or vice versa without sketching the curve?

 

[sunny]

Solve it as a quadratic instead (which is almost sketching it anyway). 3x(x - 2) gives you x intercepts of 0 and 2. Use x=-b/2a (formula for axis of symmetry) to get the min/max value, in this case its -1.

That means you have a concave up parabola with y values > 0 when x < 0 and x > 2.

Maybe someone has a better way, but this is the first that comes to my mind.

 

[pLuvia]

heres another question:

3x(x - 2) > 0

solving this, you get:

x < 0, x > 2

but how do we know x is less than 0 or vice versa without sketching the curve?

You could just test a random point between the two values you have obtained. Or you could follow the general rule:

Let a and b be two constants where a<b, and x and y be two unknowns

Consider - (x-a)(y-b)>0 then solving it x<a or y>b
Consider - (x-a)(y-b)<0 then solving it a<x<b

That's usually a rule that works, but more than on one occasion they tell you to check using some method that your answer is right, so just do the random point testing

 

[Arithela]

testing a random point works (easier than the other methods mentioned), thanks!

 

[williamc]

heres another question:

3x(x - 2) > 0

solving this, you get:

x < 0, x > 2

but how do we know x is less than 0 or vice versa without sketching the curve?

3x(x-2) > 0

Break it up so you get:

3x >0 and x-2 > 0
x>0/3 .................x>2
x> 0

You cant get x >0 coz u already have x>0 x >2
So you get x<0


I dont know the theory behind it but simply, x> 0, x>2 doesn't make sense. So x<0 and x>2.

To make sure that is right, you just expand the brackets and then graph. Testing a point would also prove x<0.