Sketching of Curves (2 Viewers)

[zenger69]

I don't do 4U Maths but i'm have trouble grasping any kind of curve sketching.

I've kinda memorised x^even numbers as parabolas
and cubic and quartic graphs.

Does anyone have notes on the others. I'm having particular trouble with Log function graphs.

The procedure I do is
1) Differentiate and find post of inflexion
2) Find the nature of these points using y''.
3) Testing concavity.
4) Then it's limits where I get lost.

For example for the function y=(lnx - 1)^3
I want to know what happens to the limit as x->0. But i get error. But does that mean the asymptote is at 0 and approaches 0?

And is there anything else i can do to help me find info about the function to draw log graphs.

 

[Affinity]

as x tends to 0.. the graph obviouslty tends to negative infinity

 

[Trefoil]

An antilog cannot be <=0, so x > 0.

Substitute in, say, x=0.001. Then x=0.0001. You get this:
x=0.001 -> y=-494.49
x=0.0001 -> y= -1064.44

Can you see how as x gets infinitesimally small, |y| gets infinitely big? That is to say:
x->0, y-> negative infinity, as Affinity said.

 

[zenger69]

Yeh it's just i've been tricked at times with horizontal asymtotes.


Like normal i test as x approaches negative infinity
x approaches positive infinity
x approaches zero.

Is there any other limits i need to test?

 

[wanton-wonton]

X approaches negative infinity.

 

[Slidey]

Yeh it's just i've been tricked at times with horizontal asymtotes.


Like normal i test as x approaches negative infinity
x approaches positive infinity
x approaches zero.

Is there any other limits i need to test?

Don't test for when x approaches zero.

Test for when the value of x makes the denominator of the function zero.

Eg if it were 1/(x^2-4), test for when x=2 or -2

 

[zenger69]

But in your example thats where my vertical asymptotes are.


Oh yeh what should i do to get horizontal asymptotes?

 

[Slidey]

For my example you would test:
x>+inf
x>-inf
x->2
x->-2

Asymptotes:
x=2
x=-2
y=0

 

[zenger69]

so how did u get y=0.

 

[Slidey]

f(x)=1/(x^2-4) = (1/x^2)/(1-4/x^2) - divide numerator and denominator by x^2 and apply the fact that lim[x->inf] of 1/x = 0

Let f(x)=y
as x->infinity, y-> 0/(1-4*0)=0
as x-> neg inf, y-> 0/(1-4*0)=0
So y=0 is a horizontal asymptote.

See with the limit
lim[x->inf] of 1/x = 0
it follows
lim[x->neg inf] of 1/x = 0
and also that
lim[x->neg inf] of 1/x^2 = {lim[x->neg inf] of 1/x} * {lim[x->neg inf] of 1/x} = 0*0 = 0

a less systematic approahc is to inspect the denominator and note that if if x is infinity, infinity-4 is still infinity, so it's basically 1/infinity = 0. Roughly speaking. Infinity isn't a number and all but this is INSPECTION!

 

[mojako]

Yeh it's just i've been tricked at times with horizontal asymtotes.


Like normal i test as x approaches negative infinity
x approaches positive infinity
x approaches zero.

Is there any other limits i need to test?

the one in y=(lnx - 1)^3 is vertical isn't it?
the y=+- infinity thing is always vertical

And is there anything else i can do to help me find info about the function to draw log graphs.

you can look at the graph of y=log x... which may be useful to memorise...
it goes down near x=0 to negative infinity, so y=(lnx - 1) will do the same, and so will y=(lnx - 1)^3

 

[zenger69]

ok thanks its helped me heaps.

 

[JamiL]

with log we know at log 1=0, so we have the x intersection at (1,0). log x >= 0 so it is asymtotic to x=0 and as x-> infinity y-> infinity. the stepness of the curve depends on the base of the log but that is realy irrelivent.
y*x=1 obvious x or y can not equal 2 zero so y=o and x=o are assytots, as y-> +or- infinity x-> +or-0
etc these should be easy graphs u can do wif ur eyes closed, this is 2u not even 3u work.
with equation u need to actually use ur brain, u should do the flowing steps.
1. y-intercpt (x=o)
2. x-intercept (y=o, if appropriate)
3. even or odd function
4. find first dirivitive
5. find all stationary points (first dirvitive = 0)
6. find second dirivitive (if appropriate)
7. cheak stationary points if they are max or min, by 2nd dirivitive or by table.
8. find any inflextion points (if appropriate)
(9 and 10 only when fractions are involved in ur equation)
9. find vertical assymtots (make denominater = 0)
10. find other assymtots (lim y as x-> infinity)
11. we are ready 2 graph

 

[zenger69]

to test the limit of y-> infinity, would I have to make x the subject?

 

[mojako]

to test the limit of y-> infinity, would I have to make x the subject?

why would u want to find the limit of x as y->infinity?

 

[blackfriday]

remember check concavity around y=x and y=-x, but this rarely happens (usually only in mofo absolute value graphs).