the sketching of curves (1 Viewer)

[hatty]

sup all

ive been practising curve sketching lately,
im not 2 sure how accurate they want the graph to be

in the HSC tips section of this forum, it says dont use calculus
in which i find it quite hard to find the stat pts, inflex pts etc without using calculus

and furthermore
in the cambridge 4u book, it teaches us the multiplication method where u just times the co-ordinates of the 2 graphs together to get a new graph, i thihnk it is so inaccurate

anyway

i get a function that i have to sketch

so i sketch it

but im not sure if i need more information

take a look at the pic i attached

now if i were to sketch it with pencil, it would look a bit neater
but is the information in that image enuf?

do i need to add all the bullshit in?

edit (i realise their are asymptotes at 1 -1, but the picture is 2 shitty)

 

[hatty]

and this 1 as well

i just used that multiplication method of the ordinates in the Cambridge book

is the information presented in my graph enuf? or do i need to work out max points

cheers

 

[Grey Council]

you need to mark in points like 1, asymptotes, turning points in some graphs. I think

 

[AGB]

isnt this the easiest way to sketch curves: find the asymptotes then just test a point in each one of the areas separated by the asymptotes??

edit: sorry my post didnt make sense (again)

 

[abdooooo!!!]

ahaha... delete it quick... no one except me has read it yet.

i just remember the shape and find the vert and horizontal asymptotes, and turning point(s) if there is any.

but have you guys notice how the new hsc curve sketching question is a bit weird... like easy. they just tell you to transform a graph from y = f(x) to y^2 = f(x) or something.

 

[AGB]

Originally posted by abdooooo!!!
ahaha... delete it quick... no one except me has read it yet.

no no no, i am actually right this time





(i think )

 

[AGB]

this is wot i mean....(look at attachment)

u find x and y asymptotes (x's are at 2 and -2 and y is at 0)

then you test a point, say 1....and u find that the answer is negative, therefore it is going to be a concave down parabola in between 2 and -2 and also below 0

then just let x = 0 to find the turning point

 

[AGB]

forgot to attach

 

[abdooooo!!!]

don't worry man... i think you got the right idea... there is no need to prove to us.

but for them y = 1/f(x) graphs you just remember the general shape ie a parabola in the middle and two hyperbola shape at the side. and find the asymptotes thats it. why do you need to test anything?

because by using your graph if x ---> 2 or -2 then y ----> minus infinity. so its a down parabola.

 

[AGB]

*nods in agreement*

(but really is quite lost )

 

[abdooooo!!!]

a simple explaination for those who don't understand this asymtotes thingy.

if you approach 2 or -2 in AGB's example from inside the range of -2 > x > 2 then it should therefore produce a upside down prabola because y is approaching minus infinity on both side inside the range.

if you approach 2 or -2 from outside the range of -2 > x > 2 the vert asymtote should be y is approching positive infinity one on the right of x = 2 and one on the left of x = -2.

then you find the horizontal asymtotes and turning point of the parabola to complete the graph.

 

[hatty]

hey AGB in ur graph

why is there a maximum at (0, -1/4 ) ??

how did u find that?

i thought all u had to do was graph
(x^2-4), and the minimum of that graph (0, -4)
becomes a maximum

 

[abdooooo!!!]

Originally posted by hatty
hey AGB in ur graph

why is there a maximum at (0, -1/4 ) ??

how did u find that?

i thought all u had to do was graph
(x^2-4), and the minimum of that graph (0, -4)
becomes a maximum

well by inspection the turning point of the parabola is (0, -1/4 ), because this turning point should occur at the centre of -2 > x > 2 which is on the y axis. so you just sub x = 0 and get this: (0, -1/4 )

you have to find all the key points you can without the use of calculus ie turning points, y and x intercept, and other critical points.

 

[AGB]

Originally posted by hatty
hey AGB in ur graph

why is there a maximum at (0, -1/4 ) ??

how did u find that?

i thought all u had to do was graph
(x^2-4), and the minimum of that graph (0, -4)
becomes a maximum

im scared to post anymore coz abdoo always picks up every little mistake i make

edit: btw, im just joking

 

[abdooooo!!!]

Originally posted by AGB
im scared to post anymore coz abdoo always picks up every little mistake i make

edit: btw, im just joking

i know... when did you ever made a mistake... its just a bunch of silly questions that you ask... hahaha.

pm'd you again back.

 

[hatty]

what if the graph is

y = 1
------
(x-2)(x+3)

its
y= (1/(x-2)(x+3)

sketch that for me PLEASE

 

[AGB]

a tiny bit of the graph got cut off

 

[abdooooo!!!]

yep and the turning point will be (-1/2, -4/25)... if i didn't make a mistake reading your graph.

 

[hatty]

but how did u find that turning point?

 

[Dash]

The derivative...