BEC'S methods questions

[Mao]

firstly, this is a negative graph (sad face)
as there is a stationary point of inflection, the graph does not change direction except for its lone turning point, and (1,16) is between the two x-intercepts, hence it shoots up on the left, the down towards the right.


the point of inflection means that one of these roots will have to be cubed.

two possibilities:

possibility 1:

subbing in (1,16), we get:


expanding gives us



possibility 2:

subbing in (1,16), we get:


expanding gives us

[bec]

ahhhh thank you!
nicely explained, i completely get it now!

ok last one...
Do i need to be able to do these without a calculator for methods 3/4?
Sketch the graph with the equation:
a) y = x⁵-x⁴-2x³
b) y = x⁶-x²
c) y = x⁵+x⁴-3x³-3x²

[Mao]

bec i type really slowly, and fix mistakes before i finally gives a post up

it's all fixed now, you probably got the first half-finished not done prototype! xD

[Collin Li]

ok last one...
Do i need to be able to do these without a calculator for methods 3/4?
Sketch the graph with the equation:
a) y = x⁵-x⁴-2x³
b) y = x⁶-x²
c) y = x⁵+x⁴-3x³-3x²

Yes, you can be expected to draw them without a calculator. The method would be to factorise them and draw the x-intercepts first, find the y-intercepts, and use differentiation to find the turning points.

[bec]

ok...but my teacher keeps telling us that "we haven't done calculus" and therefore won't accept differentiation as a method. awesome.

but it's ok, i figured out how to do them without it - thanks anyway

[Mao]

ahhhh thank you!
nicely explained, i completely get it now!

ok last one...
Do i need to be able to do these without a calculator for methods 3/4?
Sketch the graph with the equation:
a) y = x⁵-x⁴-2x³
b) y = x⁶-x²
c) y = x⁵+x⁴-3x³-3x²

i seriously dont think you'll be required to do these in exam one. They dont like going to anything more than , as not all equations in the 5th degree polynomails are solved yet (but i cant explain what or why)
it's good practice if you try to do a few of these questions, knowing extra wont hurt

and just btw, from the looks of these questions, you can easily factorise or , in which case it becomes significantly easier to draw.

[Collin Li]

ok...but my teacher keeps telling us that "we haven't done calculus" and therefore won't accept differentiation as a method. awesome.

but it's ok, i figured out how to do them without it - thanks anyway

Well, you can do it with just the x-intercepts, but you won't have any idea where (coordinate-wise) the turning points are.

[bec]

you know those questions that are easy but however many times you try, you can't get them right?

3x + 2y = 2
4x + xy + y = 2

it's bugging me. can someone show their working for it?
thanks!

[Collin Li]

What are you trying to do? Are they both related, or are they individual equations?

[bec]

sorry, minor details! it's a simultaneous eqn - solve for x and y

[Collin Li]

Substitute into:







Use the quadratic formula:





Substitute these back into the first equation: to find the corresponding solutions:



Your two solutions (in coordinate form):





Just a note: remember that and are paired, so you've got to keep them together.

A joint effort by coblin and dcc

[bec]

thank you, to both of you! why do i always forget the quadratic equation? whyyy?

i just realised i'm almost up to page 10 on this thread...forget bringing in notes as my reference material, i'm just going to print this!

[AppleXY]

this should help in remembering the quad. eqn:

http://au.youtube.com/watch?v=H_7lNT9oDzI


Enjoy.

LOL @ this thread = bound book.

[bec]

this should help in remembering the quad. eqn:

http://au.youtube.com/watch?v=H_7lNT9oDzI

hahahha that's the saddest thing i've seen in my life. i love it.

[/0]

ahhhh thank you!
nicely explained, i completely get it now!

ok last one...
Do i need to be able to do these without a calculator for methods 3/4?
Sketch the graph with the equation:
a) y = x⁵-x⁴-2x³
b) y = x⁶-x²
c) y = x⁵+x⁴-3x³-3x²

These are probably ridiculously hard addition of ordinates problems