Negative Angles

[destain]

not sure if this is a common method, but there's a method of dividing these kinda expressions which is to me, easier and faster but I kinda forgot how do it, it has something to do laying out the coeffcients and then multiplying and adding and something like that..
If someone actually understands my shitty explanation, can you remind me how to do it that way with this question LOL

[barydos]

not sure if this is a common method, but there's a method of dividing these kinda expressions which is to me, easier and faster but I kinda forgot how do it, it has something to do laying out the coeffcients and then multiplying and adding and something like that..
If someone actually understands my shitty explanation, can you remind me how to do it that way with this question LOL

It's called synthetic division, right?

[destain]

not sure about the actual proper name of it, but it's just a short cut, or short way, that's just what the kids call it LOLl

[barydos]

not sure about the actual proper name of it, but it's just a short cut, or short way, that's just what the kids call it LOLl

I guess check this out: http://www.khanacademy.org/math/algebra/polynomials/v/synthetic-division-example-2
looks like the one you wanted

if not there's another way

for (2x+1)/(x-5)
you lay out a couple (x-5)'s out
and multiply it to try and achieve 2x+1
so you first times it by 2
to get 2x-10
to get to 2x+1, we need to add 11, but by doing 11(x-5) you'll get something fancy, so we instead make it 11/(x-5) which works out!
so the answer would by 2+ 11/(x-5), this is just one of the many ways to do this division thingo LOL, dunno my explaining sounds like nonsense or not

[destain]

o yes that is it, but im pretty sure it works with more than just that as he says in the video, i learnt it while doing the polynomial chapter last year?

also in the product rule for exmaple...X x (x-5)+2
when we label it u and v, we ignore the +2 ?, it's not included in our calculation for the derivative?

[Phy124]

Yeah that's right you differentiate each separately:



You probably do this without realising quite often for easy polynomial questions, as it isn't necessary to write down the intermediate step.

Also when differentiating factorised polynomials I usually expand then differentiate, for me it's easier.

[destain]

ah i see, also........................

f(x)= x^4-6x^3+12x^2 -8x +2

a) Describe a sequence of transformations which maps the graphs of y=f(x) on to the graph of y= f(2x) -2
I answered, dilation from y by 1/2 and translation of 2 units downwards (because 2 in front of the x means dilation from y by 1/2 and -2 is just downwards by 2)

b) Find the x-axis intercepts of the graph of y = f(2x) -2
and for this, I just took f(x) and multiplied everything that has an x by 2 and took away the +2 cause of the translation 2 down?
and i made that = 0 and solved for x...I got x =0 or 2 which is WRONG!

the answer is retarded, don't get it..

c) Find the real values of p for which the equation abs( f(2x) -2 ) = p has exactly four solutions
?!?!?!?!

[barydos]

ah i see, also........................

f(x)= x^4-6x^3+12x^2 -8x +2

a) Describe a sequence of transformations which maps the graphs of y=f(x) on to the graph of y= f(2x) -2
I answered, dilation from y by 1/2 and translation of 2 units downwards (because 2 in front of the x means dilation from y by 1/2 and -2 is just downwards by 2)

b) Find the x-axis intercepts of the graph of y = f(2x) -2
and for this, I just took f(x) and multiplied everything that has an x by 2 and took away the +2 cause of the translation 2 down?
and i made that = 0 and solved for x...I got x =0 or 2 which is WRONG!

the answer is retarded, don't get it..

c) Find the real values of p for which the equation abs( f(2x) -2 ) = p has exactly four solutions
?!?!?!?!

a) Dilation of factor 1/2 from the y-axis, followed by a translation of 2 units in the negative direction of the y-axis.
b) I got x =0 or x= 1, remember it's not f(2x-2) it's f(2x)-2? possibly idk
c) Graph it out, you can see that if you draw a straight line, you'll intersect 4 times only between the local max and the x-axis. The calculator tells me that the local max is at a y-value of 27/16. So for 4 solutions or 4 intersections, p element of (0, 27/16) right?

[Phy124]

b) Find the x-axis intercepts of the graph of y = f(2x) -2
and for this, I just took f(x) and multiplied everything that has an x by 2 and took away the +2 cause of the translation 2 down?

Remember that if you have you have to do not .



I assume you did which is incorrect.

In the end you should have

[destain]

OMG THAT IS EXACTLY WHAT I DID ROFLLL, predictable stupidity ...

and with

ah i see, also........................

f(x)= x^4-6x^3+12x^2 -8x +2

a) Describe a sequence of transformations which maps the graphs of y=f(x) on to the graph of y= f(2x) -2
I answered, dilation from y by 1/2 and translation of 2 units downwards (because 2 in front of the x means dilation from y by 1/2 and -2 is just downwards by 2)

b) Find the x-axis intercepts of the graph of y = f(2x) -2
and for this, I just took f(x) and multiplied everything that has an x by 2 and took away the +2 cause of the translation 2 down?
and i made that = 0 and solved for x...I got x =0 or 2 which is WRONG!

the answer is retarded, don't get it..

c) Find the real values of p for which the equation abs( f(2x) -2 ) = p has exactly four solutions
?!?!?!?!

a) Dilation of factor 1/2 from the y-axis, followed by a translation of 2 units in the negative direction of the y-axis.
b) I got x =0 or x= 1, remember it's not f(2x-2) it's f(2x)-2? possibly idk
c) Graph it out, you can see that if you draw a straight line, you'll intersect 4 times only between the local max and the x-axis. The calculator tells me that the local max is at a y-value of 27/16. So for 4 solutions or 4 intersections, p element of (0, 27/16) right?

c) Is there a way to do it with algebra, that's how i usually do it, instead of graphing and looking for pois?

[Jenny_2108]

c) Is there a way to do it with algebra, that's how i usually do it, instead of graphing and looking for pois?

Use calculus, find the nature of turning points I guess lol
But why do you wanna do the other ways whilst sketching graph is faster and saves time?

[destain]

yeah thats what i did and it didnt turn out right somehow? and haha idk my points always don't turn out to be totally accurate if i use the graph

[Jenny_2108]

yeah thats what i did and it didnt turn out right somehow? and haha idk my points always don't turn out to be totally accurate if i use the graph

but even you sketch the graph, you still need to use calculus to find coordinates of turning point as CAS doesn't give exact values though

[Phy124]

(I didn't explain any of the steps because I think you already know why I did them, but I can if you wish)