Ho Ho... No

[blacksanta62]

Both these questions have a focus on the conjugate root theorem. 
Q1. -2-i is a solution and so is -2+i (conj root theorem). So to find the other solutions, expand (z+2+i)(z+2-i), then use this as your divisor for long division of the original polynomial, or use the method of equating coefficients.
Q2. Again conj root theorem, you get p(z)=(az+b)(z-2+i)(z-2-i). You're quadratic factor is the expansion of (z-2+i)(z-2-i). Equate coefficients to solve for a and b, or use long division.

I understand the first one perfectly but could you explain the second question a bit more Thank you

[StupidProdigy]

I understand the first one perfectly but could you explain the second question a bit more Thank you

So you're quadratic term is (z-2+i)(z-2-i), or in expanded form is z^2-4z+5 (I got those two roots by conjugate root theorem). Plug in our solutions of z=2-i and z=2+i (i.e find p(2+i) and p(2-i), solve simultaneously for a and b. It'll give a=-7 and b=6.
Now to solve, the method I would use is equating coefficients. So we know that (cz+d)(z^2-4z+5)=2z^3-7z^2+6z+5
You should be able to see that the third degree term on the left hand side will be cz^3, and this must equal 2z^3, so c=2. Also the constant term in the expansion of the left hand side will be 5d and this must equal 5 so d=1. So 2z+1 is a factor, z=-1/2. If you're confused about where I got 5d, expand the whole left hand side and you will see. I conveniently chose the constant term 5d to equate to 5.

[blacksanta62]

Hey, I need some help with a particular question:

a) Show that the graphs of x = y^2 - y and y = 1/2x + 1 intersect where x = 2 and find the coordinates of this point
            (1) x = y^2 - y
            (2) y = 1/2x + 1
Transpose equation (2) to get it in terms of x
It follows that  (3) x = 2y - 2
Because the graphs intersect we set them equal to each other (without the x's)
(1) = (3)
y^2 - y = 2y - 2
y^2 - 3y + 2 = 0
Factorise and set equal to zero
(y - 2) (y - 1) = 0
y = 2 or y = 1
sub y = 2 into equation (1)
It follows that (1) x = 2^2 - 2
x= 2
sub y = 2 into equation (3)
It follows that (3) 2(2) - 2 =x
x = 2
Therefore the coordinates for the intersection: (2,2)

Is this working correct? I don't have the solutions at all


b) Find, correct to two decimal places, the angle between the line y = 1/2x + 1 and the tanget to the graph of x = y^2 - y at the point of intersection found in a (that is, at the point where x = 2)
I'm not even sure where to begin with this, do I use my CAS?

Thank you

[Sine]

Hey, I need some help with a particular question:

a) Show that the graphs of x = y^2 - y and y = 1/2x + 1 intersect where x = 2 and find the coordinates of this point
            (1) x = y^2 - y
            (2) y = 1/2x + 1
Transpose equation (2) to get it in terms of x
It follows that  (3) x = 2y - 2
Because the graphs intersect we set them equal to each other (without the x's)
(1) = (3)
y^2 - y = 2y - 2
y^2 - 3y + 2 = 0
Factorise and set equal to zero
(y - 2) (y - 1) = 0
y = 2 or y = 1
sub y = 2 into equation (1)
It follows that (1) x = 2^2 - 2
x= 2
sub y = 2 into equation (3)
It follows that (3) 2(2) - 2 =x
x = 2
Therefore the coordinates for the intersection: (2,2)

Is this working correct? I don't have the solutions at all


b) Find, correct to two decimal places, the angle between the line y = 1/2x + 1 and the tanget to the graph of x = y^2 - y at the point of intersection found in a (that is, at the point where x = 2)
I'm not even sure where to begin with this, do I use my CAS?

Thank you

a) is correct just sub yoru answer back in into the original equations

b) two decimal places suggests CAS use.

1)  y = 1/2x + 1

tangent of x=y^2-y at x=2
implicit diff
dy/dx = 1/(2y-1)
when x = 2 y = 2
so dy/dx given y=2 = 1/3
y1-y2 = 1/3(x1-x2)
y1-2=1/3(x-2)
2) y=1/3x + 4/3



let the first angle be a and second b
m = gradient

angle 1.
a=arctan(m)
a=arctan(1/2)
26.57degrees
This is the acute angle 1) makes with any horizontal line
angle 2
b=arctan(m)
b=arctan(1/3)
b=18.43
The acute angle 2) makes with any horizontal line
Required angle is a-b = 26.57 - 18.43 = 8.13degrees (the acute angle these lines make)

[blacksanta62]

"Consider the function with rule f(x) = x/sqrt(x - 2)"
1) Find f'(x)
2) Find the coordinates and nature of all stationary points (can't do this since I can't set f'(x) = 0)
3) Find the equation of the asy(mptote (asy at x=2)
4) Find the equation of the other asymptote (How do I do this?)

I keep getting f'(x) = sqrt(x - 2) - x/2(sqrt(x-2)) / (sqrt(x-2))^2
My CAS gives me a different answer and I'm not to sure if the squared and sqrt cancel.

Thank you

[lzxnl]

"Consider the function with rule f(x) = x/sqrt(x - 2)"
1) Find f'(x)
2) Find the coordinates and nature of all stationary points (can't do this since I can't set f'(x) = 0)
3) Find the equation of the asy(mptote (asy at x=2)
4) Find the equation of the other asymptote (How do I do this?)

I keep getting f'(x) = sqrt(x - 2) - x/2(sqrt(x-2)) / (sqrt(x-2))^2
My CAS gives me a different answer and I'm not to sure if the squared and sqrt cancel.

Thank you

Your derivative is fine. But setting that to zero is simple.
sqrt(x-2) = x/(2(sqrt(x-2)))
(x-2)=x/2
x=4

The other asymptote is found as follows:

f(x) = x/sqrt(x-2) = sqrt(x^2/(x-2))=sqrt(x + 2 + 4/(x-2))
As x->infinity, the fraction vanishes, so f(x) -> sqrt(x+2)

[blacksanta62]

Your derivative is fine. But setting that to zero is simple.
sqrt(x-2) = x/(2(sqrt(x-2)))
(x-2)=x/2
x=4

The other asymptote is found as follows:

f(x) = x/sqrt(x-2) = sqrt(x^2/(x-2))=sqrt(x + 2 + 4/(x-2))
As x->infinity, the fraction vanishes, so f(x) -> sqrt(x+2)

Thanks IzxnI, helped me alot

[blacksanta62]

Can someone explain what second derivatives are useful for? In my notes I have: "Second derivatives can be taken. They are used for"
And then nothing. I know that you can use the second derivative test but that seems to have me confused as well.
Thank you

[lzxnl]

Can someone explain what second derivatives are useful for? In my notes I have: "Second derivatives can be taken. They are used for"
And then nothing. I know that you can use the second derivative test but that seems to have me confused as well.
Thank you

In VCE? Not much. They're useful for the second derivative test and acceleration (second time derivative of position) but that's about it.

In real life? Many more applications there...
For instance, you can extend linear approximation to quadratic approximations
Many physical systems require the use of second order linear/non-linear ordinary/partial differential equations (like the heat equation, Maxwell's equations etc)

[blacksanta62]

Thanks for clearing that all up!!
displacement -----> dy/dx(velocity) -------> d^2y/dx^2(acceleration).

[blacksanta62]

Can anyone give some application uses/real life uses of integral calculus. So far I have: finding the area underneath a curve and I'm pretty sure we're going to use it in methods 3/4 for probability.

Also, in an exam, will we ever be told to find the area underneath a curve with a specific technique like the rectangle technique or will we simply have to do a definite integral with the specified limits? Thank you

[blacksanta62]

In addition to the above question, when we integrate by substitution, do we want to do this with respect to u or x? 
Thank you

[JellyBeanz]

In addition to the above question, when we integrate by substitution, do we want to do this with respect to u or x? 
Thank you

When you use a U substitution, you want to integrate with respect to u and then substitute the value of u back in after you performed the integration. I think most of the time you will probably be asked to find the area underneath a curve using definite integrals, but may be asked to use another method as well.

[blacksanta62]

When you use a linear substitution, you want to integrate with respect to u and then substitute the value of u back in after you performed the integration. I think most of the time you will probably be asked to find the area underneath a curve using definite integrals, but may be asked to use another method as well.

Is linear substitution the same as U substitution? Haven't heard of linear substitution before.

[JellyBeanz]

Is linear substitution the same as U substitution? Haven't heard of linear substitution before.

I just meant substitution * lol, but yes, it is a U substitution.