My Methods 3/4 questions thread

[smile+energy]

Hi, everyone

I am doing Methods 3/4 this year, and please help me with some questions. Thanks very much

 Can anyone show me how to solve this question(step by step):
Kx-y-z=8
3x+ky+2z=2
X+3y+z=-6
For what values of k, is there
a) a unique solution?
b) no solution?
c)an infinite number of solutions?

And what about this question? I tried to substitute the x-intercepts into the equation(s) but I still got the wrong answers.
a)The function F(x)=x^3+x^2+bx-64 has x-intercepts (-2, 0) and (4, 0). Find the values of a and b.
b)The functions f(x)=x^3-2x^2+ax+10 and y=6+(a+b)x-4x^2-x^3 both have (-1, 0) as an x-intercept. Find the values of a and b.
Thanks in advance

[Stevensmay]

Hi, everyone

I am doing Methods 3/4 this year, and please help me with some questions. Thanks very much

 Can anyone show me how to solve this question(step by step):
Kx-y-z=8
3x+ky+2z=2
X+3y+z=-6
For what values of k, is there
a) a unique solution?
b) no solution?
c)an infinite number of solutions?

First create a matrix from the equations given, we are only interested in the one containing coefficients for now.


Now find the determinant of this matrix, which will tell us when solutions will occur. This was done using CAS.
Determinant is


When the determinant of a matrix is non-zero there will be one solution, if it is 0 there are no or infinite solutions.
First we will find for one solution.
Let the determinant equal zero.


Solve this to get

So when k equals either of those values there will only be one solution to the set of equations.

Therefore we can see that no or infinite solutions will occur when k is not equal to the two values above.
lzxnl finishes this part in the post below. Thanks
(I have no idea if any of this is right, it probably isn't)

[lzxnl]

Hi, everyone

I am doing Methods 3/4 this year, and please help me with some questions. Thanks very much

 Can anyone show me how to solve this question(step by step):
Kx-y-z=8
3x+ky+2z=2
X+3y+z=-6
For what values of k, is there
a) a unique solution?
b) no solution?
c)an infinite number of solutions?

And what about this question? I tried to substitute the x-intercepts into the equation(s) but I still got the wrong answers.
a)The function F(x)=x^3+x^2+bx-64 has x-intercepts (-2, 0) and (4, 0). Find the values of a and b.
b)The functions f(x)=x^3-2x^2+ax+10 and y=6+(a+b)x-4x^2-x^3 both have (-1, 0) as an x-intercept. Find the values of a and b.
Thanks in advance

Firstly...what the heck? Since when did Methods ask for a system of 3 equations in 3 unknowns?
If you have a CAS, the matrix method above would work. Then, you would substitute one of the k values into your system of equations. If two equations become multiples of each other (like, for instance, x+y=1 and 2x+2y=2), then you know you have infinitely many solutions. If not, then you have no solutions.

Alternatively you can just try get your calculator to solve the 3x3 equations with k being each of the values found above. If there are no solutions, the calculator will say "false". If there are infinitely many solutions, the calculator will give the answer in terms of a parameter (like, for instance, z=n, y=2n+1 and x=4n-3).

For your first function question, I don't see an "a" in the question.
For b, I would just go -1-2-a+10=0 (which is f(-1)) => a=7
y=6+(a+b)x-4x^2-x^3, x=-1 => 0 = 6+(7+b)*-1 - 4 + 1 = 2-(7+b) = 0
b=-5

[smile+energy]

First create a matrix from the equations given, we are only interested in the one containing coefficients for now.


Now find the determinant of this matrix, which will tell us when solutions will occur. This was done using CAS.
Determinant is


When the determinant of a matrix is non-zero there will be one solution, if it is 0 there are no or infinite solutions.
First we will find for one solution.
Let the determinant equal zero.


Solve this to get

So when k equals either of those values there will only be one solution to the set of equations.

Therefore we can see that no or infinite solutions will occur when k is not equal to the two values above.
lzxnl finishes this part in the post below. Thanks
(I have no idea if any of this is right, it probably isn't)

Your methods were right, but the answers of the equation were wrong. I know how to do it now, thanks for your help.

[smile+energy]

Firstly...what the heck? Since when did Methods ask for a system of 3 equations in 3 unknowns?
If you have a CAS, the matrix method above would work. Then, you would substitute one of the k values into your system of equations. If two equations become multiples of each other (like, for instance, x+y=1 and 2x+2y=2), then you know you have infinitely many solutions. If not, then you have no solutions.

Alternatively you can just try get your calculator to solve the 3x3 equations with k being each of the values found above. If there are no solutions, the calculator will say "false". If there are infinitely many solutions, the calculator will give the answer in terms of a parameter (like, for instance, z=n, y=2n+1 and x=4n-3).

For your first function question, I don't see an "a" in the question.
For b, I would just go -1-2-a+10=0 (which is f(-1)) => a=7
y=6+(a+b)x-4x^2-x^3, x=-1 => 0 = 6+(7+b)*-1 - 4 + 1 = 2-(7+b) = 0
b=-5

It's a clear answer for the solving systems of equations.
Sorry, I forgot to type an 'a'. For part b, the function is y=x^3+ax^2+bx-64, it has x-inter (-2,0) and (4,0).
Thanks for your help.

[IndefatigableLover]

It's a clear answer for the solving systems of equations.
Sorry, I forgot to type an 'a'. For part b, the function is y=x^3+ax^2+bx-64, it has x-inter (-2,0) and (4,0).
Thanks for your help.

So if your question is asking you to find the 'a' & 'b' values for the function where there are x-intercepts at (-2,0) and (4,0)....

Well you know from your x-intercepts, you know that x=-2 & 4. This can be expressed as .

Because this is a cubic, we must find the third intercept. To do this we will use substitute 'n' into the equation. Therefore the equation looks something like: (does this seem more familiar?)

Using the y-intercept as reference, we can work out the the value of 'n' is 8 (since )

Expand that all out you should get:
where a=6 & b=-24

(I probably worded a few things incorrectly LOL so forgive me about that but I hope you get the gist of what I'm trying to explain I guess~)

[smile+energy]

So if your question is asking you to find the 'a' & 'b' values for the function where there are x-intercepts at (-2,0) and (4,0)....

Well you know from your x-intercepts, you know that x=-2 & 4. This can be expressed as .

Because this is a cubic, we must find the third intercept. To do this we will use substitute 'n' into the equation. Therefore the equation looks something like: (does this seem more familiar?)

Using the y-intercept as reference, we can work out the the value of 'n' is 8 (since )

Expand that all out you should get:
where a=6 & b=-24

(I probably worded a few things incorrectly LOL so forgive me about that but I hope you get the gist of what I'm trying to explain I guess~)

Yeah, I get what you meant. It's another way of doing it. Thanks for your help.

[smile+energy]

Hi, there
I need some help.
1. the transformation which maps the curve with the equation y=x^3 to the curve with the equation y=(3x-6)^3+1, using matrices.
2. a function g(x) is mapped to the curve h(x)=-g(4(x+1)+3. create a matrix equation that will map g(x) to h(x).
Thanks in advance

[RKTR]

1.
y'-1 = (3x'-6)^3
x'=x/3 +2
y'=y+1


 (1/3   0)   (x)  + (2)
 (0      1 )   (y)    (1)

[Nato]

1.
y'-1 = (3x'-6)^3
x'=x/3 +2
y'=y+1


 (1/3   0)   (x)  + (2)
 (0      1 )   (y)    (1)

just wondering, if you took the 3 out from the bracket, would that change the dilation factor. Can it be done this way, and still get the same answer?

[lzxnl]

Taking the three from the bracket initially?
So y'-1=27(x'-2)^3 => (y'-1)/27 = (x'-2)^3

There is now no dilation from the y axis and the dilation now occurs from the x axis factor 1/27.

[smile+energy]

1.
y'-1 = (3x'-6)^3
x'=x/3 +2
y'=y+1


 (1/3   0)   (x)  + (2)
 (0      1 )   (y)    (1)

I'm sorry, I get lost. Could you please show me the steps for the questions?

[RKTR]

y'-1 = (3x'-6)^3        your original one is y=x^3
so compare them
x'=x/3 +2
y'=y+1

(x')     (a     b)   (x)     (e)
      =                      +
(y')     (c      d)   (y)     (f)

x'=ax+by+e  y'=cx+dy+f
x'=x/3 +2      y'=y+1

so a=1/3 b=0 c=0 d=1 e=2 f=1

final answer

(x')     (1/3    0)   (x)     (2)
      =                         +
(y')     (0        1)   (y)     (1)

[smile+energy]

y'-1 = (3x'-6)^3        your original one is y=x^3
so compare them
x'=x/3 +2
y'=y+1

(x')     (a     b)   (x)     (e)
      =                      +
(y')     (c      d)   (y)     (f)

x'=ax+by+e  y'=cx+dy+f
x'=x/3 +2      y'=y+1

so a=1/3 b=0 c=0 d=1 e=2 f=1

final answer

(x')     (1/3    0)   (x)     (2)
      =                         +
(y')     (0        1)   (y)     (1)

Thanks very much, I get it now

[smile+energy]

Can anyone please shows the steps of this question?  a function g(x) is mapped to the curve h(x)=-g(4(x+1)+3. create a matrix equation that will map g(x) to h(x). I will appreciate your help.