VCE Methods Question Thread!

[skrt skrt]

Really? I thought it was a dilation of 4 factors from the y- axis using f(ax) and dilating from the y-axis gives 1/a.

[Lear]

If i’m not mistaken I believe it’s a dilation of factor 4 from the x axis.

[Sine]

Hey

Usually there are numerous ways in which you can reach the same correct answer especially for transformation questions.
In this case both
Dilation of factor 4 from the x-axis
and
Dilation of factor 2 from the y-axis
are suitable solutions.

[brightsky]

Hi people, just need help with this reflection question.  Its question 10 in chapter 3B for the Cambridge Methods book.
state a transformation which maps the graph of y = f(x) to the graph of y = f1(x):

e) f(x)= 1/4x^2   ,  f1(x)=1/x^2


Answer is dilation of a factor of 2 from the y-axis

 

There are two possible answers here.

1. Dilation by a factor of 4 from the x-axis, which transforms y = 1/(4x^2) into y/4 = 1/(4x^2), which is equivalent to y = 1/x^2.
2. Dilation by a factor of 2 from the y-axis, which transforms y = 1/(4x^2) into y = 1/[4(x/2)^2], which is equivalent to y = 1/[4(x^2/4)], which is equivalent to y = 1/x^2.

Both are correct.

EDIT: See Sine's reply above.

[snowisawesome]

f(x) = 4-x has range [-2,6)
And you have to find the domain, and two of your options are
[-2,6), and (-2,6]
then how would you work out which one is correct?

(vcaa 2014 mm cas exam 2 mcq 2)

[Lear]

As made clear by the question, the y-value of 6 is NOT included. Therefore we need to find a domain that corresponds to a y value less (and not including) 6 to -2 (included)

If f(x)=6, it can be found out that x = -2
Like I mentioned before our domain should NOT include this value.
The other would be f(x)=-2, x=6

Ultimately our domain does NOT include -2 but does include everything above it to 6 and therefore it would be
(-2,6]

[snowisawesome]

As made clear by the question, the y-value of 6 is NOT included. Therefore we need to find a domain that corresponds to a y value less (and not including) 6 to -2 (included)

If f(x)=6, it can be found out that x = -2
Like I mentioned before our domain should NOT include this value.
The other would be f(x)=-2, x=6

Ultimately our domain does NOT include -2 but does include everything above it to 6 and therefore it would be
(-2,6]

Thank you

[Phoenix11]

Hello guys!
I was just wondering if someone could clarify a question?
So I did the problem below and I drew up the matrix with 5 rows and two columns but the answer wrote it with two rows and five columns?
So I was just wondering how we know which way to write it up(with the columns and rows)?

At a certain school there are 200 girls and 110 boys in Year 7, 180 girls and 117 boys in Year 8, 135 and 98 respectively in Year 9, 110 and 89 in Year 10, 56 and 53 in Year 11 and 28 and 33 in Year 12. Summarise this information in matrix form.

Thank you so much

[kalopsia]

Hello guys!
I was just wondering if someone could clarify a question?
So I did the problem below and I drew up the matrix with 5 rows and two columns but the answer wrote it with two rows and five columns?
So I was just wondering how we know which way to write it up(with the columns and rows)?

At a certain school there are 200 girls and 110 boys in Year 7, 180 girls and 117 boys in Year 8, 135 and 98 respectively in Year 9, 110 and 89 in Year 10, 56 and 53 in Year 11 and 28 and 33 in Year 12. Summarise this information in matrix form.

Thank you so much

Matrices are not in the course anymore, so tbh I'm not sure.
But, I don't think it really matters which way you write the matrices.

[Phoenix11]

Oh, wait they're not in the course anymore?
I had no idea because I was just doing these questions from a Essential Mathematical Methods 1 & 2 CAS/Cambridge.
Well, thank you.

[Bri MT]

Oh, wait they're not in the course anymore?
I had no idea because I was just doing these questions from a Essential Mathematical Methods 1 & 2 CAS/Cambridge.
Well, thank you.

You need to be able to apply a transformation matrix but that's it

[Phoenix11]

Oh ok!
Thank you!
 

[kalopsia]

Oh, wait they're not in the course anymore?
I had no idea because I was just doing these questions from a Essential Mathematical Methods 1 & 2 CAS/Cambridge.
Well, thank you.

Yeah, that textbook is part of the old study design.

[snowisawesome]

log10(x)+log10(x-3) = 1
i got x = -2 and x = 5 as the answers and they both satisfied the equation but the solutions said to reject x = -2 as x>3
How does this work?

[VanillaRice]

log10(x)+log10(x-3) = 1
i got x = -2 and x = 5 as the answers and they both satisfied the equation but the solutions said to reject x = -2 as x>3
How does this work?

Your solutions need to satisfy both log terms. Substituting x = -2 into the second one will give a negative number in the brackets.

Hope this helps