VCE Methods Question Thread!

[Sine]

hey how do you solve this problem.

Sketch the graph of y=x^3-3x+2 and y=-4x.
Show the coordinates of the intercepts and the intersection point (Do not show the coordinate of the maximum turning point of the cubic graph.)

What aspect are you struggling with?
What have you tried?

[amanaazim]

hey sine like how do i solve the equation and plot it in to the graph?

[Azila2004]

Helloo everyone!
Hope you all are well. I have a question on this matrices question: A linear transformation T maps the point (1,3) to the point (−2,−3) to the point (2,4) to the point (−3,−11). Find the matrix of the transformation.

I have used simultaneous equations to help solve the matrix which transforms (1,3) to (-2,-3) and (-2,-3) to (2,4), but this matrix equation does not transform (2,4) to (-3,-11). I may have read the question wrong or made a silly mistake, but any help would be greatly appreciated. Thank you! 

[Azila2004]

hey sine like how do i solve the equation and plot it in to the graph?

I'm not Sine but I'll try to make it clearer lol.

You would use the algebra to solve x if the equations have the same x-value and y-value. So, you would solve for x in x^3-3x+2=-4x as the y-values are the same. You would bring the 4x to the other side (so x^3+x+2=0) and factorise for the values of x. You would then solve for x using the Null Factor Law and use those values to find the y-coordinates to find the intersection points.

[parieeelol]

Hey, hope everyone's doing well.

I'm really stuck on this question from Methods 1+2. The question is in the image below, but it doesn't offer any formulas or any numbers in the examples - so how is the graph derived from the vessel? What calculations would be used?

Would appreciate any help. Thank you in advance!

[Opengangs]

Hey, hope everyone's doing well.

I'm really stuck on this question from Methods 1+2. The question is in the image below, but it doesn't offer any formulas or any numbers in the examples - so how is the graph derived from the vessel? What calculations would be used?

Would appreciate any help. Thank you in advance!

It's mainly more visualising than anything. The width of the vessel is inversely proportional to the change in height \(h\). In other words, the wider the vessel is at some height, the slower it is for the height to change since it has to travel more to cover the same height.

In part (a), we see that the width of the vessel is the same until you get to the neck. So the volume increases linearly as \(h\) increases. In part (b), we see that the height will increase slower at the start but gradually increase faster due to the shape of the vessel. The same idea works with part (c) and part (d).

[kat05]

Hi there!
Could someone please help me out with this VCAA Exam 2 Question from 2011. It's 4c and I have been stuck on it for ages. I don't get where the formula from distance comes from?? I would greatly appreciate if someone could please explain this to me!!

[fun_jirachi]

Hey there!

Consider a general point on the parabolic river (x, x²-1). The distance from the campsite to this point will be calculated as follows:

Since the person travels at 2km an hour, the time it takes him to move to the river will be half that, which you see in the left part of the equation given in part c.

Now noting the y-coordinate of the general point, we have that the time is proportional to \(\frac{3}{4}-(x^2-1)\) ie. \(T = k\left(\frac{3}{4}-(x^2-1)\right)\).

From here, we can simplify this to get the part on the right-hand side. Thus, the total time will be just the sum of each of these two expressions, which we see in the equation in part c).

Hope this helps

[Azila2004]

Hello everyone!

I wrote down a question earlier but nobody seemed to respond to it, so I'll just paste it again. I also have another which I have no idea how to do.

1. A linear transformation T maps the point (1,3) to the point (−2,−3) to the point (2,4) to the point (−3,−11). Find the matrix of the transformation.

2. If g(x) = x^2 +4x -6, find h for which g(x-h) has: i. two positive solutions ii. one positive and one negative solution.

I'm not sure exactly how to do these, so help would be appreciated 

[TigerMum]

1. A linear transformation T maps the point (1,3) to the point (−2,−3) to the point (2,4) to the point (−3,−11). Find the matrix of the transformation.

Yes, I remember this question, it has a typo. It is supposed to say "A linear transformation T maps the point (1,3) to the point (-2,-3) and the point (2,4) to the point (-3,-11)." You should be able to find the transformation matrix now.

2. If g(x) = x^2 +4x -6, find h for which g(x-h) has: i. two positive solutions ii. one positive and one negative solution.

Recall that for a function \(g(x)\), \(g(x-h)\) is given by a horizontal translation h units to the right. Therefore, to find the values of h for which \(g(x-h)\) has two positive solutions or one positive solution, we need only to consider the solutions to the function g(x) and their horizontal position relative to the origin.
We will solve for \(g(x)=0\) using the quadratic formula, \(x=\frac{-b\pm\sqrt{b^2 -4ac}}{2a}\).


To find the solutions for \(g(x-h)\), we remember that the graph of \(g(x-h)\) is just \(g(x)\) shifted h units to the right. See if you couple this with our results for the solutions to g(x)=0 to find the values of h for which g(x-h) has two positive solutions or one positive solution.
Hope that helped you get started.

[Azila2004]

Yes, I remember this question, it has a typo. It is supposed to say "A linear transformation T maps the point (1,3) to the point (-2,-3) and the point (2,4) to the point (-3,-11)." You should be able to find the transformation matrix now.Recall that for a function \(g(x)\), \(g(x-h)\) is given by a horizontal translation h units to the right. Therefore, to find the values of h for which \(g(x-h)\) has two positive solutions or one positive solution, we need only to consider the solutions to the function g(x) and their horizontal position relative to the origin.
We will solve for \(g(x)=0\) using the quadratic formula, \(x=\frac{-b\pm\sqrt{b^2 -4ac}}{2a}\).


To find the solutions for \(g(x-h)\), we remember that the graph of \(g(x-h)\) is just \(g(x)\) shifted h units to the right. See if you couple this with our results for the solutions to g(x)=0 to find the values of h for which g(x-h) has two positive solutions or one positive solution.
Hope that helped you get started.

Thank you so much! This really helped me a lot.

[Evolio]

Hello! 
 
Part a
So for part a), you need to determine the expressions for length, width and height so that you can multiple all of them to give you the volume.
It says squares will be cut out from the corners, however, we do not know what the length and the width of the squares will be. So, we can label the length (and width) of the side of the square as x. This means that the height will be x, as if you picture a rectangle with squares cut out on the edges, and if we fold up the remaining sides to make a swimming pool, the height will be x.
The width will be 8-2x since two squares have been cut out from each side (as they have been cut from each corner).
The length will be 12-2x.
Although, I just realised when I multiplied (8-2x)(12-2x)x, it doesn't give the answer above....Hmm, I'll update this post if I think I've done it incorrectly. But, basically, you need to find the dimensions of the pool and then multiply them and make sure the expression equals the one in the question.

b)
Well, x cannot be negative as you can't have a negative length and neither can it equal 0. It has to be larger than 0. Also, you need to consider the dimensions found above. 12-2x cannot equal 0 or be negative and 8-2x cannot be 0 or negative.
So: x has to be less than 4. Thus the domain would be 0<x<4.

c)
For this question, you need to use the volume expression they gave you above.
Steps to solve this problem:
1. Differentiate the volume expression (given in part a) using your CAS
2. Make the differentiated expression=0 and solve for x using your CAS. Also, in your working out, you should state that V'(x) (whatever the differentiated expression is) =0 at stationary points i.e maximum when you answer this question.
3. When you solve this, you get two values. Reject the negative value as x cannot be negative. So: you have your x value at the maximum volume
4. Since they asked you to specifically prove the nature, you can use a gradient table and show that it is a maximum that way.

d)
Using the x value you got from part c (for the maximum), you substitute this value into the length, width and height expressions, thus giving you the dimensions of the pool.

I'm not a hundred percent sure about part a) though because the answer I got is different to the volume expression they've given.
Please correct me if I'm wrong! It's been a while, hehe.

[TigerMum]

I'm not a hundred percent sure about part a) though because the answer I got is different to the volume expression they've given.
Please correct me if I'm wrong! It's been a while, hehe.

Yeah it looks like an error to me, your method is correct.

[kat05]

Hey there!

Consider a general point on the parabolic river (x, x²-1). The distance from the campsite to this point will be calculated as follows:

Since the person travels at 2km an hour, the time it takes him to move to the river will be half that, which you see in the left part of the equation given in part c.

Now noting the y-coordinate of the general point, we have that the distance is proportional to \(\frac{3}{4}-(x^2-1)\) ie. \(D = k\left(\frac{3}{4}-(x^2-1)\right)\).

From here, we can simplify this to get the part on the right-hand side. Thus, the total time will be just the sum of each of these two expressions, which we see in the equation in part c).

Hope this helps

Hi!
Thank you for the reply I am just confused at the point you write D=k(3/4-(x^2-1)), is that distance or time? And what exactly is K and how do we know it goes there?

[fun_jirachi]

Hi!
Thank you for the reply I am just confused at the point you write D=k(3/4-(x^2-1)), is that distance or time? And what exactly is K and how do we know it goes there?

Hey!

That is my mistake, I've gone back and edited my previous post to reflect the change. Good pickup!

k is some constant of proportionality - ie. we just use it to denote some placeholder value that lies within the real numbers to describe a relationship between two things that change with respect to each other. You might be more familiar with 'T is proportional to some f(x)' or '\(T \propto f(x)\)' - this is just another way of writing the exact same thing. We know it goes there because in the question it explicitly states the time taken to move upstream is directly proportional to the difference in y-coordinates

Hope this helps