VCE Methods Question Thread!

[Rishi97]

f(x) = x-3/2-x
What are the asymptotes?
How would I go about solving this question? If its on cas, pls give me a brief description on where to go

Thanks

[Eugenet17]

Try long division to form a normal hyperbola equation

[IndefatigableLover]

f(x) = x-3/2-x
What are the asymptotes?
How would I go about solving this question? If its on cas, pls give me a brief description on where to go

Thanks

You should use long division to change the form which the equation is in into one that you would be able to recognise it in.

Once you change it you can see that there are asymptotes for y=-1 & x=2
(Use long division)

        ______
-x+2|x-3

        -1____
-x+2|x-3
         x-2
         ____
            -1



EDIT: Beaten by Eugenet17 and sorry for the dodgy working..

[lzxnl]

How would I find n in the equation attached?

Dear me. Where the heck did you get this question?

We want to find a way to simplify . Let the first cube root be x and the second cube root be y.
Now somehow we're going to have to get rid of those cube roots. In other words, we want to convert into .
This can be done if you multiply by . Note how we already have ? We want to convert this expression into something related to .


From our definition of x and y,
What about the left hand side? Well, you're given that in your equation, so we can multiply by to give , or .
Hence, instead of solving , we can solve
Now, let's see what is. Can you see that is simply ? is the product of the two cube roots and they're conjugates, so apply difference of two squares to get .
Thus, our equation becomes

Verified with CAS

Seriously though, they would NOT ask you this in an exam.

[Zealous]

f(x) = x-3/2-x
What are the asymptotes?
How would I go about solving this question? If its on cas, pls give me a brief description on where to go

Thanks

Or here's an alternative method of expressing it in a standard hyperbola form (for cool people =p):



Therefore the asymptotes are x=2 and y=-1.

Dear me. Where the heck did you get this question?

I must say that is some very interesting working out. =)

[nerdmmb]

Didn't know there was a 1&2 Methods Question Thread :S

My Question:
Find the equation of the straight line(s) which pass through the point (1,-2) and is (are) tangent to the parabola with equation y=x²

Thanks!

[Orb]

Didn't know there was a 1&2 Methods Question Thread :S

My Question:
Find the equation of the straight line(s) which pass through the point (1,-2) and is (are) tangent to the parabola with equation y=x²

Thanks!

I'm not sure whether there's a linear way to approach this question, but I was able to solve it with calculus.

So the parabola with equation y=x^2, differentiate it (becomes 2x) to find the gradient.

Basically the gradient at any given point is 2 x the co-ord.

The line has a constant gradient, but when the line touches the parabola the gradient should be equal.

----

Now we can create a second point, at (x,x^2) (can be applied anywhere)

Using the formula rise/run, x^2-(-2) / x-1 = 2x (gradient of parabola)

x^2+2 / x-1 = 2x
x^2-2x-2=0

Using the quadratic formula, we get:

x= 1 +- root 3

And then just sub the x values in to find the co-ords!

Edit: P.S I love food too XD

[IndefatigableLover]

I'm not sure whether there's a linear way to approach this question, but I was able to solve it with calculus.

So the parabola with equation y=x^2, differentiate it (becomes 2x) to find the gradient.

Basically the gradient at any given point is 2 x the co-ord.

The line has a constant gradient, but when the line touches the parabola the gradient should be equal.

----

Now we can create a second point, at (x,x^2) (can be applied anywhere)

Using the formula rise/run, x^2-(-2) / x-1 = 2x (gradient of parabola)

x^2+2 / x-1 = 2x
x^2-2x-2=0

Using the quadratic formula, we get:

x= 1 +- root 3

And then just sub the x values in to find the co-ords!

Edit: P.S I love food too XD

Same question was posted here Re: Parabola and Linear Equation Questions and I managed to solve it without calculus but we got different results in the end (for the quadratic formula part :|)

[Orb]

Same question was posted here Re: Parabola and Linear Equation Questions and I managed to solve it without calculus but we got different results in the end (for the quadratic formula part :|)

Your 2+root 12 should be correct except you forgot the divided by 2a at the bottom which gets you 2 + 2root 3 over 2
= 1 + root 3

[IndefatigableLover]

Your 2+root 12 should be correct except you forgot the divided by 2a at the bottom which gets you 2 + 2root 3 over 2
= 1 + root 3

I rechecked my working with my CAS and it seems to be correct? The main difference at hand is what we used the quadratic formula for (you used x^2-2x-2=0 whilst I used m^2-4m-8=0) which is why we have different values at the end..  (I don't even know if I did the question properly LOL)

[lzxnl]

If r = 

Can anyone show why ?

I'm going to do this two ways. As you're a spesh student, this second way is probably easier. However, the first method is what is expected of a Methods student.

1.


But from the given equation.
Hence

2. Implicit diff

[Orb]

I rechecked my working with my CAS and it seems to be correct? The main difference at hand is what we used the quadratic formula for (you used x^2-2x-2=0 whilst I used m^2-4m-8=0) which is why we have different values at the end..  (I don't even know if I did the question properly LOL)

I had a look at yours, here's my theory:
Maybe it stems from one of these two following points?

1. You solved for m, I solved for x.

2. To solve your method you plug it into y=m(x-1)-2

Is this just y= mx+c in the form of (1,-2)?
If so, it should be -2=m+c instead.

[IndefatigableLover]

I had a look at yours, here's my theory:
Maybe it stems from one of these two following points?

1. You solved for m, I solved for x.

2. To solve your method you plug it into y=m(x-1)-2

Is this just y= mx+c in the form of (1,-2)?
If so, it should be -2=m+c instead.

1. Although we solved for different things, we should still end up with the same answer (or equation for the line really) so I don't think it's that.

2. I used the rule which is finding an equation of a tangent. I subbed in the point (1,-2) as it is known that it lies on the tangent and I'm just solving for the unknown gradient of the line which I can then put back and work out the equation of the tangent since the rule I originally used is finding the equation of the tangent.
It wouldn't really work out if I subbed (1,-2) into y=mx+c but yeah that's just how I thought it through :S

(Dunno if any of that made sense LOL but hopefully it did)

[M_BONG]

I'm going to do this two ways. As you're a spesh student, this second way is probably easier. However, the first method is what is expected of a Methods student.

1.


But from the given equation.
Hence

2. Implicit diff

Perhaps wrong thread but can you show me now you implicitly differentiated that? I forgot how to do it

[alchemy]

Dear me. Where the heck did you get this question?

It was from an ancient (and unrelated) math paper. Mind you, the only reason I posted it is because I've come across similar instances in spesh that I didn't know how to deal with. For example, if we were asked to find sin(pi/8) using half angle formula while keeping our answer in the simplest form. However, apparently VCAA lets you use the CAS for these questions.