VCE Methods Question Thread!

[1729]

heyy guys,
Can I someone please help me with this:
how can
(x- 5/2) (2x^2 +2x+6)
be equal to
(2x-5) (x^2 + x+ 3)

Thank you!

Consider how \(\left(2x^2+2x+6\right)\left(x-\frac{5}{2}\right)\) is equivalent to \(2\left(x^2+x+3\right)\left(x-\frac{5}{2}\right)\) by factoring the \(2\) out of the quadratic.

Now we can multiply the \(2\) to the linear factor, becoming \(2\left(x-\frac{5}{2}\right)\left(x^2+x+3\right)=\left(2x-5\right)\left(x^2+x+3\right)\)

Hope this helps.

Hi guys!!! 
Could someone please help me with this
Sketch: y = (x − 1)(x^2 + 1) (Note: There is no turning point or ‘flat point’ of this cubic.)

What does "There is no turning point or ‘flat point’ of this cubic" mean here?

And also:
If we simply just have:
Sketch
y=(x+2)(x^2-x+7)

There are so many ways to sketch this. How do we do it?


Thank you!

Apologies, I missed this post.

I'm not too sure if I'm understanding your first question. A cubic function can have either 0, 1 or 2 turning points. If it has 0 turning points it means that the derivative of the cubic (which is a quadratic) never equals 0.

To graph \(\left(x+2\right)\left(x^{2}-x+7\right)\) consider how it does not have a stationary point, is strictly increasing and also remember to consider it's axis intercepts of course.

[PizzaMaster]

Thank you so much that makes sense!
Just a quick follow up question:
how do we know if a cubic graph is meant to have a stationery point, or turning points or both?

and the basic equation of a cubic is ax3 + bx2 + cx + d
what effects do a, b, c and d have on your graph?
For example if a graph has no bx^2 term and only has ax3 + cx + d
what effect does that have on the graph..

Thank you!

[Billuminati]

Thank you so much that makes sense!
Just a quick follow up question:
how do we know if a cubic graph is meant to have a stationery point, or turning points or both?

and the basic equation of a cubic is ax3 + bx2 + cx + d
what effects do a, b, c and d have on your graph?
For example if a graph has no bx^2 term and only has ax3 + cx + d
what effect does that have on the graph..

Thank you!

I failed methods myself so I ain’t the most reliable source. If you derive the cubic and the derivative function (which will be a quadratic ie a parabola) has no x-intercepts, then the cubic has no stationary points since f’(x) can never be 0. If there’s no bx² term, it just means b = 0. I can only say that (0,d) will be the y-intercept from the basic form

[PizzaMaster]

Thank you for your reply!

however, is it possible to explain to me assuming I have no knowledge of differentiation

[Billuminati]

Thank you for your reply!

however, is it possible to explain to me assuming I have no knowledge of differentiation

You can also graph it using a CAS or try factorising to get a quadratic factor eg f(x) = 5x(x-3)² has a turning point at (3,0) which also happens to be an x-intercept. The standard way to tackle these problems is to derive then solve f’(x) = 0, you can’t really get around calculus in methods

[1729]

Thank you for your reply!

however, is it possible to explain to me assuming I have no knowledge of differentiation

You need to have some knowledge of differentiation to determine whether a cubic has 0, 1 or 2 turning points.

Just adding to Billuminati's response, if you want to know how many turning points a quadratic has then you can just take the discriminant of the derivative. \(f\left(x\right)=ax^3+bx^2+cx+d\) then \(f'\left(x\right)=3ax^{2}+2bx+c\)

The discriminant of the derivative is equal to, \(\Delta _{f'}=4b^2-12ac\).
\(\Delta _{f'}<0\Longrightarrow \text{0 stationary points}\)
\(\Delta _{f'}>0\Longrightarrow \text{2 stationary points}\)
\(\Delta _{f'}=0\Longrightarrow \text{1 stationary point}\)

Note: 2 stationary points implies that the stationary points are turning points, and 1 stationary point means that the stationary point is an inflection point.

So I guess what you could do, without knowledge of differentiation, is just plug in \(a,\:b,\:c\) for a cubic in the form \(ax^3+bx^2+cx+d\) into the expression \(4b^2-12ac\) to determine the amount of turning points.

This is a little formulaic, which I don't really like but I guess if you do not know differentiation, just use this.

[beep boop]

Hi all,

Can someone please help me w/ q17 part g?

Much appreciated,
beep boop

[1729]

Hi all,

Can someone please help me w/ q17 part g?

Much appreciated,
beep boop

Hi beepboop

Assuming you were able to correctly answer the previous parts of the question.

Part F.
Consider how \(f^{-1}\) is just a reflection over \(y=x\), the shaded regions essentially remain the same, however, the location of the \(x\) coordinate of the point of tangency is different.

If the shaded regions \(f\) makes with it's tangent are at a minimum when the point of tangency is at \(x=10^{-\frac{1}{3}}\), then the shaded regions \(f^{-1}\) makes with it's tangent is at a minimum when the point of tangency is at \(f\left(10^{-\frac{1}{3}}\right)\).

This is because a reflection in \(y=x\) maps the coordinates of \(f\) as such \(\left(a,f\left(a\right)\right)\rightarrow \left(f\left(a\right),a\right)\)

Part G.
The tangent of \(f\) at \(x=1\) would be \(f'\left(1\right)\left(x-1\right)+f\left(1\right)=3-3x\).
However, \(f^{-1}\) is not actually differentiable at \(x=1\), therefore, having a vertical tangent of \(x=1\). We know that the angle formed with the \(x\) axis and the tangent of \(f\) in the clock-wise negative direction is \(\left|\tan^{-1}\left(-3\right)\right|=\tan^{-1}\left(3\right)\) and the angle formed with \(x\) axis in clockwise negative direction is \(\frac{\pi}{2}\) as it's vertical.

Therefore the acute angle formed is \(\frac{\pi}{2}-\tan^{-1}\left(3\right)\).

The graph below may help visualising this, I essentially took the approach of subtracting red angle with green.

[PizzaMaster]

Guyss
I don't get it
is square root 16
4
or
+- 4
Also,
when we have 9^1/2
is that 3 or +- 3

Thank you!

[1729]

Guyss
I don't get it
is square root 16
4
or
+- 4
Also,
when we have 9^1/2
is that 3 or +- 3

Thank you!

Hi, this is actually a very good question which many students fail to get right.

\(x^2=16\) has two solutions, \(x=4\) and \(x=-4\)

But \(x=\sqrt{16}\) has only one solution, \(x=4\)

To help understand why this is the case, try and consider it visually... the graph of \(y=\sqrt{x}\) is strictly positive (or 0)

Pretty much the square root of a number by itself is always positive. Also, \(x^{\frac{1}{2}}=\sqrt{x}\).

[PizzaMaster]

Heyyy That makes sense!
Thanks alot!!! Help is greatly appreciated.
I have another question;
what is the best way to approach simultaneous exponential equations
eg. 3000 = a x b^7
     6000 = 2a x b^10
I am used to adding or subtracting to solve simultaneous equations.
However, how about here?
Do we add/subtract/multiply/divide
and just in general, what's the best way to approach simultaneous exponential equations

Thank you!

[hvo-ade]

Divide 2 equations to eliminate a and apply the index law with b.

[ket_w]

Hey guys, this is probably quite a strange scenario I have but I hope I can get some advice 

I'm currently in year 8 (yes, quite young to even be on ATAR Notes) though I am quite adept in mathematics I'd say. Currently I'm in this extension program at my school (similar to SEAL) and my mathematics teacher still doesn't know what to do with me. It's been a great hassle me having to try and beg him to teach me something besides basic year 8 mathematics and I already knew year 10 advanced stuff so there was very little point in even paying attention to what he was teaching.

Fast forward a few months and I've been told to basically go through the units 1-2 methods textbook and try to complete what I can. It's been quite difficult as I haven't been put in a methods class, I've just had the book thrown at me. Though, I have managed to get through around 6 chapters so far of the methods book and I'm aiming to hopefully finish another 2-3 in the next few weeks.

I've spoken to the mathematics department leaders at my school and they are suggesting that I could potentially start my scored methods studies next year (I would be in year 9). I told them that I would only do this if it wasn't to be scored (basically doing the content without it actually contributing to my ATAR), however I am starting to wonder if maybe it wouldn't be too bad of an idea.

How risky do you think this would be if I did this scored? 

[1729]

How risky do you think this would be if I did this scored? 

If your school is suggesting that you complete units 1/2 methods next year then I think it's a great idea. Unit 1/2 does not contribute to ATAR anyways, and if you complete units 1 and 2 well, then proceed to unit 3/4 methods the following year (which contributes to ATAR).

Worst case scenario, you do unit 1/2 methods next year and you don't feel comfortable with proceeding to 3/4 then just repeat it!

[beep boop]

Hi All,

Can someone please help me w/ q9 a?

Much appreciated,
beep boop