VCE Methods Question Thread!

[Bri MT]

Welcome to the forums Azila!

Quick note: you'll be used to seeing the symbol for discriminant as upper case delta (the triangle) whereas fun_jirachi has shown it as lowercase delta (like a weird incomplete 8 ).

Thank you fun_jirachi for the supportive & detailed answer

Edit: the symbol has since been updated to be what you'd be used to

[^^^111^^^]

Hello guys!

I’m not actually doing VCE yet but am studying some (rather basic) Maths Methods 1/2. When I say basic, it’s rudimentary compared to what everyone else has written here.

The discriminant tells you the number and nature of the roots, which is understandable. But is it also used to find out the number of solutions (intercepts) with another function? If so, how do you use it.

Eg. For what real values of t will y=tx+1 NOT intercept y=2x^2+5x+11? I saw someone use the discriminant to solve this, but I don’t exactly understand the concept. Also, I know there are other ways to solve this, but not enough. Any other methods to solve this would be great for more proficiency.


Also, are there any tips for getting batter at using the CAS calculator?


Thanks!

Just a quick fun fact, there is something known as the cubic discriminant, but it is very complicated and there are better ways to find the x-intercepts of cubic or higher degree functions (using the factor and remainder theorems). But don't worry cubic discriminant is not covered in the VCE Methods study design and you will hardly ever need to use it in University (I believe).

Just something for you to think about, don't let that confuse u

[yeshelly]

NEED HELP PLEASE

Find the coordinates of the x-intercept of the line which passes through the point (8, −2), and is parallel
to the line 2y − 4x = 7.

Can someone help me start this?

[whys]

NEED HELP PLEASE

Find the coordinates of the x-intercept of the line which passes through the point (8, −2), and is parallel
to the line 2y − 4x = 7.

Can someone help me start this?

You need to first find the equation of the line. If it is parallel to the given equation, then the gradients will be the same. Thus, you can rearrange that equation in the form y = mx + c and take the m-value as the gradient, and continue on to solve for the x-intercepts, also using the given point to determine the equation of the line. Hopes this helps and that this starting is enough to get you in the right direction!

[colline]

Also, are there any tips for getting batter at using the CAS calculator?

Check out the YouTube channel TI Australia, they do CAS tutorials that are specific for the VCE curriculum. There are also many other videos on YouTube which teaches you how to use the CAS in the most efficient way possible. At the end of the day the best way would just to do a lot of long application questions that rely the CAS so that you can get used to it.

Jacaranda's MathQuest also has a calculator companion for the TI-Nspire and the Classpad which might be worth taking a look at, but of course you'll have to buy it unless you can get a free copy somehow.

[SmartWorker]

NEED HELP PLEASE

Find the coordinates of the x-intercept of the line which passes through the point (8, −2), and is parallel
to the line 2y − 4x = 7.

Can someone help me start this?

Steps:
1. Rearrange to make y the subject: y= 2x + 7/2
2. Parallel lines have the same gradient. Therefore, the gradient of new line = 2
3. Equation of new line. y = 2x+c, label as (1)
4. Sub (8,-2) into (1)
5. Solve for c, -2=16+c. c=-18
6. New equation: y = 2x -18

[TheEagle]

Hey

I don't understand how the domain of gof* is the same as the domain of f* ??

[TigerMum]

Hey

I don't understand how the domain of gof* is the same as the domain of f* ??

It's because we are restricting the domain of f such that gof exists, that is, we are finding a restriction of f (we call this f*) such that the range of f* is a subset of the domain of g. This means that our domain for f* is the domain of gof*, because it is the domain we defined for gof* to exist. Hopefully this makes sense.

[TheEagle]

It's because we are restricting the domain of f such that gof exists, that is, we are finding a restriction of f (we call this f*) such that the range of f* is a subset of the domain of g. This means that our domain for f* is the domain of gof*, because it is the domain we defined for gof* to exist. Hopefully this makes sense.

Sounds pretty intuitive right until the last step where you said "our domain for f* is the domain of gof*"
It trips me out
I thought the range of f* will be our input for gof* and therefore the domain

[TigerMum]

Sounds pretty intuitive right until the last step where you said "our domain for f* is the domain of gof*"
It trips me out
I thought the range of f* will be our input for gof* and therefore the domain

Ah, I think I see where you are having trouble. The range of f* is essentially the "domain" or "input" that we put into g. gof* is the combination of two functions: g(f*(x)), and so the domain of gof* is defined as the x-values that we put into the first function, f*, not the values that come out of f* that we put into g.
We made our restriction to f* that meant the range of f* was a subset of g (and in doing so, making gof* exist) so we can put this very restriction onto the function gof, (making it gof*) so that it is defined.

[Geoo]

I'm having a hard time getting to the answer with this one (i've never done functional equations before.)

For the function f(x)=3x-7 is the functional equations below true?
a) f(x+y)=f(x)+f(y)
Here is what I did.
f=3(x+y)-7 = 3x+3y-7 and F= (3x-7) + (y) = 3x+y-7.
The answer:
f (x + y) = 3x + 3y - 7
f (x) + f (y) = 3x + 3y - 14
Two things, how is the 3y and -14 gotten? There was nothing to multiply with.

With the same function except the functional equation changed to f(x-y)=f(x) -f(y)
So I did.
f (x - y) = 3x - 3y - 7
f (x) - f (y) = 3x -7 -y
The answer:
f (x - y) = 3x - 3y - 7
f (x) - f (y) = 3x - 3y
What am I missing?

[whys]

I'm having a hard time getting to the answer with this one (i've never done functional equations before.)

For the function f(x)=3x-7 is the functional equations below true?
a) f(x+y)=f(x)+f(y)
Here is what I did.
f=3(x+y)-7 = 3x+3y-7 and F= (3x-7) + (y) = 3x+y-7.
The answer:
f (x + y) = 3x + 3y - 7
f (x) + f (y) = 3x + 3y - 14
Two things, how is the 3y and -14 gotten? There was nothing to multiply with.

With the same function except the functional equation changed to f(x-y)=f(x) -f(y)
So I did.
f (x - y) = 3x - 3y - 7
f (x) - f (y) = 3x -7 -y
The answer:
f (x - y) = 3x - 3y - 7
f (x) - f (y) = 3x - 3y
What am I missing?

f(x+y)=f(x)+f(y)
Is this true?
f(x)=3x-7

f(x+y) = 3(x+y)-7
f(x+y) = 3x+3y-7

f(x) + f(y) = 3x-7+3y-7
f(x) + f(y) = 3x+3y-14

Thus f(x) + f(y) ≠ f(x+y)

Next question:
f(x-y) = f(x) - f(y)
Is this true?
f(x-y) = 3(x-y)-7
f(x-y) = 3x-3y-7

f(x) - f(y) = 3x-7-(3y-7)
f(x) - f(y) = 3x-7-3y+7
f(x) - f(y) = 3x-3y

Thus f(x-y) ≠ f(x) - f(y)

You are making the mistake of subtracting/adding 'y' instead of f(y). f(y) = 3y-7 (sub y into the equation)

[Geoo]

Thank you! I didn't know you could sub the function into the y since it had an x. So thank you!

[whys]

Thank you! I didn't know you could sub the function into the y since it had an x. So thank you!

No worries! You can sub anything instead of 'x' with the given notation: f(whatever you want to sub)

[MoonChild1234]

hi, i'm having a bit of trouble with this question, especially cii. i got c=1 but im not sure as the answer is c=1 and c>2

i've had a look at the solutions but im still a bit confused

thank you!