VCE Methods Question Thread!

[alchemy]

Why do you need a CAS for that? Y coordinate of centre plus radius!

Just curious. Thought it might be some bug or something....
Goal for this holidays is to become more used to the cas, so these little things tick me off.

[alchemy]

Find all the points on the graph of y=x^2 such that the lines tangent to the curve at those points pass through (0,-2).
I got the answer to this using a really long and tedious method. What's quick way?

[Splash-Tackle-Flail]

In the Math Quest 12, it says "for the composite function f(g(x)) to be defined, the range of g must be a subset of (or equal to) the domain of f, that is ran g (subset symbol) dom f." Is there a reason for this as in an explanation to why. And wouldn't the function be defined as long as there still points that satisfy the domain of the input function (in this case g(x) ) and the domain of the composite function? (hope that made sense)

[keltingmeith]

Find all the points on the graph of y=x^2 such that the lines tangent to the curve at those points pass through (0,-2).
I got the answer to this using a really long and tedious method. What's quick way?

Welp, here's how I'd do it:

Since a tangent line is a linear relationship, we can express it in the form y=mx+c. Since (0, -2) is contained within, we can express it as . So, our line MUST be of the form y=mx-2.

From this, we know the line will touch the parabola exactly once - that's the point of a tangent line, after all. So, we take mx-2=x^2 ===> 0=x^2-mx+2. Since this should only have one solution, we can use the discriminant - remember, we have exactly one solution if the discriminant is 0. So,



Now, remember that m is the gradient at each point. So, we find the derivative and solve backwards:



Finally, we sub these into the original graph to get our coordinates and .

But Eulerfan, how the fuck did you know to do all that?!

Truth is, I didn't. Sometimes, the best way to approach a question is to just keep doing things you THINK might work, and see if it does. So, I thought "we have a point, let's try and get a line". After I had the line, I knew that it must touch the parabola once, hence why I then used that line in the discriminant and continued on as one might think.

[keltingmeith]

In the Math Quest 12, it says "for the composite function f(g(x)) to be defined, the range of g must be a subset of (or equal to) the domain of f, that is ran g (subset symbol) dom f." Is there a reason for this as in an explanation to why. And wouldn't the function be defined as long as there still points that satisfy the domain of the input function (in this case g(x) ) and the domain of the composite function? (hope that made sense)

This vTextbook video explains the whole thing quite beautifully, I definitely suggest giving it a watch and coming back if there's still something you don't understand.

[Splash-Tackle-Flail]

This vTextbook video explains the whole thing quite beautifully, I definitely suggest giving it a watch and coming back if there's still something you don't understand.

This tutorial is actually incredible- it all makes sense now! thx

[alchemy]

Welp, here's how I'd do it:

Since a tangent line is a linear relationship, we can express it in the form y=mx+c. Since (0, -2) is contained within, we can express it as . So, our line MUST be of the form y=mx-2.

From this, we know the line will touch the parabola exactly once - that's the point of a tangent line, after all. So, we take mx-2=x^2 ===> 0=x^2-mx+2. Since this should only have one solution, we can use the discriminant - remember, we have exactly one solution if the discriminant is 0. So,



Now, remember that m is the gradient at each point. So, we find the derivative and solve backwards:



Finally, we sub these into the original graph to get our coordinates and .

But Eulerfan, how the fuck did you know to do all that?!

Truth is, I didn't. Sometimes, the best way to approach a question is to just keep doing things you THINK might work, and see if it does. So, I thought "we have a point, let's try and get a line". After I had the line, I knew that it must touch the parabola once, hence why I then used that line in the discriminant and continued on as one might think.

That's a beautiful way of thinking about it. Thank you.
When I have some time I'll type up my odd method of doing it that yours puts to shame lol

[psyxwar]

Find all the points on the graph of y=x^2 such that the lines tangent to the curve at those points pass through (0,-2).
I got the answer to this using a really long and tedious method. What's quick way?

So I know EulerFan answered this already but I thought I'd share a different method that might be a bit more intuitive (or not )



Let the point(s) where the above conditions are satisfied be

Thus the gradient of our hypothetical tangent line(s) is 2a and our y-intercept is -2. Our tangent line must then be:



But we know that is a point on the tangent line (because it is the point where it intersects the graph). Thus we can sub it back into the tangent equation:






Then we just sub them into the graph and get the coordinates as required.

and

[SE_JM]

Hello, I'm having a trouble with this question on transformations:

The question is: Sketch the graph of y=18 (x+1)²-2(x+1)⁴by applying suitable transformations to y=9x²-x⁴

I'm not sure whether the dilation factor of 2 must be applied first, and then the horizontal translation, hence mapping of (x,y) --> (2x-1,y)
OR
whether the horizontal translation must be applied first and then the dilation factor hence the mapping (x,y) ---> (2(x-1),y)

I Feel that I keep getting confused with questions like this.... Could someone clear this up for me?

Thank you I appreciate it

[lzxnl]

Thanks!

Could I  ask one more question?

Regarding the very first question i asked: Finding inverse of x² and x⁴,

Since the question doesn't explicitly say find the inverse function, does that mean my answer is right? (as + - x^1/2 and + - x^1/4)?

Thank you

As mentioned above, a many-to-one function doesn't have an inverse function. It does, however, have an inverse relation. The inverse relation of y = x^4 is just simply x = y^4. You take the appropriate fourth root ONLY when you know the domain of the original function. So if y = x^4 was defined for negative x only, then the inverse would be y = - x^(1/4)

Hello, I'm having a trouble with this question on transformations:

The question is: Sketch the graph of y=18 (x+1)²-2(x+1)⁴by applying suitable transformations to y=9x²-x⁴

I'm not sure whether the dilation factor of 2 must be applied first, and then the horizontal translation, hence mapping of (x,y) --> (2x-1,y)
OR
whether the horizontal translation must be applied first and then the dilation factor hence the mapping (x,y) ---> (2(x-1),y)

I Feel that I keep getting confused with questions like this.... Could someone clear this up for me?

Thank you I appreciate it

Let's rewrite the equation as y/2 = 9(x+1)^2 - (x+1)^4
So...looks to me all you have to do is a dilation of factor 2 from the x axis and a translation of 1 unit to the left.

The dilation is parallel to the y axis. It doesn't affect the x values.
The mapping is (x,y) => (x-1, 2y)

[SE_JM]

As mentioned above, a many-to-one function doesn't have an inverse function. It does, however, have an inverse relation. The inverse relation of y = x^4 is just simply x = y^4. You take the appropriate fourth root ONLY when you know the domain of the original function. So if y = x^4 was defined for negative x only, then the inverse would be y = - x^(1/4)

Let's rewrite the equation as y/2 = 9(x+1)^2 - (x+1)^4
So...looks to me all you have to do is a dilation of factor 2 from the x axis and a translation of 1 unit to the left.

The dilation is parallel to the y axis. It doesn't affect the x values.
The mapping is (x,y) => (x-1, 2y)

Thank you lzxnl!!!

[catherine_mimi]

Hello guys,
I have a question I was unsure about the textbook's answer
Please help!


The total amount of snowfall S(t) m, measured during a period of t hours, can be approximated by the equation
S(t)=2-16/(t+8), where t is bigger or equal to 0.
What is the maximum amount if snow possible?
My answer: The maximum amount of snow approaches 2 metres i.e. lim S(t)=2
                                                                                               t--> infinity
Textbook answer: lim  S(t)=2
                       t-->0
Thanks,

[SE_JM]

I was just wondering,
Are we expected to solve this by hand? (it involves solving for 4 unknowns)

Find the equation of the quartic function for which the graph passes through the points with coordinates:
(-1,43) , (0,40), (2,70), (6, 1618), (10,670)

Even if we don't need to know, could someone please solve it for me? I tried several times and got some really bizarre answers

[keltingmeith]

Hello guys,
I have a question I was unsure about the textbook's answer
Please help!


The total amount of snowfall S(t) m, measured during a period of t hours, can be approximated by the equation
S(t)=2-16/(t+8), where t is bigger or equal to 0.
What is the maximum amount if snow possible?
My answer: The maximum amount of snow approaches 2 metres i.e. lim S(t)=2
                                                                                               t--> infinity
Textbook answer: lim  S(t)=2
                       t-->0
Thanks,

Considering S(0) exists (making the use of limits incredibly redundant) and it ISN'T 2, I think they're in the wrong here.   I'd take your answer.

EDIT: in fact, I just realised that the limit as t goes to 0 doesn't exist, so the textbook is DEFINITELY wrong there.

I was just wondering,
Are we expected to solve this by hand? (it involves solving for 4 unknowns)

Find the equation of the quartic function for which the graph passes through the points with coordinates:
(-1,43) , (0,40), (2,70), (6, 1618), (10,670)

Even if we don't need to know, could someone please solve it for me? I tried several times and got some really bizarre answers

4 unknowns...? No, sorry, it's 5 unknowns. The general quartic:

[tex]f(x)=ax^4+bx^3+cx^2+dx+e[/tex

Of course, that trusty (0, 40) immediately shows us that e is 40 - so, yay for that! Normally with anything more than 2, you're not expected to do them by hand. However, you may be asked for 3 unknowns. It's just too long to worth testing it, tbh. The method for solving n number of unknowns is always the same - reduce. Like so:

f(-1)=43
f(0)=40
f(2)=70
f(6)=1618
f(10)=670

So:
40=e (all equations will now have e on the LHS)

3=a-b+c-d
30=16a+8b+4c+2d
1578=1296a+216b+36c+6d
630=10000a+1000b+100c+10d

Now, just slowly multiply and subtract each row (i.e, elementary row operations) until you've solved it. I will not go through the elimination process, because those numbers are massive, and I don't feel like wasting half an hour.

[Orb]

Considering S(0) exists (making the use of limits incredibly redundant) and it ISN'T 2, I think they're in the wrong here.   I'd take your answer.

4 unknowns...? No, sorry, it's 5 unknowns. The general quartic:

[tex]f(x)=ax^4+bx^3+cx^2+dx+e[/tex

Of course, that trusty (0, 40) immediately shows us that e is 40 - so, yay for that! Normally with anything more than 2, you're not expected to do them by hand. However, you may be asked for 3 unknowns. It's just too long to worth testing it, tbh. The method for solving n number of unknowns is always the same - reduce. Like so:

f(-1)=43
f(0)=40
f(2)=70
f(6)=1618
f(10)=670

So:
40=e (all equations will now have e on the LHS)

3=a-b+c-d
30=16a+8b+4c+2d
1578=1296a+216b+36c+6d
630=10000a+1000b+100c+10d

Now, just slowly multiply and subtract each row (i.e, elementary row operations) until you've solved it. I will not go through the elimination process, because those numbers are massive, and I don't feel like wasting half an hour.

^^
You'll never get a question like that on a VCAA exam, unless it's with a calculator.