VCE Methods Question Thread!

[keltingmeith]

How would you do this question?

In the expansion of , the coefficient of the second term is −192. Find the value of n.

Remember the binomial theorem:



So, inputting your brackets in, we get:



So, evaluating the first few terms:



So, the second term is

So, we simply equate the coefficients, and solve for n:



And by inspection, we see that n=6. Note that normally for these questions, you would be expected to use a calculator - don't worry about trying to solve something so complicated by hand.

[knightrider]

Remember the binomial theorem:



So, inputting your brackets in, we get:



So, evaluating the first few terms:



So, the second term is

So, we simply equate the coefficients, and solve for n:



And by inspection, we see that n=6. Note that normally for these questions, you would be expected to use a calculator - don't worry about trying to solve something so complicated by hand.

Thanks so much Eulerfan101 

i dont understand this part.

From to    what happened to the a with how did you get rid of it when equating the coefficents. Also in isnt the only coefficient how did you get   as a coefficient as well.

Also how is

[keltingmeith]

Thanks so much Eulerfan101 

i dont understand this part.

From to    what happened to the a with how did you get rid of it when equating the coefficents. Also in isnt the only coefficient how did you get   as a coefficient as well.

Also how is

I got rid of the a because a is our variable - they were concerned with the coefficient, not the variable. If I wanted to include the a, I'd have to do it on both sides:



As for your second question:

[knightrider]

I got rid of the a because a is our variable - they were concerned with the coefficient, not the variable. If I wanted to include the a, I'd have to do it on both sides:



As for your second question:

Thankyou so much Eulerfan101  i understand now  Really appreciate your help

[Splash-Tackle-Flail]

Could someone explain dilation from the y axis? I understand it when its from the x axis, but why is each x value of the basic graph multiplied by 1/n instead of by n? e.g. how is (2x+1)^3 dilated from the y axis by a factor of 1/2, wouldn't it be 2?

[brightsky]

Could someone explain dilation from the y axis? I understand it when its from the x axis, but why is each x value of the basic graph multiplied by 1/n instead of by n? e.g. how is (2x+1)^3 dilated from the y axis by a factor of 1/2, wouldn't it be 2?

Just remember that you always replace 'the variable affected' with '(the variable affected)/(dilation factor)'. If you dilate by a factor of k from the x-axis, you affect the y-variable, so you replace y with y/k. Hence, y=f(x) becomes y/k = f(x) which is the same as y = k*f(x). If you dilate by a factor of k from the y-axis, you affect the x-variable, so you replace x with x/k. Hence, y = f(x) becomes y = f(x/k). Hope this makes sense!

[Splash-Tackle-Flail]

Ah ok, so the process makes sense then to answer these types of questions, but would there be a conceptual explanation to it? As in why do we divide the variable affected by the dilation factor, and why each x value of the basic graph multiplied by 1/n instead of by n?

[keltingmeith]

Ah ok, so the process makes sense then to answer these types of questions, but would there be a conceptual explanation to it? As in why do we divide the variable affected by the dilation factor, and why each x value of the basic graph multiplied by 1/n instead of by n?

Consider the transformation (x', y')--->(ax, by). In this set of transformations on the function y=f(x), we are dilating by a from the y-axis, and by b from the x-axis. In order to put these transformations into the function, however, we need them in terms of x and y. So, we take the equations:

x'=ax
y'=by

And solve for x/y:

x'/a=x
y'/b=y

And sub into the function:

y'/b=f(x'/a)
y'=bf(x'/a)

[brightsky]

Ah ok, so the process makes sense then to answer these types of questions, but would there be a conceptual explanation to it? As in why do we divide the variable affected by the dilation factor, and why each x value of the basic graph multiplied by 1/n instead of by n?

Consider the function y = f(x). Suppose we wanted to dilate the graph of this function by a factor of k from the y-axis. What would the equation of the resultant function be?

To find out, we shall use the method of trial and error. We know that we must put a 'k' somewhere in the equation. Let us guess that the resultant function will be y = f(k*x). At first, this guess seems perfectly reasonable. Let us test whether or not this equation actually works. Consider a random point (a, f(a)) on the graph of y = f(x). After the dilation, the point becomes (ka, f(a)). We know for sure that this must be a point on the graph of the new function. Hence, if our guess at the equation of the new function were true, when we plug this coordinate into the equation, we should get LHS = RHS. Let us see if we do:

LHS = f(a)
RHS = f(k*ka) = f(k^2 a)

Woops! LHS does not equal to RHS. This suggests that our initial guess is wrong. Let us try a second time. Let us now guess that the resultant function will be y = f(x/k). At first, this guess seems quite random. Why would you put the k in the denominator? Let us test if it works. As before, let us substitute the coordinate (ka, f(a)), which we know must be a point on the graph of the new function, into the equation and see whether LHS = RHS.

LHS = f(a)
RHS = f((ka)/k) = f(a)

Hey, LHS = RHS! This suggests that y = f(x/k) is in fact that equation of the resultant function, and that our rather odd second guess is in fact correct.

[Splash-Tackle-Flail]

Consider the function y = f(x). Suppose we wanted to dilate the graph of this function by a factor of k from the y-axis. What would the equation of the resultant function be?

To find out, we shall use the method of trial and error. We know that we must put a 'k' somewhere in the equation. Let us guess that the resultant function will be y = f(k*x). At first, this guess seems perfectly reasonable. Let us test whether or not this equation actually works. Consider a random point (a, f(a)) on the graph of y = f(x). After the dilation, the point becomes (ka, f(a)). We know for sure that this must be a point on the graph of the new function. Hence, if our guess at the equation of the new function were true, when we plug this coordinate into the equation, we should get LHS = RHS. Let us see if we do:

LHS = f(a)
RHS = f(k*ka) = f(k^2 a)

Woops! LHS does not equal to RHS. This suggests that our initial guess is wrong. Let us try a second time. Let us now guess that the resultant function will be y = f(x/k). At first, this guess seems quite random. Why would you put the k in the denominator? Let us test if it works. As before, let us substitute the coordinate (ka, f(a)), which we know must be a point on the graph of the new function, into the equation and see whether LHS = RHS.

LHS = f(a)
RHS = f((ka)/k) = f(a)

Hey, LHS = RHS! This suggests that y = f(x/k) is in fact that equation of the resultant function, and that our rather odd second guess is in fact correct.

Consider the transformation (x', y')--->(ax, by). In this set of transformations on the function y=f(x), we are dilating by a from the y-axis, and by b from the x-axis. In order to put these transformations into the function, however, we need them in terms of x and y. So, we take the equations:

x'=ax
y'=by

And solve for x/y:

x'/a=x
y'/b=y

And sub into the function:

y'/b=f(x'/a)
y'=bf(x'/a)

That actually makes sense-thanks! I could answer the questions in the exercises but it's good to know what I'm doing, and why I need to do it, instead of just what I need to do

[Zues]

Just remember that you always replace 'the variable affected' with '(the variable affected)/(dilation factor)'. If you dilate by a factor of k from the x-axis, you affect the y-variable, so you replace y with y/k. Hence, y=f(x) becomes y/k = f(x) which is the same as y = k*f(x). If you dilate by a factor of k from the y-axis, you affect the x-variable, so you replace x with x/k. Hence, y = f(x) becomes y = f(x/k). Hope this makes sense!

how does a dilation from the x affect the y and vice versa?

[brightsky]

how does a dilation from the x affect the y and vice versa?

Think about what happens to the point (x,y) after you dilate it by a factor of k FROM the x-axis. It becomes (x,ky). The x-coordinate stays as is. It is the y-coordinate that changes.

[Zues]

Think about what happens to the point (x,y) after you dilate it by a factor of k FROM the x-axis. It becomes (x,ky). The x-coordinate stays as is. It is the y-coordinate that changes.

Would you show me an example of how each x and y is affected in two different equations with its respective dilations please? Thanks

Hello everyone, hope you all are having a good morning.


As someone mentioned on the Chemistry thread (Zues), asking whether to do the Checkpoints after every chapter or not. Would it be better for me now, as I fly through the chapters in my own time to do the Checkpoint questions, or should I do them right before my sacs as my class revisits these chapters?

Thanks

I'm saving it for before class sacs because you will most likely remember it, and if you've worked ahead you might forget some stuff by the time the sacs comes + doing seen questions will make you think your a boss but you might not do that well? Make sense

[Zues]

Let's approach these graphs logically - first, graph f(x). Now, to sketch f(|x|), we know that any negative x numbers that go in, they're going to become positive, so you have to treat all negative x-values like positive x-values. So, you should see the function mirror itself across the y-axis. To sketch |f(x)|, any negative outputs (or negative y-values) are going to become positive. So, you should see the negative portions flip over the x-axis. This explains the flip in your second attachment.

There is no such thing as a "normal" domain. You have either the domain defined by the function, or the maximal domain. Let's consider the function f:[1,2]-->R, f(x)=sqrt(x). In this case, the domain of the function is [1, 2], however the MAXIMAL domain of sqrt(x) is x>=0, since those are all the possible numbers you could put into the function. If no domain is given for a function, its domain is implied to be its maximal domain. Sometimes, the specified domain is the maximal domain.

Hey Euler back on this question.

first attachment: i'm not sure how to graph these type of questions? i know how to graph absolute value functions but not those with it to the power. (answer is second attachment, im not sure how they drew that, looks kinda like a parabola, why?)
third attachment: the equation is y = |e^x - 1|, i see how theres a reflection in the x axis hence the new asymptote is y=1. lets say the equation was y = |e^x - 1| + 3, would the new asymptote be y=  4 now?, how do we know if the graph is above or below the asymptote? e.g. originally it was below but after the translation etc a part of it is above and shows asymptotic behaviour being above it.

[keltingmeith]

Just like brightsky did before me, I'm going to take this in steps:

1. Put it in terms of a function we know.
2. Apply appropriate transformations, including the modulus composition.
3. Apply any final transformations.

Hey Euler back on this question.

first attachment: i'm not sure how to graph these type of questions? i know how to graph absolute value functions but not those with it to the power. (answer is second attachment, im not sure how they drew that, looks kinda like a parabola, why?)

Consider . Let's do this in steps:

1. Sketch e^x. This is our basic function, we know how to sketch this.
2. Now is the appropriate time to apply the modulus transformation - so, we sketch g(|x|) (g(x)=e^x), using what I showed you earlier.
3. Now we just need to apply the translations - which is one in the position x direction, and 4 in the positive x direction.

third attachment: the equation is y = |e^x - 1|, i see how theres a reflection in the x axis hence the new asymptote is y=1. lets say the equation was y = |e^x - 1| + 3, would the new asymptote be y=  4 now?, how do we know if the graph is above or below the asymptote? e.g. originally it was below but after the translation etc a part of it is above and shows asymptotic behaviour being above it.

1. This one's a lot easier than before - once again, we start with y=e^x.
2. Now, we translate it one down, giving us f(x)=e^x-1. Now, we do |f(x)| as I showed earlier, and obtain our graph.
3. In your example, yes, you would then translate the entire graph up 3 - leading to a y=4 asymptote.

The reason we know that it approaches the asymptote from the bottom, is because before the reflection it approached it from the top. Then, when you've reflected it, it starts to approach from the bottom, because reflections do things like that. Remember how the shape changes between y=e^x and y=-e^x?