VCE Methods Question Thread!

[AlphaZero]

So does that mean dilation by 1/2 from x, reflection in y, reflection in x, right 4, up 15/2 is incorrect b/c the order is incorrect? I'm still a bit confused because when I apply these transformations I still get to the final function.

When I say the order of transformations matters, I mean that you cannot just reorder a given sequence of transformations and expect it to perform the same map.

The sequence of transformations you just provided is correct. The one you provided earlier, where you had \(3\) units up instead of \(15/2\) units up is incorrect because it doesn't result in the correct image function.

You can apply transformations in whatever order you want. If you want to do translations first, go ahead. If you want to do reflections first, go ahead. But, the sequence of transformations you provide must result in the same image function, and so the values and directions you apply your transformations in need to be adjusted accordingly. I showed an example of this in sequence 3 above.

[persistent_insomniac]

When I say the order of transformations matters, I mean that you cannot just reorder a given sequence of transformations and expect it to perform the same map.

The sequence of transformations you just provided is correct. The one you provided earlier, where you had \(3\) units up instead of \(15/2\) units up is incorrect because it doesn't result in the correct image function.

You can apply transformations in whatever order you want. If you want to do translations first, go ahead. If you want to do reflections first, go ahead. But, the sequence of transformations you provide must result in the same image function, and so the values and directions you apply your transformations in need to be adjusted accordingly. I showed an example of this in sequence 3 above.

Ok it makes sense now. Thankyou!

[EllingtonFeint]

Okay, I’ve spent legit 20 min trying to figure out how to upload a photo that is a jpeg that is under the size Li it and I swear this happens to me every few weeks but ugh it kills me every time.
Please someone suggest an easy way to upload photos from an iPhone. Is there an app I can download??

Anyway, my q probs won’t make send but here it is anyway...
Why is it that when there is a translation of 3 units to the left when you are applying translations to sketch graphs and it's mapped like (x,y) -> (x-1, y) Why is the equation y=(x+1)^2  ? As in plus NOT minus?? I mean it's not hugely important but I was just kinda curious.

[Bri MT]

Okay, I’ve spent legit 20 min trying to figure out how to upload a photo that is a jpeg that is under the size Li it and I swear this happens to me every few weeks but ugh it kills me every time.
Please someone suggest an easy way to upload photos from an iPhone. Is there an app I can download??

Anyway, my q probs won’t make send but here it is anyway...
Why is it that when there is a translation of 3 units to the left when you are applying translations to sketch graphs and it's mapped like (x,y) -> (x-1, y) Why is the equation y=(x+1)^2  ? As in plus NOT minus?? I mean it's not hugely important but I was just kinda curious.

My methods teacher used to explain this to us as:
- when you find a "y" value you don't need to bring anything to the other side of the equation
(Eg y = mx +c the c and m stay on the right)
- but when you solve for an x value you need to bring this to the other side, which means using opposite operations
(Eg y = mx + c, x = (y-c)/m )

Hope this helps

[AlphaZero]

Why is it that when there is a translation of 3 units to the left when you are applying translations to sketch graphs and it's mapped like (x,y) -> (x-1, y) Why is the equation y=(x+1)^2  ? As in plus NOT minus?? I mean it's not hugely important but I was just kinda curious.

Suppose we a graph of a function  \(y=f(x)\)  and some point  \((x,\,y)\).

Let's say we apply a translation of \(3\) units in the positive \(x\)-direction (to the right).  That is,  \((x',\,y')=(x+3,\,y)\).  This makes intuitive sense.  We want to move a point  \((x,\,y)\)  \(3\) units to the right, and so we add \(3\) to the \(x\)-coordinate.

But why does this transformation map the graph of  \(y=f(x)\)  onto the graph of  \(y=f(x-3)?\)  Miniturtle mentioned a way to help you remember that transformations involving changing \(x\) isn't 'intuitive', but it doesn't carry any mathematical reasoning.

Firstly, we have \[\begin{cases}x'=x+3\\ y'=y\end{cases}\implies \begin{cases}x=x'-3\\y=y'\end{cases}\quad\text{(rearrange both equations for }x\ \text{and }y)\] and so we can write  \(y=f(x)\)  with our expressions for \(x\) and \(y\) in terms of \(x'\) and \(y'\). That is, \[y=f(x)\implies y'=f(x'-3)\] and so if we drop the dashes, we have \[y=f(x)\ \overset{\text{3 units right}}{\longrightarrow}\ y=f(x-3)\]To summarise (and generalise), a translation of \(h\) units in the positive \(x\)-direction has the following affects:\[(x,\,y)\longrightarrow(x+h,\,y)\quad\text{and}\quad y=f(x)\longrightarrow y=f(x-h)\]

[aspiringantelope]

The graph of 3y + 4 = 12 would make an angle with the positive direction of the x-axis (to the nearest degree)

Not really sure how to find the angle and what this question is asking for if anyone can visually show me?

Thanks!

[undefined]

The graph of 3y + 4 = 12 would make an angle with the positive direction of the x-axis (to the nearest degree)

Not really sure how to find the angle and what this question is asking for if anyone can visually show me?

Thanks!

Are you sure there's no 'x' in that equation? Because that would just make a straight line (y=8/3) and wouldn't even cut the x axis

[AlphaZero]

The graph of 3y + 4 = 12 would make an angle with the positive direction of the x-axis (to the nearest degree)

Not really sure how to find the angle and what this question is asking for if anyone can visually show me?

I think you meant to write  \(3y+4x=12\),  so I'll go with that  (otherwise, no angle is even formed).

The question is also underspecified. There are two possible answers depending on whether the angle is measured clockwise (red) or anticlockwise (black).

Regardless, we can form a triangle with the coordinate axes as shown above, or the fact that  \(\tan(\theta)=m\), to find either of the angles the line makes with the positive direction of the \(x\)-axis.

[DBA-144]

Pretty sure that "positive side of the x axis" refers to it being measure anticlockwise?

[scientificllama]

lol I'm terrible at math

Question: Show that the triangle formed by the points A(1,0), B(2,7) and C(5,3) contains a right angle using gradients. Hence find the areas of the triangle.

I got that the right angle is at C and the area is 25√2/2 (which is totally wrong)
Thanks for the help!

[S200]

lol I'm terrible at math

Question: Show that the triangle formed by the points A(1,0), B(2,7) and C(5,3) contains a right angle using gradients. Hence find the areas of the triangle.

I got that the right angle is at C and the area is 25√2/2 (which is totally wrong)
Thanks for the help!

Hmm. Just from sorta looking at it (without graphing) I think the right angle is midway between A and B... so \(\triangle ACB\) is Isoceles, and hence  \(\angle BAC = \angle ABC \)

The Area calculations

So the area of a triangle is obviously \(\frac{1}{2}\text{base} \times{\text{height}}\), so the area of this whole triangle can be worked out by multiplying the length AC with the Length A(Midpoint-B)

But yeah... that's without graphing.

Your best approach to these types of questions is to use the two-point gradient method (\(\frac{y_2 - y_1}{x_2 - x_1} \)) to find the equations of the three lines which then enables you to visualize the graph.

[AlphaZero]

Pretty sure that "positive side of the x axis" refers to it being measure anticlockwise?

The question says "positive direction of the \(x\)-axis". I don't see how that refers to anything regarding which angle should be taken. It only refers to the section of the line segment  \(y=0\)  we should be considering. See Question 2b of the VCAA 2016 Exam 1 for a possible way the question should be worded.

...

Hmm. Just from sorta looking at it (without graphing) I think the right angle is midway between A and B... so \(\triangle ACB\) is Isoceles, and hence  \(\angle BAC = \angle ABC \)

The Area calculations

So the area of a triangle is obviously \(\frac{1}{2}\text{base} \times{\text{height}}\), so the area of this whole triangle can be worked out by multiplying the length AC with the Length A(Midpoint-B)

But yeah... that's without graphing.

Your best approach to these types of questions is to use the two-point gradient method (\(\frac{y_2 - y_1}{x_2 - x_1} \)) to find the equations of the three lines which then enables you to visualize the graph.

Here's the approach I believe was intended by the question. No graphing or finding equations of lines needed.

First find the gradients of each of the line segments of the triangle. \begin{align*}m_{AB}&=\frac{7-0}{2-1}=7\\
m_{AC}&=\frac{3-0}{5-1}=\frac34\\
m_{BC}&=\frac{7-3}{2-5}=\frac{-4}{3} \end{align*} Clearly, we have  \(\overline{AC}\!\perp\!\overline{BC}\),  and so  \(\angle ACB=90^\circ\).

Hence, the area of  \(\triangle ABC\)  is given by \begin{align*}A&=\frac12 \overline{AC}\!\times\!\overline{BC}\\
&=\frac12\sqrt{(5-1)^2+(3-0)^2}\sqrt{(2-5)^2+(7-3)^2}\\
&=\frac12\sqrt{16+9}\sqrt{9+16}\\
&=\frac12\!\times\!5\!\times\! 5\\
&=\frac{25}{2}\ \ \text{units}^2 \end{align*}

[aspiringantelope]

For \(f\left(x\right)=2x+1\) when \(x\ ∈\ ℝ\)
For what values of t is \(f\left(t\right)=t\)
Unsure on what to do
Make 2x + 1 = t ?
Doesn't seem to eliminate the other variable though.

[MB_]

For \(f\left(x\right)=2x+1\) when \(x\ ∈\ ℝ\)
For what values of t is \(f\left(t\right)=t\)
Unsure on what to do
Make 2x + 1 = t ?
Doesn't seem to eliminate the other variable though.

f(t)=2t+1
Set that equal to t
t=2t+1
Solve for t

[aspiringantelope]

f(t)=2t+1
Set that equal to t
t=2t+1
Solve for t

Ah I see !!
Thanks a lot!!