VCE Methods Question Thread!

[aspiringantelope]

Question 1: No, I don't believe there is a way to change the independent variable on the graphing page. If you are solving a problem where a function \(f\) has \(t\) as the independent variable, then just sketch \(y=f(x)\) on the graphing page. That is, just pretend \(x\) is \(t\) for the minute. For example, suppose you're given \(d(t)=t^2-2t+1\). Define that on your calculator page, and to graph it, just input \(d(x)\) instead.

Question 2: I'm not sure what you mean by "how do I graph an equation[?]" Could you provide some more details so I could identify the point of confusion?

Question 3: Could you elaborate on what you mean by "solv[ing] a \(y=ax^2+bx+c\) with three different equations"? Do you mean using 3 pieces of information to solve for \(a\), \(b\) and \(c\)?

Yes for \(y=ax^2+bx+c\)  , you get three pieces of information or just like 3 points of the parabola and you have to find a, b and c.
I know how to do it but not on the calculator. >.<

[AlphaZero]

Yes for \(y=ax^2+bx+c\)  , you get three pieces of information or just like 3 points of the parabola and you have to find a, b and c.
I know how to do it but not on the calculator. >.<

You just need to solve 3 equations simultaneously. On the TI-nspire CAS, press  \(\texttt{Menu}\to \texttt{3}\to \texttt{7}\to \texttt{1}\).  In the dialog box, set the number of equations and the variables you wish to solve for separated by commas. Press OK and type your equations in the boxes provided.
Here's an example.

Let  \(f(x)=ax^2+bx+c\),  where \(a,b,c\in\mathbb{R}\).

Given that the graph of \(f\) passes through the points \((1,2)\), \((2,2)\) and \((3,3)\), find the values of \(a\), \(b\) and \(c\).

Your inputs should look like this:\[\boxed{\text{Define }f(x)=a\cdot x^2+b\cdot x+c}\\ \boxed{\texttt{solve}\!\left(\begin{cases}f(1)=2\\ f(2)=2\\ f(3)=3\end{cases},\ \{a,b,c\}\right)}\]

[Jimmmy]

Hey All!

Could someone explain to me what transformation would be required to map graph to graph

I got a dilation by a factor of 3 from the y-axis, but the answers say dilation by a factor of 1/3 from the y-axis.

Thanks!

[AlphaZero]

Hey All!

Could someone explain to me what transformation would be required to map graph to graph

I got a dilation by a factor of 3 from the y-axis, but the answers say dilation by a factor of 1/3 from the y-axis.

Thanks!

Hey there, let's just look at what dilations actually do.

A dilation by factor \(a\) from the \(x\)-axis is achieved by multiplying all the \(y\)-values by \(a\).  That is,  \(y'=ay\).

A dilation by factor \(b\) from the \(y\)-axis is achieved by multiplying all the \(x\)-values by \(b\).  That is,  \(x'=bx\).

In this question, we have  \(x'=\dfrac{1}{3}x\),  and so the required transformation is:
> a dilation by factor \(\dfrac13\) from the \(y\)-axis.
Small side note: this is not the only possible transformation.

One could write  \(y=\sqrt{\dfrac{x}{3}}\iff \sqrt{3}\,y=\sqrt{x}\)  giving  \(y'=\sqrt{3}\,y\),  and so another possible answer is:

> a dilation by factor \(\sqrt{3}\) from the \(x\)-axis.

[Jimmmy]

Hey there, let's just look at what dilations actually do.

A dilation by factor \(a\) from the \(x\)-axis is achieved by multiplying all the \(y\)-values by \(a\).  That is,  \(y'=ay\).

A dilation by factor \(b\) from the \(y\)-axis is achieved by multiplying all the \(x\)-values by \(b\).  That is,  \(x'=ax\).

In this question, we have  \(x'=\dfrac{1}{3}x\),  and so the required transformation is:
> a dilation by factor \(\dfrac13\) from the \(y\)-axis.
Small side note: this is not the only possible transformation.

One could write  \(y=\sqrt{\dfrac{x}{3}}\iff \sqrt{3}\,y=\sqrt{x}\)  giving  \(y'=\sqrt{3}\,y\),  and so another possible answer is:

> a dilation by factor \(\sqrt{3}\) from the \(x\)-axis.

Hi Alpha,

Thanks for the response! The \(x'=\dfrac{1}{3}x\) is where I tripped up, the dashes really mess me up sometimes  You simplified it well.

This LaTex code really confuses me too...I'm pretty much just copying & pasting square roots and powers every time....

And also; thanks for basically reading my mind and clarifying that the same equation can be reached by dilations in both the x or y-axis.

[aspiringantelope]

You just need to solve 3 equations simultaneously. On the TI-nspire CAS, press  \(\texttt{Menu}\to \texttt{3}\to \texttt{7}\to \texttt{1}\).  In the dialog box, set the number of equations and the variables you wish to solve for separated by commas. Press OK and type your equations in the boxes provided.
Here's an example.

Let  \(f(x)=ax^2+bx+c\),  where \(a,b,c\in\mathbb{R}\).

Given that the graph of \(f\) passes through the points \((1,2)\), \((2,2)\) and \((3,3)\), find the values of \(a\), \(b\) and \(c\).

Your inputs should look like this:\[\boxed{\text{Define }f(x)=a\cdot x^2+b\cdot x+c}\\ \boxed{\texttt{solve}\!\left(\begin{cases}f(1)=2\\ f(2)=2\\ f(3)=3\end{cases},\ \{a,b,c\}\right)}\]

Hi! I did the exact thing as you wrote but it said "false"

I've also got another question
Show that the graph equation \(y=-2x^2+4x-1\) can be written as \(y=-2\left(x-1\right)^2+1\)
So I do \(y=-2\left(x^2-2x+\frac{1}{2}\right)\)
\(y=-2\left(x^2-2x+\frac{1}{2}+\frac{1}{2}-\frac{1}{2}\right)\)
\(y=-2\left(x^2-2x+1-\frac{1}{2}\right)\)
\(y=-2\left(x-1\right)^2-\frac{1}{2}\)
Am I supposed to times -2 to that half if i'm taking it out? It just doesn't seem logical because this has nothing to do with the -2 >.<

Also, another quick question that I keep messing up
If I've got something like
\(0=-2\left(x-1\right)^2+1\) and I'm trying to solve for x or the intercept when i've put y=0
and I've got \(-1=-2\left(x-1\right)^2\)
Ok now is where I get confused
Do I root -1 or divide it  by -2?
And does this work all time?
So should I root a number first or take the coefficient on the outside of the brackets pls?
I need an answer >.<

[AlphaZero]

...

Please send screenshots/photos of your CAS calculator inputs so I can figure out what went wrong. (I can't really guess lol).
You made an algebraic slip from your the second last line to the last line. Please take more care in your working and try not to skip steps. \begin{align*}y&=-2\left(x^2-2x+1-\frac12\right)\\
&=-2\left((x-1)^2-\frac12\right)\\
&=-2(x-1)^2+1,\ \ \text{as required.}\end{align*}
You need to divide by \(-2\) first.

Let \(y=0\) \begin{align*}0&=-2(x-1)^2+1\\
-1&=-2(x-1)^2\\
\frac12&=(x-1)^2\\
\frac{\pm 1}{\sqrt2}&=x-1\\
&\vdots \end{align*} I'm not sure what you mean by:

And does this work all time?
So should I root a number first or take the coefficient on the outside of the brackets pls?
I need an answer >.<

Could you please elaborate?

[Evolio]

Hi guys.
I was wondering how to differentiate e^(x/2)?
Like, how come the answer is not 1/2e^(x/2)?

[MB_]

Hi guys.
I was wondering how to differentiate e^(x/2)?
Like, how come the answer is not 1/2e^(x/2)?

The answer is that

[matthewzz]

Hi! I've got a questions about 'show that' or 'prove that' questions. I'm running into a few in the textbook and not sure what I need to include for me to be able to say that it's correct, or in an exam situation, get full marks.

This question in particular, I just make x=-1 and then the whole equation = 0, which got me to 0=0 but I'm not sure if I even proved anything with this?

Thanks in advance

[RuiAce]

Hi! I've got a questions about 'show that' or 'prove that' questions. I'm running into a few in the textbook and not sure what I need to include for me to be able to say that it's correct, or in an exam situation, get full marks.

This question in particular, I just make x=-1 and then the whole equation = 0, which got me to 0=0 but I'm not sure if I even proved anything with this?

Thanks in advance

For that question, if you let the whole equation equal to 0 and then sub in \(x=-1\), you're falling into the trap of "assuming what you're trying to prove".

A workaround is to do something like say let \( P(x) = x^3+(k-1)x^2+(k-9)x-7\) first. And then manually compute \(P(-1)\), verifying it is equal to 0. (As opposed to literally setting \(P(x) = 0\) beforehand and then subbing \(-1\) in to just get a \(0=0\) statement.)

Although once you do that, you can just say that by the factor theorem, it is divisible by \(x+1\) for all values of \(k\).

[aspiringantelope]

Please send screenshots/photos of your CAS calculator inputs so I can figure out what went wrong. (I can't really guess lol).
You made an algebraic slip from your the second last line to the last line. Please take more care in your working and try not to skip steps. \begin{align*}y&=-2\left(x^2-2x+1-\frac12\right)\\
&=-2\left((x-1)^2-\frac12\right)\\
&=-2(x-1)^2+1,\ \ \text{as required.}\end{align*}
You need to divide by \(-2\) first.

Let \(y=0\) \begin{align*}0&=-2(x-1)^2+1\\
-1&=-2(x-1)^2\\
\frac12&=(x-1)^2\\
\frac{\pm 1}{\sqrt2}&=x-1\\
&\vdots \end{align*} I'm not sure what you mean by:Could you please elaborate?

Hey Sorry! But I've just redone it and realised that it now works. Just a few questions to consolidate
Should I always add the *  for like a*x^2 or can i just do ax^2.
Another question is the after the three equations, Do you always have to put { these brackets } ?

Thank you for your completing of the square for me!
Also for the x intercept, I got \(1\pm\sqrt{\frac{1}{2}}\) is that the same as yours?

I've got another one sorry -
Without solving, show that the straight line with rule y = x + 2 and the parabola y =2x^2+6x-1 have two points of intersection
Not sure how to prove without solving >.< Does anyone have a simple way? Thanks

[AlphaZero]

Hey Sorry! But I've just redone it and realised that it now works. Just a few questions to consolidate
Should I always add the *  for like a*x^2 or can i just do ax^2.
Another question is the after the three equations, Do you always have to put { these brackets } ?

You need to put a multiplication symbol between letters. Your CAS calculator treats the input  \(\texttt{ax}\)  as a single variable instead of the intended  \(a\cdot x\).  This would be the reason why you were getting  \(\texttt{false}\)  as your output. In general, you should leave the syntax of your input as is. Some other ways of writing simultaneous equations are understood by the software, but to be honest, it would be a bit too time consuming trying to figure out what inputs work rather than just to use the one provided when you press  \(\texttt{Menu}\to \texttt{3}\to \texttt{7}\to \texttt{1}\).  (Besides, I'm not sure why you would want to remove the curly brackets since they're placed there for you anyway).

Edit: if you only have one variable to solve for, curly brackets are not required. For example:
\[\texttt{solve}\Big(x^2-1=0,x\Big)\]

Thank you for your completing of the square for me!
Also for the x intercept, I got \(1\pm \sqrt{\frac12}\) is that the same as yours?

Yes, it is:  \(1\pm\dfrac{1}{\sqrt{2}}=1\pm \sqrt{\dfrac12}\).  The former does look a little more pleasing though.

I've got another one sorry -
Without solving, show that the straight line with rule y = x + 2 and the parabola y =2x^2+6x-1 have two points of intersection
Not sure how to prove without solving >.< Does anyone have a simple way? Thanks

Show that the discriminant of the resulting quadratic is greater than zero.

[Evolio]

The answer is that

Apparently the answer is...

[AlphaZero]

Apparently the answer is...

\(e^{x/2}\) is not the same as \(e^{x^{1/2}}\)\begin{align*}\frac{d}{dx}\Big[e^{x^{1/2}}\Big]&=\frac{d}{dx}\Big[x^{1/2}\Big]e^{x^{1/2}}\\
&=\frac12 x^{-1/2}e^{x^{1/2}}\end{align*}