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October 06, 2026, 04:44:51 pm
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Author Topic: VCE Methods Question Thread!  (Read 6267084 times)  Share 

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alchemy

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Re: VCE Methods Question Thread!
« Reply #4800 on: May 18, 2014, 01:40:14 pm »
+3
VCAA is obsessed with Tasmania Jones
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Tasmania Jones

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Re: VCE Methods Question Thread!
« Reply #4801 on: May 18, 2014, 03:26:40 pm »
+4
Quote from: alchemy on May 18, 2014, 01:40:14 pm
VCAA is obsessed with Tasmania Jones

Yeah VCAA loves me  ;)
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2014: Mathematical Methods CAS | Chinese SL

Bluegirl

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Re: VCE Methods Question Thread!
« Reply #4802 on: May 18, 2014, 05:16:48 pm »
0
"A curve has equation  y=2x^3 + bx^2 +cx+d. The curve has a stationary point at (-1,25) and passes through the point (1,-7). Find the values of b, c and d.


and
"Sketch the graph of y=f(x) if:
f'(x) < 0 for x<1; f'(1) =0; f'(x) >0 for x> 1; f(1) = -4

How would I go about doing these?
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lzxnl

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Re: VCE Methods Question Thread!
« Reply #4803 on: May 18, 2014, 05:23:20 pm »
+1
Quote from: Bluegirl on May 18, 2014, 05:16:48 pm
"A curve has equation  y=2x^3 + bx^2 +cx+d. The curve has a stationary point at (-1,25) and passes through the point (1,-7). Find the values of b, c and d.


and
"Sketch the graph of y=f(x) if:
f'(x) < 0 for x<1; f'(1) =0; f'(x) >0 for x> 1; f(1) = -4

How would I go about doing these?

For the first one, you know that when x=-1, y=25. Also, you know when x=-1, the derivative is zero. Then, you also know when x=1, y=-7. That's three pieces of information, so you can find b, c and d

As for the second one, you have f(1) = -4. It's flat at x=1. Then, x>1 the graph points upwards, and when x<1 the graph has a negative slope.
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2012
Mathematical Methods (50) Chinese SL (45~52)

2013
English Language (50) Chemistry (50) Specialist Mathematics (49~54.9) Physics (49) UMEP Physics (96%) ATAR 99.95

2014-2016: University of Melbourne, Bachelor of Science, Diploma in Mathematical Sciences (Applied Maths)

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Accepting students for VCE tutoring in Maths Methods, Specialist Maths and Physics! (and university maths/physics too) PM for more details

Bluegirl

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Re: VCE Methods Question Thread!
« Reply #4804 on: May 18, 2014, 05:36:04 pm »
0
Quote from: lzxnl on May 18, 2014, 05:23:20 pm
For the first one, you know that when x=-1, y=25. Also, you know when x=-1, the derivative is zero. Then, you also know when x=1, y=-7. That's three pieces of information, so you can find b, c and d

As for the second one, you have f(1) = -4. It's flat at x=1. Then, x>1 the graph points upwards, and when x<1 the graph has a negative slope.

Thankyou :) Seems to make more sense when someone else words it haha
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lzxnl

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Re: VCE Methods Question Thread!
« Reply #4805 on: May 18, 2014, 06:00:13 pm »
+1
That's Methods for you; the hard bit isn't doing the maths, it's translating the words into maths.
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2012
Mathematical Methods (50) Chinese SL (45~52)

2013
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Accepting students for VCE tutoring in Maths Methods, Specialist Maths and Physics! (and university maths/physics too) PM for more details

Mieow

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Re: VCE Methods Question Thread!
« Reply #4806 on: May 18, 2014, 06:13:35 pm »
0
how do i differentiate:
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alchemy

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Re: VCE Methods Question Thread!
« Reply #4807 on: May 18, 2014, 06:18:09 pm »
+1
Quote from: Mieow on May 18, 2014, 06:13:35 pm
how do i differentiate:


Use product rule, and chain rule for differentiating the second equation.
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Mieow

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Re: VCE Methods Question Thread!
« Reply #4808 on: May 18, 2014, 06:21:32 pm »
0
I've tried but I keep getting the wrong answer for some reason. There's probably something wrong with the way I'm working it out
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Zealous

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Re: VCE Methods Question Thread!
« Reply #4809 on: May 18, 2014, 06:29:28 pm »
+6
Quote from: Mieow on May 18, 2014, 06:21:32 pm
I've tried but I keep getting the wrong answer for some reason. There's probably something wrong with the way I'm working it out
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vce:
2013: Further [50] (+Premier's) | Methods [48]
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jessss0407

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Re: VCE Methods Question Thread!
« Reply #4810 on: May 18, 2014, 08:42:43 pm »
0
Hi guys! Whenever I enter "3.5=cos(x/6)+4" to solve for x over the domain of [0, 12] I keep getting "invalid answer" or "false". Does anyone know how to solve this? Thanks!
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Zealous

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Re: VCE Methods Question Thread!
« Reply #4811 on: May 18, 2014, 08:49:24 pm »
+2
Quote from: jessss0407 on May 18, 2014, 08:42:43 pm
Hi guys! Whenever I enter "3.5=cos(x/6)+4" to solve for x over the domain of [0, 12] I keep getting "invalid answer" or "false". Does anyone know how to solve this? Thanks!
Make sure your CAS is in radians! Very important for Methods.

How?
Spoiler
Main Menu -> [5] Settings -> [2] Document Settings -> Angle

What you'd be inputting into your CAS (Ti-Nspire):

« Last Edit: May 18, 2014, 08:52:54 pm by Zealous »
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vce:
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Jason12

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Re: VCE Methods Question Thread!
« Reply #4812 on: May 18, 2014, 10:16:06 pm »
0
i defined a function on ti-nspire cx cas as v(t)=10^3(90-t)^3 and then tried to graph it but nothing comes up. windows fit nope. help?
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IndefatigableLover

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Re: VCE Methods Question Thread!
« Reply #4813 on: May 18, 2014, 10:24:16 pm »
0
Quote from: Jason12 on May 18, 2014, 10:16:06 pm
i defined a function on ti-nspire cx cas as v(t)=10^3(90-t)^3 and then tried to graph it but nothing comes up. windows fit nope. help?
Replace the 't' with a 'x' and you should be fine :)
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Jason12

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Re: VCE Methods Question Thread!
« Reply #4814 on: May 18, 2014, 10:45:39 pm »
0
Quote from: IndefatigableLover on May 18, 2014, 10:24:16 pm
Replace the 't' with a 'x' and you should be fine :)

it worked but the lines don't properly show up just parts of a line when i graph the derivative. However graphing just f(x) works fine
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