Author Topic: VCE Methods Question Thread! (Read 5783167 times) Tweet Share

hobbitle · « **Reply #3765 on:** January 16, 2014, 02:30:42 am »

Quote from: Only Cheating Yourself on January 16, 2014, 01:04:11 am

I've heard transformations before, what does it actually mean?

Shifting of graphs around the axis (in the positive or negative x/y direction, reflections about the x or y axis, etc).

M_BONG · « **Reply #3766 on:** January 17, 2014, 06:06:28 pm »

Hey!

Can anyone lend me a hand here?

Solve for y : $x= e^{2y}-2e^{y}+1$

My working:
Move +1 to the left hand side and natural log both sides
1. $In (x-1) = In(e^{2y}-2e^{y})$
$In (x-1) = \frac{In (e^{2y})}{In(2e^{y})}$
$In (x-1)= \frac{2y\log_{e}e}{y\log_{e}2e{}}$
Cancel RHS : $\frac{2}{\log_{e}2e}$

That is where I got up to

Actual answer:
y= $\log_{e}(x^{\frac{1}{2}}+1)$

Thanks!

LOLs99 · « **Reply #3767 on:** January 17, 2014, 06:12:10 pm »

Quote from: Zezima. on January 17, 2014, 06:06:28 pm

Hey!

Can anyone lend me a hand here?

Solve for y : $x= e^{2y}-2e^{y}+1$

My working:
Move +1 to the left hand side and natural log both sides
1. $In (x-1) = In(e^{2y}-2e^{y})$
$In (x-1) = \frac{In (e^{2y})}{In(2e^{y})}$
$In (x-1)= \frac{2y\log_{e}e}{y\log_{e}2e{}}$
Cancel RHS : $\frac{2}{\log_{e}2e}$

That is where I got up to

Actual answer:
y= $\log_{e}(x^{\frac{1}{2}}+1)$

Thanks!

U started off wrongly. Let e^y=A and solve from there.

Hope it helps

Phy124 · « **Reply #3768 on:** January 17, 2014, 06:13:45 pm »

Quote from: Zezima. on January 17, 2014, 06:06:28 pm

Hey!

Can anyone lend me a hand here?

Solve for y : $x= e^{2y}-2e^{y}+1$

$e^{2y}-2e^{y}+(1-x) = 0$

$(e^y)^2-2e^y+(1-x)=0$

$e^y = \frac{-(-2)\pm\sqrt{(-2)^2-4(1)(1-x)}}{2(1)}$

$e^y = \frac{2\pm\sqrt{4x}}{2}$

$e^y = \frac{2 \pm 2 \sqrt{x}}{2}$

$e^y = 1 \pm \sqrt{x}$

$y = \log_e\left(1 \pm \sqrt{x}\right)$

edit: messed up a +-

Toki · « **Reply #3769 on:** January 17, 2014, 06:59:42 pm »

Quote from: Nato on January 07, 2014, 09:16:05 pm

need help with this one:
The graph of $y= 2(x+3)^{3}+ 1$ has been reflected in the x-axis, shifted 3 units to the right and 1 unit up.
what is the resulting equation.

well i started off with
$(x,y)\rightarrow (x,-y)\rightarrow (x+3,-y)\rightarrow (x+3,-(y+1))\rightarrow (x',y')$

so from here
$x+3=x'$
$x=x'-3$

and for the y
$-(y+1)=y'$
$y=-y'-1$

and then i subbed this into the above equation

$-y'-1=2(x'-3+3)^{3}+1$
and then after rearranging i got
$y=-2x^{3}-2$ which doesn't seem to be correct.

can someone please tell me where i may have went wrong?
cheers.

What do you call a question where you have to use a similar method to this? (Our class never did this last year)

Daenerys Targaryen · « **Reply #3770 on:** January 17, 2014, 09:17:24 pm »

Quote from: Toki on January 17, 2014, 06:59:42 pm

What do you call a question where you have to use a similar method to this? (Our class never did this last year)

Transformations I believe.
If you're doing it, do it alpha order such that: Dilation, Reflection, Translation (moving up/down/left/right)

Yacoubb · « **Reply #3771 on:** January 18, 2014, 05:50:46 pm »

Can someone take me through the steps of using simultaneous equations with no calculator to solve the quartic function that runs through the points (-1,43), (0,40), (2,70), (6,1618) and (10,670).

The quartic general equation:

y = ax⁴ + bx³ +cx² + dx + e

We now that e = 40

My 4 simultaneous equations are:
a - b + c - d = 3...(1)
8a + 4b + 2c + d= 15...(2)
216a + 36b + 6c + d= 263...(3)
1000a + 100b + 10c + d= 63...(4)

Help would be appreciated

Daenerys Targaryen · « **Reply #3772 on:** January 18, 2014, 06:36:44 pm »

Quote from: Yacoubb on January 18, 2014, 05:50:46 pm

Can someone take me through the steps of using simultaneous equations with no calculator to solve the quartic function that runs through the points (-1,43), (0,40), (2,70), (6,1618) and (10,670).

The quartic general equation:

y = ax⁴ + bx³ +cx² + dx + e

We now that e = 40

My 4 simultaneous equations are:
a - b + c - d = 3...(1)
8a + 4b + 2c + d= 15...(2)
216a + 36b + 6c + d= 263...(3)
1000a + 100b + 10c + d= 63...(4)

Help would be appreciated

If doing it by hand;
I would transpose (1) and make d the subject, then sub it into (2)(3)(4) to eliminate d
Then again transpose one of these three to make c the subject and sub it into the remaining two.

Do simultaneous equations on these two to find the two letters, sub it into the equation with 3 letters, then the remaining equation

You can also set up a matrix (unfortunately i dont know the latex for that so I won't explain that), and also a rref function on your calc which can do it too (but again the matrix thing applies here) or go solve((1) and (2) and (3) and (4),a,b,c,d) or that Simul.Linear thing

Blondie21 · « **Reply #3773 on:** January 19, 2014, 01:58:19 am »

Quote from: Yacoubb on January 18, 2014, 05:50:46 pm

8a + 4b + 2c + d= 15...(2)
216a + 36b + 6c + d= 263...(3)
1000a + 100b + 10c + d= 63...(4)

~~Hey Yacoubb, I think you just need to have a closer look at your simultaneous equations! Remember: y = ax⁴ + bx³ +cx² + dx + e~~

edit: wait sorry!! I see what you've done.. ignore this!

Daenerys Targaryen · « **Reply #3774 on:** January 19, 2014, 10:01:13 am »

Quote from: Blondie21 on January 19, 2014, 01:58:19 am

Hey Yacoubb, I think you just need to have a closer look at your simultaneous equations! Remember: y = ax⁴ + bx³ +cx² + dx + e

Ah yes, good spot!

~V · « **Reply #3775 on:** January 19, 2014, 10:40:40 am »

Been doing fine with all other questions but how do i do this log question? Someone take me through the steps?
Express in simplest form:
log₃(27)+1
log₄(16)+3

psyxwar · « **Reply #3776 on:** January 19, 2014, 10:42:48 am »

Quote from: supviv on January 19, 2014, 10:40:40 am

Been doing fine with all other questions but how do i do this log question? Someone take me through the steps?
Express in simplest form:
log₃(27)+1
log₄(16)+3

log₃(27)=3, therefore log₃(27)+1=4
log₄(16)=2, therefore log₄(16)+3=5

~V · « **Reply #3777 on:** January 19, 2014, 10:46:38 am »

Quote from: psyxwar on January 19, 2014, 10:42:48 am

log₃(27)=3, therefore log₃(27)+1=4
log₄(16)=2, therefore log₄(16)+3=5

But the answer was log₃(81)=4
and log₄(1024)=5
I dont know where the 81 and 1024 came from :/

~V · « **Reply #3778 on:** January 19, 2014, 10:50:28 am »

Quote from: supviv on January 19, 2014, 10:46:38 am

But the answer was log₃(81)=4
and log₄(1024)=5
I dont know where the 81 and 1024 came from :/

Woops i typed the question out wrong ==' its actually log₃(27)+1 and log₄(16)+3

psyxwar · « **Reply #3779 on:** January 19, 2014, 11:00:19 am »

Quote from: supviv on January 19, 2014, 10:46:38 am

But the answer was log₃(81)=4
and log₄(1024)=5
I dont know where the 81 and 1024 came from :/

oh they did it expressing it in log form.

log₃(27)+1

1=log₃(3)

therefore: log₃(27)+1= log₃(27)+log₃(3)=log₃(3*27)=log₃(81)

Search the forums now!

We have moved!

Author Topic: VCE Methods Question Thread! (Read 5783167 times) Tweet Share

hobbitle

Re: VCE Methods Question Thread!

M_BONG

Re: VCE Methods Question Thread!

LOLs99

Re: VCE Methods Question Thread!

Phy124

Re: VCE Methods Question Thread!

Toki

Re: VCE Methods Question Thread!

Daenerys Targaryen

Re: VCE Methods Question Thread!

Yacoubb

Re: VCE Methods Question Thread!

Daenerys Targaryen

Re: VCE Methods Question Thread!

Blondie21

Re: VCE Methods Question Thread!

Daenerys Targaryen

Re: VCE Methods Question Thread!

~V

Re: VCE Methods Question Thread!

psyxwar

Re: VCE Methods Question Thread!

~V

Re: VCE Methods Question Thread!

~V

Re: VCE Methods Question Thread!

psyxwar

Re: VCE Methods Question Thread!

Recent Posts

User:
Password: