Author Topic: Questions thread (Read 19040 times) Tweet Share

m@tty · « **Reply #120 on:** December 13, 2009, 10:54:02 pm »

b) is just a dilation by factor of 2 from the x-axis
$y=\frac{y'}{2} \implies \ y'=2y \ \ (x,y) \to (x,2y)$
c) is first a dilation of factor 2 from the x-axis, then a translation of 1 unit to the left
$\begin{align} y&=\frac{y'}{2} \implies y'=2y \\ x&=x'+1 \implies \ x'=x-1 \\ (&x,y) \to (x-1,2y)\end{align}$

I think I got all the dashes in the right place :S

Hielly · « **Reply #121 on:** December 14, 2009, 12:10:02 am »

Quote from: m@tty on December 13, 2009, 10:54:02 pm

b) is just a dilation by factor of 2 from the x-axis
$y=\frac{y'}{2} \implies \ y'=2y \ \ (x,y) \to (x,2y)$
c) is first a dilation of factor 2 from the x-axis, then a translation of 1 unit to the left
$\begin{align} y&=\frac{y'}{2} \implies y'=2y \\ x&=x'+1 \implies \ x'=x-1 \\ (&x,y) \to (x-1,2y)\end{align}$

I think I got all the dashes in the right place :S

thanks, what about the 18 ? and how would sketching it with the dilation look like.. i know you multiply all the x or y values by the dilation factor. but the answer didnt multiply it?

vcesuccess · « **Reply #122 on:** December 14, 2009, 07:04:50 pm »

Hielly, you doing all your 3/4 next year?

Hielly · « **Reply #123 on:** December 15, 2009, 11:26:28 am »

More questions!

For Q2. Why isn't 3 the answer as well?

For Q5. What is the easiest way to do these problems. I used simultaneous equations and it takes quite awhile, not to mention how many careless mistakes you can make!

Thanks!!

Cataclysmic · « **Reply #124 on:** December 16, 2009, 10:25:08 am »

For question 3:
1 solution if:
$b^{2}-4ac = 0$

$So\; (a+b)^{2}-4(a)(b) = 0$
$a^{2}+2ab+b^{2}-4ab = 0\; [1]$
$a^{2}-2ab+b^{2} = 0$
$(a-b)^{2} = 0$
$a = b$
$Subbing\; a = b\; into\; [1]$
$you\; get b^{2}+2b^{2} + b^{2} - 4b^{2} = 4b^{2}-4b^{2} = 0.$

4:
General cubic equation form:
$y = ax^{3}+bx^{2}+cx+d$
No vertical translations so d = 0
$y = ax^{3}+bx^{2}+cx$
$y = x(ax^{2}+bx+c)$
$x=1, y = 3/4$
3/4 = a+b+c [1]
x = -2, y = -3
$-3 = -2(4a-2b+c) [2] -3 = -8a+4b-2c$
x = 2, y = 3
$3 = 2(4a+2b+c) [3] 3 = 8a+4b+2c$

[2]+[3]
0 = 0a+ 8b + 0. therefore b = 0

so $y = x(ax^{2}+c)$
when x = 1, y = 3/4
$\frac{3}{4} = (a+c)$ [4]

x = 2, y = 3
$3 = 2(4a+c) => 3 = 8a+2c$ [5]

[4]*8 = 6 =8a+8c
[5]-[4]*8
-3 = -6c
c = 1/2

so $y = x(ax^{2}+0.5)$
if c = 0.5, then a = 0.75-0.5 = 0.25 ( using [4])

$y = 0.25x^{3} + 0.5x$

for 5, not sure. I don't see a faster way than what you did.
for 1, lol i suck at that stuff, nfi.

Damo17 · « **Reply #125 on:** December 16, 2009, 10:37:25 am »

Quote from: Hielly on December 15, 2009, 11:26:28 am

More questions!

For Q2. Why isn't 3 the answer as well?

because if a=3 then x-1 is a factor, so a=4 is the only answer.

Search the forums now!

We have moved!

Author Topic: Questions thread (Read 19040 times) Tweet Share

m@tty

Re: Questions thread

Hielly

Re: Questions thread

vcesuccess

Re: Questions thread

Hielly

Re: Questions thread

Cataclysmic

Re: Questions thread

Damo17

Re: Questions thread

Recent Posts

User:
Password: