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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: lolbox on March 05, 2010, 01:54:14 am

Title: maths help thread
Post by: lolbox on March 05, 2010, 01:54:14 am: hey guys, would appreciate help with some of these questions (maths is definitely not my forte unfortunately) :)

- its pretty easy i think

The graph of the function f is obtained from the graph of y=1/x by the following transformations applied in the order given

a dilation by factor of 0.5 from the y-axis
a reflection in the y axis
a translation of 3 units parallel to the x axis
a translation of 1 unit parallel to the y axis

write down the rule for f and hence state the domain and range of f
Title: Re: maths help thread
Post by: TrueTears on March 05, 2010, 01:56:09 am: (x,y) -> (0.5x,y) -> (-0.5x,y) -> ( -0.5x+3,y+1)

[for the last transformation i assume you mean in the positive direction.]

x' = -0.5x+3

y ' = y+1

(x'-3)/(-0.5) = x

y'-1=y

=> y'-1 = 1/((x'-3)/(-0.5))
Title: Re: maths help thread
Post by: the.watchman on March 05, 2010, 06:23:25 am: Here's another way (sorry TT :D):

1) Dilation by factor $\frac{1}{2}$ from the y-axis

$y=\frac{1}{x \div 0.5}=\frac{1}{2x}$

2) Reflection in the y-axis

$y=\frac{1}{2(-x)} = -\frac{1}{2x}$

3) Translation three units right, one unit up (same assumption as TT)

$y-1 = -\frac{1}{2(x-3)}$

$y = 1-\frac{1}{2(x-3)}$

SO $f(x) = 1-\frac{1}{2(x-3)}$

I think this way is a little quicker in determining the effect of transformations, although it is important to learn TT's way for more complex transformations.
All the best!
Title: Re: maths help thread
Post by: Cthulhu on March 05, 2010, 01:44:05 pm: Beating TrueTears at his own game? +1 for you sir!
Title: Re: maths help thread
Post by: TrueTears on March 05, 2010, 01:53:01 pm: assumption of y = f(x) is invalid
Title: Re: maths help thread
Post by: the.watchman on March 05, 2010, 02:15:20 pm: Quote from: TrueTears on March 05, 2010, 01:53:01 pm
assumption of y = f(x) is invalid

Erm ... the question does say the transformed function is f ... :P
Title: Re: maths help thread
Post by: TrueTears on March 05, 2010, 02:18:18 pm: Let y = f(x) valid
Title: Re: maths help thread
Post by: lolbox on March 09, 2010, 12:57:08 am: cheers
I've got a few more q's :)

(1) Let p(x) = (x^2 + a) (x + b) (x - c) where a, b and c are three distinct positive real numbers. What is the number of real solutions to the equation p(x) = 0

Not sure how to do this at all so instead of just saying the answer, I'd appreciate a bit of working out to guide me for future q's. thanks
Title: Re: maths help thread
Post by: /0 on March 09, 2010, 02:25:34 am: When solving quadratic equations, you are normally able to factor them, for example:
$(x-3)(x+5)=0$
Now, either $x-3=0$ , OR $x+5=0$ , it's just the null factor theorem

With $(x^2+a)(x+b)(x-c)=0$ ,

$x^2+a=0$ OR $x+b=0$ OR $x-c=0$

$x = \pm \sqrt{-a}$ or $x = -b$ or $x=c$

However, since $a$ is a positive number, $-a$ will be negative, and the square root of a negative number is not on the real number line, so $x = \pm \sqrt{-a}$ is not a real solution.

So we have $x=-b$ and $x=c$ , so 2.
Title: Re: maths help thread
Post by: the.watchman on March 09, 2010, 05:58:26 am: http://vcenotes.com/forum/index.php/topic,23815.0.html

Although that was a bit of a mess :P