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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: d0minicz on January 21, 2009, 02:02:28 pm

Title: Families of Functions
Post by: d0minicz on January 21, 2009, 02:02:28 pm: For y= f(x) = 1 / X^2 , sketch the graph of each of the following:
a) y= f(2x)
b) y= 2f(x)
c) y= f(x/2)
d) y= 3f(x)
e) y= f(5x)
f) y=f(x/4)

How would I do these? thanks
Title: Re: Families of Functions
Post by: kurrymuncher on January 21, 2009, 02:09:12 pm: Just draw the graph of f(x)=1 / X^2

a) y= f(2x) means that the original graph has been dilated by a factor of 1/2 from the y axis, so draw the graph as if the x values have been halfed.

b) y= 2f(x) means that the graph has been dilated by a factor of 2 away from the x axis, so draw the graph as if the y values have been doubled.

c)y= f(x/2) means that the graph has been dilated by a factor of 2 away from the y axis, so draw it as if the x values have been doubled. It is 2 because the dilation is 1/(1/2)

just apply the same thing to the other questions
Title: Re: Families of Functions
Post by: d0minicz on January 21, 2009, 02:13:14 pm: Thanks for that, but how would I go about finding the exact equation of the dilated graph?
Title: Re: Families of Functions
Post by: Mao on January 21, 2009, 02:14:39 pm: $f(2x)=\frac{1}{(2x)^2}=\frac{1}{4x^2}$

$2f(x)=2\cdot \frac{1}{x^2} = \frac{2}{x^2}$

etc
Title: Re: Families of Functions
Post by: d0minicz on March 09, 2009, 07:50:05 pm: Express the area of an equilateral triangle as a function of:

a) the length s of each side

b) the altitude h

thanks !
Title: Re: Families of Functions
Post by: Flaming_Arrow on March 09, 2009, 08:15:35 pm: ok we know if a triangle is equilateral the interior angles of the traingle is 60

find height interms of sides

draw a right angle triangle with hypotenuse s and find the opposite

so $s \sin 60^o = o$

we know that $\sin 60^o = \frac{\sqrt{3}}{2}}$

so height is $\frac{\sqrt{3}s}{2}}$

Area of a triangle is $\frac{1}{2} \mbox{base * height}$

$\frac{\sqrt{3}s}{2}} \cdot s \cdot \frac{1}{2} \implies \frac{\sqrt{3}s^2}{4}}$
Title: Re: Families of Functions
Post by: Flaming_Arrow on March 09, 2009, 08:24:52 pm: 2nd part

from the previous part we know that h= $\frac{\sqrt{3}s}{2}}$

now solve for s

$s= \frac{2h}{\sqrt{3}}$

sub into the area formula from the previous part

$\frac{\sqrt{3}}{4}\cdot (\left \frac{2h}{\sqrt{3}})^2$

that yields

$\frac{\sqrt{3}}{4}\cdot \frac{4h^2}{3}$

$\implies \frac{\sqrt{3}h^2}{3}$
Title: Re: Families of Functions
Post by: d0minicz on March 11, 2009, 06:09:27 pm: Let $\ a$ be a positive number, let $\ f : [2, \infty)$ $\rightarrow R$ , $\ f (x) = a - x$ and let $\ g: (-\infty , 1] \rightarrow R$ , $\ g(x) = x^2 + a$ .
Find all values of $\ a$ for which $\ f \circ g$ and $\ g \circ f$ both exist.

thanks =]
Title: Re: Families of Functions
Post by: TrueTears on March 11, 2009, 07:29:42 pm: $ran f \subseteq (- \infty, 1]$ , $ran g \subseteq [2, \infty)$

So 1. we need to find a so that any value of $f:[2 ,\infty) \rightarrow R, f(x)=a-x$ will only be a subset of $(-\infty, 1]$

The function f(x) is a straight line and always decreasing, so the maximum value will be at its left endpoint, at f(2) = a-2. Hence, we can say $ranf = (-\infty, a-2]$ . For it to have the correct range, we require $a-2 \leq 1 \Rightarrow a \leq 3$

and 2. we need to find a so that any value of $g:(-\infty, 1] \rightarrow R, g(x) = x^2+a$ will only be a subset of $[2, \infty)$

For $g(x) = x^2+a$ , a determines the vertical translation of the standard parabola with vertex at (0,0). We can say $rang = [a, \infty)$ . So to have the right range, $a \geq 2$ .

Hence the answer is $2 \leq a \leq 3.$